Final simple FLT post, answers
fjm...@my-dejanews.com
are my replies to the five persons who replied to my post of Jan 21st. Nico
Benschop replied >What you show is quite easily seen algebraically. If >x^n
+ y^n = z^n for some x,y,z>0 and a special exponent n,then for no other n
>can equality hold at those x,y,z. Because f(x,y,z,n) = x^n + y^n - z^n (with
>x+y>z since x^n+y^n=z^n for n=2) is a monotone and exponential function of
>n, which can only once cross the zero value for increasing n (starting at
n=1). >However, this does not prove anything about the necessary inequality
for n>2.
My reply. Since f(x,y,z,n) will only = 0 when n=2 does that not show that
the inequalities of n>2 make the equation invalid for all values of n<>2? I
mean the equation (or function) works when n=2 and doesn't work when n<>2. By
assigning values (I used the 3,4,5 triangle roots) to f(x,y,z,n) and solving,
at n=1 f=2; at n=2 f=0; at n=3 f= -34; at n=4 f = -288 and f gets more
negative as n increases never again to cross the zero point again as you say.
These values of f are the differences of what z^n is and should be and is
only 0 when n=2. Your point further verifies my conjecture about the x^n +
y^n = z^n equation only working when n=2, making it invalid for all other
values of n.
Keith Ellul had much to say
>Oh my. I would truly hate to know how little of an understanding you had of >FLT before…. My reply is That's cute, real cute, but I am not at all ashamed to admit that I have no understanding of FLT in the customary way, nor can I understand any of the attempts at proving it. But I am proud to possible be the first one in 360 years to be thinking of it in a different way which could result in a very simple way to verify what Fermat said and so anyone would be able to understand it.
>Ummm…no…not even close to sufficient…
Time will tell. All I need is one person to understand what I am trying to
say.
>Here is the first problem with your logic. Why must x,y,z remain the same?
This is not a problem with my logic. x,y,z must be the same to visually see
the results. The value of n is the only variable. I could have used any
Pythagorean triplet or for that matter any three values of x,y,z that form a
right triangle. And when you use these same values with a different value of
n, you will not get the same visual results. I have done this many times and
have seen the results and they are different. When n does not equal 2 you
will never see a right triangle.
>What you are trying to show is that if x^2 + y^2 = z^2, then x^n + y^n = z^n >
>has no solutions
You got that right. Thanks, I needed that. x^n + y^n = z^n has no solutions
when you compare apples with apples which is why I must keep x,y,z the same.
N=2 forms right triangles, n<>2 will not
>Similarly, I can show…….etc. But this doesn't mean that x^2 + y^2 = z^2 has >no solutions, it just means that the solutions are different from the solutions of >x + y = z.
x^2 + y^2 = z^2 has no solutions?? Blasphemy!! I am glad you agree that they
are different, tho, but can you show how? I can. You have it backwards, tho.
The solutions of x+y=z (which is when n=1) and all solutions when n>2 are
different from when n=2. This is all that I am showing.
>Ok, all that you have managed to show here (and you could have done this a >much,much, shorter and easier-to-understand way) is that 3^2n+4^2n=5^2n >has no solutions other than n=1. Furthermore, the fact that you showed this >appears to be accidental rather than by design. If you are trying to prove >something for the general case, you can't plug numbers in and show that it >works for 1 example. You have to show that it works for all numbers that you >could plug in, and, since we're dealing with the natural numbers here, and >there are infinitely many naturals, you can't do it by trying all the cases.
Again you agree with me. But I have shown a way to see that this is true by
reason of whether or not right triangles are formed by the only right
triangle generating equation namely x^n+y^n=z^n when n=2. I have done this
for many naturals and they always follow my example of the 3,4,5 roots. I am
sure you recognize this as a pythagorean triplet and if one p-triplet works,
they all will work. It's a simple axiomatic truth. It even works with non
p-triplets as long as they form a right triangle. I came across my conjecture
by plugging in numbers and I was able to recognize the results and
implications. It was by no means accidental and I seen the results
immediately.
Ull...@math.okstate.edu wrote:
>If there have been no replies it's because people are tired of pointing out
>obvious errors. The "proof" below is in the "Huh?????" category-it's hard to >say what's wrong with the details because the outline simply makes no sense.
I am not offering a proof of FLT at all; a conjecture, yes, but not a proof.
All that needs to be shown is that the equation in question is different when
n=2 from when n<>2. People are trying to prove the equation for n>2 and it is
not necessary to do so. The equation is invalid when n<>2 since x^n+y^n will
not equal z^n and I am just giving a couple of ways of showing this
invalidity.
>What "controversy"?
The controversy is that people have been trying to prove FLT for the last 360
years and everyone has a mindset on proving it one way instead of just
realizing that the equation is valid only when n=2.
>"is Pythagorean in nature"?
Yes, FLT is Pythagorean in nature. Its only valid equation is x^2+y^2=z^2 and
that came from Pythagorus and the Egyptians a long time ago.
>Yes, they are totally different. That's why this "proof" makes no sense: If you >want to prove FLT, etc, etc. So what? The question is whether (2) has any >solutions, not whether it has any solutions that also satisfy (1)
First, I believe you have (1) and (2) reversed but that's okay. Again, this
is not a proof because there is nothing to prove. To show, yes, but not to
prove. I am showing that my (2) has solutions that form right triangles (and
this is important) as it should and my (1) does not have solutions that will
form right triangles (and this again is important) as it should to be the
same as my (2). And, the only variable involved is n and when n<>2 it causes
the equation to be invalid. This is verified by the trig equations mentioned
and simply by seeing that x^n+y^n does not equal z^n when n<>2. The fact that
any nth power will not equal the sum of two other nth powers (except 2) is
proven by the invalidity of the equation when n is not 2. Why prove that (1)
has solutions when it is an invalid equation.
>Now honestly. I have no idea who "curmudgeon" is, I haven't seen what he >wrote. Was the proof the two of you talked about the same as the proof above? >Assuming yes: honestly now, what he said about it is more less….etc, etc.
Yes, the proof, as you call it, was the same. I have seen many of your posts
in sci.math and have seen about six of his during the same time period. This
is not his real name, it is his email name.
Clive Tooth wrote:
>First…I have no idea who Mr. Wolfshelt is. Probably many people reading >your post do not know. The statement of FLT is well known and fairly easily >understood. I would recommend that you forget Mr. Wolfshelt and stick to the >usual statement of FLT.
I am sure many people know who Mr. Wolfskehl (name properly spelled) was. His
name and Gottinberg University are synonymous with FLT. I seem to be the only
one sticking to the original FLT statement and everyone else is trying to
prove it. Didn't he say that no cube is the sum of two other cubes?
>Ok…You have posted a supposed proof of FLT. You may genuinely think it >has some value. You have, in all sincerity, asked for comments on it. You have >not had many comments - for or against. I have to tell you that to anybody >with some mathematical training - perhaps having studied mathematics at >university-your proof is nonsense.
I respectfully disagree that the proof, as you call it, is nonsense. It is no
proof and may not even be a conjecture. But it does show that one type of the
equation in question is invalid.
>It is not just that there is a mistake in your proof. Your proof appears to be >meaningless. Suppose somebody posted a recipe…etc,etc. I think to most >mathematicians, your proof of FLT is…etc, etc. It is not a proof of FLT. It is >not close to a proof of FLT. It does not look like a proof of anything. It cannot >be corrected.
True, it may be meaningless as a proof, but then not being a proof, it could
be meaningful. You are very correct about it not looking like a proof.
Everyone has a mindset as to what should be proven. No one seems to realize
that, if an equation is invalid, why prove it? All I have done is to prove
that x^n+y^n=z^n is invalid when n<>2 when comparing it to x^n+y^n=z^n when
n=2 and have given ways to visually see this. My facts need no correction,
just understanding. Thank you, Clive, I appreciated your opinions and the
opinions of everyone else.
Daniel Weiss said:
>Possibly Paul Wolfskehl: donor of the 1908 Wolfskelf prize endowment for >proving the Fermat conjecture, see…etc.
Thank you Mr. Weiss. I see that you do know the whole story and can spot a
misspelled name. And you called it Fermats' conjecture which is good, too,
for that is all his statement was. He thought he had his conjecture proved
but failed to show it, perhaps, because he died before he could explain it.
Sure hope that doesn't happen to me.
Finally, I wish to again thank all who replied. You have all strengthened my
belief in my thoughts and my persistence (even more than JSHs') will cause me
to further improve upon it and try to explain it understandably. I am also
trying to see if my thinking is itself totally invalid in which case I will
give up my interest in FLT. But I just can not yet see that is the case. If
there is any interest still and anyone wishes to discuss this more, do so
please either in sci.math or my email address is fjm...@thegrid.net.
Frank J Manus
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Keith Ellul
> Keith Ellul had much to say
> >Oh my. I would truly hate to know how little of an understanding you
> had of >FLT before…. My reply is That's cute, real cute,
> but I am not at all ashamed to admit that I have no understanding of
> FLT in the customary way, nor can I understand
> any of the attempts at proving it. But I am proud to possible be the
> first one in 360 years to be thinking of it in a different way which
> could result in a very simple way to verify what Fermat said and so
> anyone would be able to understand it.
Ok, sorry, this was a joke, and not very nice. No offense intended.
> >Ummm…no…not even close to sufficient…
>
> Time will tell. All I need is one person to understand what I am trying to
> say.
I undeerstand what you are trying to say. You are wrong. Sorry, but you
are. If nobody understand what you are saying, try saying it in standard
notation. It will make the errors easier to spot and will make things
easier for everyone involved.
> >Here is the first problem with your logic. Why must x,y,z remain the same?
>
> This is not a problem with my logic. x,y,z must be the same to visually see
> the results. The value of n is the only variable. I could have used any
> Pythagorean triplet or for that matter any three values of x,y,z that form a
> right triangle. And when you use these same values with a different value of
> n, you will not get the same visual results. I have done this many times and
> have seen the results and they are different. When n does not equal 2 you
> will never see a right triangle.
Oh, but thisd is the problem with your logic. I'll explain better below...
> >What you are trying to show is that if x^2 + y^2 = z^2, then x^n + y^n = z^n >
> >has no solutions
>
> You got that right. Thanks, I needed that. x^n + y^n = z^n has no solutions
> when you compare apples with apples which is why I must keep x,y,z the same.
> N=2 forms right triangles, n<>2 will not
>
> >Similarly, I can show…….etc. But this doesn't mean that x^2 + y^2 =
> z^2 has >no solutions, it just means that the solutions are different
> from the solutions of >x + y = z.
>
> x^2 + y^2 = z^2 has no solutions?? Blasphemy!! I am glad you agree that they
> are different, tho, but can you show how? I can. You have it backwards, tho.
> The solutions of x+y=z (which is when n=1) and all solutions when n>2 are
> different from when n=2. This is all that I am showing.
You seem confused. You are only showing that x^2 + y^2 = z^2 has no solutions
in common with x^3 + y^3 = z^3. I'm sure you can see that this is
VERY different from saying that no solutions of x^3 + y^3 = z^3 exist.
What if the solutions are different? Ok, let me try to exaplain:
(1) x^2 + y^2 = z^2
(2) x^3 + y^3 = z^3
What you have showed (or tried to show) is that if (x1, y1, z1) is a solution
to (1), then it is not a solution for (2). Fine. But you have not showed
that some arbitrary (x2, y2, z2) is not a solution for 2. What if (x2, y2, z2)
is a solution for (2) but not for (1)???
> >Ok, all that you have managed to show here (and you could have
>> done this a >much,much, shorter and easier-to-understand way) is that
>> 3^2n+4^2n=5^2n >has no solutions other than n=1. Furthermore, the fact
>> that you showed this >appears to be accidental r
>> ather than by design. If you are trying to prove >something for the
>> general case, you can't plug numbers in and show that it >works for 1
>> example. You have to show that it works for all numbers that you >could
>> plug in, and, since we're dealing with the na
>> tural numbers here, and >there are infinitely many naturals, you can't
>> do it by trying all the cases.
>
> Again you agree with me. But I have shown a way to see that this is true by
> reason of whether or not right triangles are formed by the only right
> triangle generating equation namely x^n+y^n=z^n when n=2. I have done this
> for many naturals and they always follow my example of the 3,4,5 roots. I am
> sure you recognize this as a pythagorean triplet and if one p-triplet works,
> they all will work. It's a simple axiomatic truth. It even works with non
> p-triplets as long as they form a right triangle. I came across my conjecture
> by plugging in numbers and I was able to recognize the results and
> implications. It was by no means accidental and I seen the results
> immediately.
No, no no! (no, really, no!) Your error is in the assumption that
a solution to x^3 + y^3 = z^3 must form a right triangle... why? Yes,
any solution to x^2 + y^2 = z^2 will give the lengths of sides of a right
triangle... but this says nothing about the solutions x^3 + y^3 = z^3.
This shouldn't be all that difficult to understand. Yes, x^3 + y^3
= z^3 is different from x^2 + y^2 = z^2. So? So they are different
equations... that doesn't mean that one of them doesn't have a solution!
All you have shown is that an (x, y, z) triplet cannot be a solution to
BOTH equations... but that is not the same as saying that there are no
solutions to the second equation. It is entirely possible that some (x,
y, z) triplet is a solution to the second equation but not to the first!
-Keith!
---------------------------------------------------------------
Keith Ellul 4th Year Pure Math / Computer Science
kbe...@golden.net University of Waterloo
---------------------------------------------------------------
Clive Tooth
> Time will tell. All I need is one person to understand what I am trying to
> say.
Maybe.
But that one person might be James Harris, or Archimedes Plutonium, or
Nathan, or ...
:))
Good luck Frank...
--
Clive Tooth
http://www.pisquaredoversix.force9.co.uk/
End of document
Nico Benschop
>
> Problems with getting on dejanews yesterday delayed this but here I
> am. Here are my replies to the five persons who replied to my post of
> Jan 21st.
>
> Nico Benschop replied:
>> then for no other n can equality hold at those x,y,z.
>> Because f(x,y,z,n) = x^n + y^n - z^n (with x+y>z since x^n+y^n=z^n
>> for n=2) is a monotone and exponential function of n, which can only
>> once cross the zero value for increasing n (starting at n=1).
>> However, this does not prove anything about the necessary inequality
>> for n>2.
>
> that the inequalities of n>2 make the equation invalid for all values
> of n<>2? I mean the equation (or function) works when n=2 and doesn't
> work when n<>2. By assigning values (I used the 3,4,5 triangle roots)
> to f(x,y,z,n) and solving, at n=1 f=2; at n=2 f=0; at n=3 f=-34; at
> cross the zero point again as you say.
> These values of f are the differences of what z^n is and should be
> and is only 0 when n=2. Your point further verifies my conjecture
> about the x^n + y^n = z^n equation only working when n=2, making it
> invalid for all other values of n.
>
From your above reply, and from those to other pespondents, I see that
you understand that all you showed (by a geometric argument) is the
fact that triples {x,y,z} solving x^2+y^2=z^2,
-- and {a,b,c} solving a^n+b^n=c^n for n<>2,
are necessarily different: {x,y,z} <> {a,b,c}.
Many, including me, have pointed out that this does NOT imply FLT
(there is no solution {a,b,c} in positive integers for n>2).
It only shows that IF there were such solution, it must be different
from any {x,y,z} triple for n=2. You see the big distance to FLT ?-)
Surely, it is depressing to realize that your clue does not prove
FLT (= there is NO {a,b,c} solution for n>2) -- But what the heck,
you learned something after all...
--
Ciao, Nico Benschop
bens...@iae.nl -- http://www.iae.nl/users/benschop
Planar
}My reply. Since f(x,y,z,n) will only = 0 when n=2 does that not show that
}the inequalities of n>2 make the equation invalid for all values of n<>2? I
}mean the equation (or function) works when n=2 and doesn't work when n<>2. By
}assigning values (I used the 3,4,5 triangle roots) to f(x,y,z,n) and solving,
}at n=1 f=2; at n=2 f=0; at n=3 f= -34; at n=4 f = -288 and f gets more
}negative as n increases never again to cross the zero point again as you say.
}These values of f are the differences of what z^n is and should be and is
}only 0 when n=2. Your point further verifies my conjecture about the x^n +
}y^n = z^n equation only working when n=2, making it invalid for all other
}values of n.
So you have proved that x^n + y^n = z^n has no solution for n=1.
Don't you see any problem with that ?
--
Planar
D A S Vivash
>y^n = z^n equation only working when n=2, making it invalid for all other
>values of n.
So x^n + y^n = z^n has no integer solutions when n=1 ?
Now there's a conjecture I'd like to disprove !!
David
fjm...@my-dejanews.com
>
> So x^n + y^n = z^n has no integer solutions when n=1 ?
>
> Now there's a conjecture I'd like to disprove !!
>
> David
>
Hi David. How would you disprove it? I can prove it by showing that, when
using my sample 3,4,5 roots, x^1+y^1 does not = z^1 (3+4 does not = 5). Or am
I doing something wrong. And the same for when n>2 but this is not the case
when n = 2. And this will work with any valid root sets every time. Frank
fjm...@my-dejanews.com
>
> So you have proved that x^n + y^n = z^n has no solution for n=1.
> Don't you see any problem with that ?
>
> --
> Planar
No, it falls in my category of invalid forms of the equation when n<>2.
Please tell me what the problem is. Using my sample roots of 3,4,5 in the
above equation does 3^1 + 4^1 = 5^1? I don't think it does! Is the form when
n=1 not invalid? And this applies also when n>2 but not when n=2. Frank
fjm...@my-dejanews.com
Clive Tooth <cl...@pisquaredoversix.force9.co.uk> wrote:
> fjm...@my-dejanews.com wrote:
>
> > Time will tell. All I need is one person to understand what I am trying to
> > say.
>
> But that one person might be James Harris, or Archimedes Plutonium, or
> Nathan, or ...
> :))
>
> Good luck Frank...
>
Ha1 Ha! That's good! But I don't need any luck. Just time to be convinced my
thinking is wrong so I can forget about this thing. I think I am getting
there, tho, from the help of this group, but I just can't keep from wondering
why people are trying to prove an invalid equation? I see that James may have
given up. I am looking forward to the day when I can do that. Thanks Clive.
Frank
Frank
fjm...@my-dejanews.com
>
> From your above reply, and from those to other pespondents, I see that
> you understand that all you showed (by a geometric argument) is the
> fact that triples {x,y,z} solving x^2+y^2=z^2,
, -- and {a,b,c} solving a^n+b^n=c^n for n<>2,
> are necessarily different: {x,y,z} <> {a,b,c}
^^^^^^^^^^^^^^^^^^^^(1) (1) In my examples, triplets x,y,z are the same as
a,b,c and I used the 3,4,5 triplet for simplicity only since they all work. I
am only showing that in one form of the equation (when n=2) it is solveable
and therefore valid. In the other form (when n<>2) using the same triplet, it
is not solveable and therefore invalid. Does not this inherent invalidity of
when n<>2 have any substance or meaning to anyone but me? Why attempt to
solve an invalid equation? Not to belabor a point, does it not show that the
sum of two integer cubes can not be an integer cube? I can show that many
irrational cubes are the sum of two integer cubes, that's easy. But there
roots will not form right triangles, whereas roots when n=2 always form right
triangles, whether integer or irrational. Can not the inability/ability to
form right triangles have any significance as to the validity of various
forms of the equation?
> Many, including me, have pointed out that this does NOT imply FLT
> (there is no solution {a,b,c} in positive integers for n>2)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^(2) (2) Very true, but there are many
solutions when a,b are positive integers and c is irrational (by solving for
c) and these are specifically the ones I am talking about. These solutions
(roots) will not form right triangles, the x,y,z roots (integer or
irrational) will.
> It only shows that IF there were such solution, it must be different
> from any {x,y,z} triple for n=2. You see the big distance to FLT ?-)
>
> Surely, it is depressing to realize that your clue does not prove
> FLT (= there is NO {a,b,c} solution for n>2) -- But what the heck,
> you learned something after all...
> --
> Ciao, Nico Benschop
> bens...@iae.nl -- http://www.iae.nl/users/benschop
Nico, you are helping me to think I may soon see the error in my thiking, but
not quite yet. Please keep trying. Thank you. Frank
Ed Hook
|>
|>
|> >
|> > So x^n + y^n = z^n has no integer solutions when n=1 ?
|> >
|> > Now there's a conjecture I'd like to disprove !!
|> >
|> > David
|> >
|>
|> Hi David. How would you disprove it? I can prove it by showing that, when
|> using my sample 3,4,5 roots, x^1+y^1 does not = z^1 (3+4 does not = 5). Or am
|> I doing something wrong. And the same for when n>2 but this is not the case
|> when n = 2. And this will work with any valid root sets every time. Frank
|>
To answer your second question: you _are_ doing something wrong.
Moreover, numerous people have told you quite plainly that you're
doing something wrong. But you persist ...
The argument that you keep making claims to "prove" that x^n + y^n = z^n
has no solutions in positive integers x, y, z for n different from 2
_because_ that same equation _does_ have lots of solutions when n = 2,
none of which satisfy the Fermat equation when the exponent isn't 2.
People keep pointing out that (a) this is *trivial* and (b) it has
_absolutely_nothing_ to do with proving Fermat's Last Theorem.
Since you claim that n = 1 is ruled out for exactly the same reason
as n > 2, I'd think that the following (selected pretty much at random)
would give you pause:
1 + 2 = 3
42 + 69 = 111
2 + 7 = 9
And so forth ... Each of those lines exhibits a solution in positive
integers x, y, z of the equation x + y = z. If you agree with that
assertion _and_ you agree that your argument holds for n = 1 to just
exactly the same extent that it holds for n > 2, then you're pretty
much forced to acknowledge that your position is (not to be uncharitable)
flawed ...
|> -----------== Posted via Deja News, The Discussion Network ==----------
|> http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
--
Ed Hook | Copula eam, se non posit
MRJ Technology Solutions, Inc. | acceptera jocularum.
NAS, NASA Ames Research Center | I can barely speak for myself, much
Internet: ho...@nas.nasa.gov | less for my employer
D A S Vivash
>
>
>
>>
>> So x^n + y^n = z^n has no integer solutions when n=1 ?
>>
>> Now there's a conjecture I'd like to disprove !!
>>
>> David
>>
>
>Hi David. How would you disprove it? I can prove it by showing that, when
>using my sample 3,4,5 roots, x^1+y^1 does not = z^1 (3+4 does not = 5). Or am
>I doing something wrong. And the same for when n>2 but this is not the case
>when n = 2. And this will work with any valid root sets every time. Frank
>
Maybe you're confused. Obviously saying "x^n + y^n = z^n has no integer
solutions when n=1" is complete nonsense. It's got infinitely many solutions.
Take x=1, y=1, z=2 as an example. That's one solution. Now add one to x, and
one to z. Another solution! Repeat.
I'm not saying every x, y and z will give solutions (that's hardly ever the
case with equations, only identities), I'm saying there are SOME solutions (in
this case infinitely many). When n>2 there are NO solutions [fermat] and when
n=2 there are infintely many [Euclid?]
David
Nico Benschop
>
> >
> > So x^n + y^n = z^n has no integer solutions when n=1 ? ...(#)
> >
> > > Now there's a conjecture I'd like to disprove !! ...(*)
> > >
> > > David
> Hi David. How would you disprove it? I can prove it by showing that,
> when using my sample 3,4,5 roots, x^1+y^1 does not = z^1 (3+4 does not
> = 5). Or am I doing something wrong. And the same for when n>2 but
> this is not the case when n = 2. And this will work with any valid
> root sets every time. Frank
^^^
for n=2 (!)
Re(#,*): A classical case of quoting out-of-context ;-)
Earlier in the text, Frank assumed that{ x,y,z} is a solution for n=2.
Nico Benschop
>
> Nico wrote:
> >
> > From your above reply, and from those to other pespondents, I see
> > that you understand that all you showed (by a geometric argument)
> > is the fact that triples {x,y,z} solving x^2+y^2=z^2,
> ^^^^^^^^^^^^^^^^^ (1)
> > you learned something after all... -- Ciao, Nico Benschop
>
^^^
who?-) [*I* did my best;-]
I must admit I did not in detail follow your geometric reasoning, since
your whole line of 'proof' concentrates only on showing {x,y,z} triples
for n=2 and n<>2 are distinct [no problem there, easier full-proof by
algebraic function f(x,y,z;n) as I showed you] -- and then "stopping*!
Clearly only Pythagoras triples have to do with rectangular triangles
(n=2). Generalized: orthogonality <--> inner vector product=0. No news..
But you overlook the fact that this proves *nothing* about the very
existence of such integers {x,y,z} for n>2 (to be disproved for FLT).
fjm...@thegrid.net
ho...@nas.nasa.gov wrote:
> To answer your second question: you _are_ doing something wrong.
^^^^^^^^^^^^^^^^^^^^^^^^^^ What?? And I am not asking to help me with a
solution. I really want to know.
> Moreover, numerous people have told you quite plainly that you're
> doing something wrong. But you persist ...
Only 5 people, but they all have learned only one way (unless that's all
there is) to solve FLT. I never learned how and therefore am able to come up
with a possibly new, unbiased way of showing that Fermays' conjecture was
true.
>
> The argument that you keep making claims to "prove" that x^n + y^n = z^n
> has no solutions in positive integers x, y, z for n different from 2
^^^^^^^^^^^^^^^^^^^^^^^^ Not only this but that it is an invalid equation.
Now if this means the same thing, than you are right and I am closer to
realizing that my thinking is wrong.
> _because_ that same equation _does_ have lots of solutions when n = 2,
> none of which satisfy the Fermat equation when the exponent isn't 2.
> People keep pointing out that (a) this is *trivial* and (b) it has
^^^^^^^^^^ ^^^^^
> _absolutely_nothing_ to do with proving Fermat's Last Theorem
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ When you say "trivial" I
assume you mean unimportant. But this is the crux of my argument and I am not
trying to prove FLT. I am just trying to show that no solution is possible
because the equation becomes invalid when n<>2.
> Since you claim that n = 1 is ruled out for exactly the same reason
> as n > 2, I'd think that the following (selected pretty much at random)
> would give you pause:
^^^^^^^^^^^^
> 1 + 2 = 3
> 42 + 69 = 111
> 2 + 7 = 9
>
> And so forth ... Each of those lines exhibits a solution in positive
> integers x, y, z of the equation x + y = z. If you agree with that
> assertion _and_ you agree that your argument holds for n = 1 to just
> exactly the same extent that it holds for n > 2, then you're pretty
> much forced to acknowledge that your position is (not to be uncharitable)
> flawed ...
No pause at all!! The three sets of integers you give above are supposed to
be x,y,z roots useable in the equation to compare apples with apples. They
are not valid roots, simply the sum of two integers. For example, in your
first line, 1+2 does = 3 but 1^2 + 2^2 does not = 3^2, nor does 1^3 + 2^3 =
3^3, etc.,etc. You must start with valid roots for when n=2 and then when
using them when n<>2 these valid roots cause the equation to become invalid.
What you have done is solve for the 3rd root by adding the other two instead
of getting the square root of the sum of the others' squares. Using any of
your examples when n=2 and solving for the 3rd root will give valid roots
(one that will form right triangles). But not when n<>2. Then these roots
will only form scalene triangles with no 90 degree angle. Does this not show
that the n=2 form of the equation is totally different from the n<>2 form? I
can't help from thinking that it does and I sure hope someone will prove me
wrong so I can put this to rest. Thanks for your response, Ed. Frank J Manus
(fjm...@thegrid.net)
> Ed Hook | Copula eam, se non posit
> MRJ Technology Solutions, Inc. | acceptera jocularum.
> NAS, NASA Ames Research Center | I can barely speak for myself, much
> Internet: ho...@nas.nasa.gov | less for my employer
>
-----------== Posted via Deja News, The Discussion Network ==----------
Sander Vesik
> >
> > So you have proved that x^n + y^n = z^n has no solution for n=1.
> > Don't you see any problem with that ?
> >
> > --
> > Planar
> No, it falls in my category of invalid forms of the equation when n<>2.
> Please tell me what the problem is. Using my sample roots of 3,4,5 in the
> above equation does 3^1 + 4^1 = 5^1? I don't think it does! Is the form when
> n=1 not invalid? And this applies also when n>2 but not when n=2. Frank
But 3^1+2^1=5^1
In general, (x+n)^1 + (y+m)^1 = (x+y+n+m)^1 holds. Always.
At least for integers.
> -----------== Posted via Deja News, The Discussion Network ==----------
> http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
--
Sander
There is no love, no good, no happiness and no future -
all these are just illusions.
Clive Tooth
> No pause at all!! The three sets of integers you give above are supposed to
> be x,y,z roots useable in the equation to compare apples with apples. They
> are not valid roots, simply the sum of two integers. For example, in your
> first line, 1+2 does = 3 but 1^2 + 2^2 does not = 3^2, nor does 1^3 + 2^3 =
> 3^3, etc.,etc. You must start with valid roots for when n=2 and then when
> using them when n<>2 these valid roots cause the equation to become invalid.
> What you have done is solve for the 3rd root by adding the other two instead
> of getting the square root of the sum of the others' squares. Using any of
> your examples when n=2 and solving for the 3rd root will give valid roots
> (one that will form right triangles). But not when n<>2. Then these roots
> will only form scalene triangles with no 90 degree angle. Does this not show
> that the n=2 form of the equation is totally different from the n<>2 form? I
> can't help from thinking that it does and I sure hope someone will prove me
> wrong so I can put this to rest. Thanks for your response, Ed. Frank J Manus
> (fjm...@thegrid.net)
Frank,
I am probably going to hate myself for doing this. But I am going to
have one last try at convincing you of the error of your ways. I will
try a different approach.
I will state (one formulation of) Fermat's Last Theorem:
"When n is an integer, n>2, the equation x^n+y^n=z^n has no solution in
positive integers x, y, z."
Frank: Do you accept this is a valid formulation of FLT?
As you have rightly pointed out: in a right angled triangle with sides
x, y and hypotenuse z we always have x^2+y^2=z^2.
Imagine a cuboid (otherwise called a brick, or a rectangular
parallelepiped) with sides x, y and z. Let the length of its space
diagonal (from one corner to the opposite corner) be d. It is a fact
that x^2+y^2+z^2=d^2. I do not know if you knew this.
So, for example, a 3x4x12 brick has a diagonal of 13, because
3^2+4^2+12^2 = 9+16+144 = 169 = 13^2.
I will now formulate Clive's Last Theorem (CLT):
"When n is an integer, n>2, the equation x^n+y^n+z^n=d^n has no solution
in positive integers x, y, z, d."
My question to you, Frank, is this:
Do you think that the truth or falsity of CLT has anything to do with
the fact that the formula for the length of the space diagonal of a
cuboid is x^2+y^2+z^2=d^2 ?
[ I hope nobody responds with an opinion of the truth or falsity of CLT.
]
Planar
>No pause at all!! The three sets of integers you give above are supposed to
>be x,y,z roots useable in the equation to compare apples with apples. They
>are not valid roots, simply the sum of two integers. For example, in your
>first line, 1+2 does = 3 but 1^2 + 2^2 does not = 3^2, nor does 1^3 + 2^3 =
>3^3, etc.,etc. You must start with valid roots for when n=2 and then when
>using them when n<>2 these valid roots cause the equation to become invalid.
OK, I think I get it. You are trying (with success) to prove:
There does not exist x, y, z in N such that for all n in N,
x^n + y^n = z^n
But this is not FLT. FLT is the following.
There does not exist x, y, z in N such that there exists n in N
such that n > 2 and x^n + y^n = z^n
Do you understand the difference between the two ?
--
Planar
Planar
>No, it falls in my category of invalid forms of the equation when n<>2.
>Please tell me what the problem is.
What do you mean by the words "invalid forms of the equation" ?
> Using my sample roots of 3,4,5 in the
>above equation does 3^1 + 4^1 = 5^1? I don't think it does! Is the form when
>n=1 not invalid? And this applies also when n>2 but not when n=2. Frank
So you have proved that 3^n + 4^n = 5^n only has solutions when n=2.
Unfortunately, FLT is a lot more general than that.
--
Planar
fjm...@thegrid.net
PMA9...@shef.ac.uk (D A S Vivash) wrote:
> >> So x^n + y^n = z^n has no integer solutions when n=1 ?
> >>
> >> Now there's a conjecture I'd like to disprove !!
> >>
> >> David
> >>
fjmanus wrote this to David:
> >Hi David. How would you disprove it? I can prove it by showing that, when
> >using my sample 3,4,5 roots, x^1+y^1 does not = z^1 (3+4 does not = 5). Or am
> >I doing something wrong. And the same for when n>2 but this is not the case
> >when n = 2. And this will work with any valid root sets every time. Frank
> >
>
> Maybe you're confused. Obviously saying "x^n + y^n = z^n has no integer
> solutions when n=1" is complete nonsense. It's got infinitely many solutions
True it is complete nonsense, but I never said that. There are an infinite
number of solutions to the equation as YOU are using it but that is what's
wrong. The equation does not stand alone which is what you have done with it.
And then you are randonly plugging in values for x and y, and then adding
them. You must use a constant set of roots (pythagorean) in all forms of the
x^n+y^n=z^n equation from n=1 to infinity. Then, with these constant roots,
x^1+y^1=z^1 is the first equation of an infinite set of equations---(
x^1+y^1=z^1 does not stand alone..). Your (x^1+y^1=z^1) is actually and only
(x+y=z), the sum of two integers. My ( x^1+y^1=z^1) is the first equation in
an infinite series of equations using a common root set. I am using a common
root set only to show that it is true.
>
> Take x=1, y=1, z=2 as an example. That's one solution. Now add one to x, and
> one to z. Another solution! Repeat.
>
> I'm not saying every x, y and z will give solutions (that's hardly ever the
> case with equations, only identities), I'm saying there are SOME solutions (in
> this case infinitely many). When n>2 there are NO solutions [fermat] and when
> n=2 there are infintely many [Euclid?]
>
> David
But every x,y and z WILL give solutions. The way you are using the equation,
every x+y will equal a *z*, for all you are doing is adding two numbers.
Also, when n=2 the solutions are pythagorean, when n>2 they are invalid, and
when n=1 they are Euclidean.
Frank
Edward C. Hook
|> In article <78o065$1d5$1...@sun500.nas.nasa.gov>,
|> ho...@nas.nasa.gov wrote:
|>
|> > To answer your second question: you _are_ doing something wrong.
|>
|> ^^^^^^^^^^^^^^^^^^^^^^^^^^ What?? And I am not asking to help me with a
|> solution. I really want to know.
See, I'm not sanguine about how this is going to turn out ... I
_thought_ that what I said below was an *explanation* of your
basic mistake ...
|>
|> > Moreover, numerous people have told you quite plainly that you're
|> > doing something wrong. But you persist ...
|>
|> Only 5 people, but they all have learned only one way (unless that's all
|> there is) to solve FLT. I never learned how and therefore am able to come up
|> with a possibly new, unbiased way of showing that Fermays' conjecture was
|> true.
Except that you seem to fundamentally confused about just exactly
what Fermat's Last Theorem says *and* what sorts of things constitute
a mathematical proof.
|>
|> >
|> > The argument that you keep making claims to "prove" that x^n + y^n = z^n
|> > has no solutions in positive integers x, y, z for n different from 2
|>
|> ^^^^^^^^^^^^^^^^^^^^^^^^ Not only this but that it is an invalid equation.
What do you mean by the statement that "it is an invalid equation" ??
There's nothing wrong with it _qua_ equation -- it just so happens that
it has no solutions in positive integers x,y,z for n > 2. But, if you
look for solutions in _real_ numbers (for example), you'll find that this
equation has _lots_ of solutions for whatever value of n ...
|> Now if this means the same thing, than you are right and I am closer to
|> realizing that my thinking is wrong.
|>
|> > _because_ that same equation _does_ have lots of solutions when n = 2,
|> > none of which satisfy the Fermat equation when the exponent isn't 2.
|> > People keep pointing out that (a) this is *trivial* and (b) it has
|>
|> ^^^^^^^^^^ ^^^^^
|>
|> > _absolutely_nothing_ to do with proving Fermat's Last Theorem
|>
|> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ When you say "trivial" I
|> assume you mean unimportant. But this is the crux of my argument and I am not
|> trying to prove FLT. I am just trying to show that no solution is possible
|> because the equation becomes invalid when n<>2.
|>
No, by "trivial" I mean that's it's a result that's very easy to prove
and one which doesn't shed much light on anything. Let me try to write
out an actual proof of what I understand to be your major claim (or, in
fact, a generalization of it):
Frank's Theorem. If n, m are different positive integers and x, y, z
are nonzero real numbers satisfying x^n + y^n = z^n, then x^m + y^m is
different from z^m.
Proof: Assume, without loss of generality, that x <= y. If x^n + y^n = z^n,
then 1 + (x/y)^n = (z/y)^n or z/y = (1 + (x/y)^n)^{1/n}. Defining the
function f_n by f_n(r) = (1 + r^n)^{1/n}, we have f_n(x/y) = z/y with
x/y \in (0,1) and we want to show that z/y is _not_ equal to f_m(x/y).
But, if r \in (0,1) and n < m, we have 1 + r^n > 1 + r^m; then we have
(1 + r^n)^{1/n} > (1 + r^m)^{1/n} > (1 + r^m)^{1/m} ,
where the first inequality holds because the n'th root function is
monotone increasing and the second holds because 1 + r^m > 1 (just
take logs to verify it). This shows that, whenever r \in (0,1), it is
true that f_n(r) =/= f_m(r) when n,m are different.
There you have it -- as a corollary, your main point follows: no solution
(in positive integers -- though that's really irrelevant) of the Fermat
equation x^n + y^n = z^n for n =/= 2 can also satisfy the Pythagorean
equation x^2 + y^2 = z^2.
So, what you say is _true_ ... it just doesn't have _anything_ to do
with Fermat's Last Theorem, which is:
Fermat's Last Theorem [Wiles _et_al._]: The equation x^n + y^n = z^n
has no solutions with x, y, z positive integers, when n > 2.
The fact that an alleged solution to Fermat's equation could not _also_
be a solution of the Pythagorean version does NOT demonstrate the truth
of Fermat's Last Theorem ... because there's absolutely no reason to
suppose that the two equations have to have the same solutions. You
appear to believe that, but no one else does. And that's what everyone
has been trying to tell you.
|> > Since you claim that n = 1 is ruled out for exactly the same reason
|> > as n > 2, I'd think that the following (selected pretty much at random)
|> > would give you pause:
|> ^^^^^^^^^^^^
|> > 1 + 2 = 3
|> > 42 + 69 = 111
|> > 2 + 7 = 9
|> >
|> > And so forth ... Each of those lines exhibits a solution in positive
|> > integers x, y, z of the equation x + y = z. If you agree with that
|> > assertion _and_ you agree that your argument holds for n = 1 to just
|> > exactly the same extent that it holds for n > 2, then you're pretty
|> > much forced to acknowledge that your position is (not to be uncharitable)
|> > flawed ...
|>
|> No pause at all!! The three sets of integers you give above are supposed to
|> be x,y,z roots useable in the equation to compare apples with apples. They
|> are not valid roots, simply the sum of two integers. For example, in your
But, you see, that was the _whole_point_ -- your argument applies to the
case n = 1 just as it does to n > 2. So you are claiming that the equation
x + y = z can have _no_ solutions in positive integers x, y, z. Hence, the
randomly-chosen counterexamples to your claim that I displayed up above
_should_ put an end to the discussion. Your argument "proves" something
that is patently false - _that's_ why the above should give you pause ...
|> first line, 1+2 does = 3 but 1^2 + 2^2 does not = 3^2, nor does 1^3 + 2^3 =
|> 3^3, etc.,etc. You must start with valid roots for when n=2 and then when
|> using them when n<>2 these valid roots cause the equation to become invalid.
|> What you have done is solve for the 3rd root by adding the other two instead
|> of getting the square root of the sum of the others' squares. Using any of
|> your examples when n=2 and solving for the 3rd root will give valid roots
|> (one that will form right triangles). But not when n<>2. Then these roots
|> will only form scalene triangles with no 90 degree angle. Does this not show
|> that the n=2 form of the equation is totally different from the n<>2 form? I
Yes, it (sort of) shows that the nontrivial solutions, if there are any,
of the Fermat equation have to be different for different exponents. I say
"sort of", because nothing you've said really amounts to a proof, as that
term is usually understood hereabouts.
Again, though, this observation is a far cry from a proof of Fermat's
Last Theorem -- if you wanted to turn it into one of those, you'd have
to prove (in addition) that any solution of the Fermat equation with
n > 2 is _necessarily_ a solution of the Pythagorean equation as well.
I don't think that you're likely to be able to do that, though :-(
|> can't help from thinking that it does and I sure hope someone will prove me
|> wrong so I can put this to rest. Thanks for your response, Ed. Frank J Manus
|> (fjm...@thegrid.net)
|>
--
Charles H. Giffen
>
> fjm...@thegrid.net wrote:
>
> > No pause at all!! The three sets of integers you give above are supposed to
> > be x,y,z roots useable in the equation to compare apples with apples. They
> > are not valid roots, simply the sum of two integers. For example, in your
> > 3^3, etc.,etc. You must start with valid roots for when n=2 and then when
> > using them when n<>2 these valid roots cause the equation to become invalid.
> > What you have done is solve for the 3rd root by adding the other two instead
> > of getting the square root of the sum of the others' squares. Using any of
> > your examples when n=2 and solving for the 3rd root will give valid roots
> > (one that will form right triangles). But not when n<>2. Then these roots
> > will only form scalene triangles with no 90 degree angle. Does this not show
> > that the n=2 form of the equation is totally different from the n<>2 form? I
> > wrong so I can put this to rest. Thanks for your response, Ed. Frank J Manus
> > (fjm...@thegrid.net)
>
3^3 + 4^3 + 5^3 = 6^3
End of story for CLT.
--Chuck Giffen
fjm...@thegrid.net
Sander Vesik <san...@haldjas.folklore.ee> wrote:
> fjm...@my-dejanews.com wrote:
>
> > >
> > > So you have proved that x^n + y^n = z^n has no solution for n=1.
> > > Don't you see any problem with that ?
> > >
>
> > No, it falls in my category of invalid forms of the equation when n<>2.
> > above equation does 3^1 + 4^1 = 5^1? I don't think it does! Is the form when
> > n=1 not invalid? And this applies also when n>2 but not when n=2. Frank
>
>
> In general, (x+n)^1 + (y+m)^1 = (x+y+n+m)^1 holds. Always.
> At least for integers.
> Sander
>
> There is no love, no good, no happiness and no future -
> all these are just illusions.
Yes, but ther x,y,z roots here are 3,2,5 and they are not valid roots, they
are not pythagorean. All you show here is an equation which is the sum of two
integers and you are right that it holds. But you are not comparing apples
with apples. The starting point is the valid equation when n=2. Any other
value of n (always using constant valid roots) cause the equation to become
invalid. You are using the equation when n=1 alone, not in the set of all
equations, and, in your case, the equation is only x+y=z, not x^1+y^1=z^1.
Yours is taken out of context
Frank
fjm...@thegrid.net
Nico Benschop <bens...@iae.nl> wrote:
> fjm...@my-dejanews.com wrote:
> >
> > >
> > > So x^n + y^n = z^n has no integer solutions when n=1 ? ...(#)
> > >
>
> > > > Now there's a conjecture I'd like to disprove !! ...(*)
> > > >
> > > > David
>
> > Hi David. How would you disprove it? I can prove it by showing that,
> > when using my sample 3,4,5 roots, x^1+y^1 does not = z^1 (3+4 does not
> > = 5). Or am I doing something wrong. And the same for when n>2 but
> > this is not the case when n = 2. And this will work with any valid
> > root sets every time. Frank
> for n=2 (!)
Definitely for n=2 and only for n=2 as I have been saying. In the equation
using n=2, with my sample roots (3,4,5), does not 3^2+4^2=5^2? It sure does!
But, when these roots are used in the equation when n=1, 3+4 does not = 5.
And further, when n=3, 3^3+4^3 does not equal 5^3 (27+64 does not =125);
when n=4, 3^4+4^4 does not =5^4 (81+256 does not equal 625); etc., etc.
>
> Re(#,*): A classical case of quoting out-of-context ;-)
I assume you mean that others are using the n=1 form of the equation out of
context with my parameters.
>
> Earlier in the text, Frank assumed that{ x,y,z} is a solution for n=2.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I did not assume (x,y,z) is a solution for
n=2. The values of (x,y,z) are (3,4,5) respectively and these are root values
to be used as an example only in the equation from 1 to infinity. But only
when n=2 will any of the solved equations (using these sample roots) be
valid. Did I mention the right triangle yet? The fact that a right triangle
can be formed with these roots only when n=2 is only used to show that n=2
stands alone.
Frank Manus
>
> --
> Ciao, Nico Benschop
> bens...@iae.nl -- http://www.iae.nl/users/benschop
>
-----------== Posted via Deja News, The Discussion Network ==----------
fjm...@thegrid.net
Nico Benschop <bens...@iae.nl> wrote:
> I must admit I did not in detail follow your geometric reasoning, since
> your whole line of 'proof' concentrates only on showing {x,y,z} triples
> for n=2 and n<>2 are distinct [no problem there, easier full-proof by
> algebraic function f(x,y,z;n) as I showed you] -- and then "stopping*!
But I don't need to go any further for there is nothing to solve. I think
these thoughts suffice to show that no nth power (except 2) is the sum of
two nth powers (except 2). Even Wiles did not show this and that was all
Fermat said. I wish you would follow all the details of my reasoning and look
past trying to solve FLT as I have done. I don't think anyone has ever put
in figures, turned the crank, and enjoyed seeing what came out like we all
have done during our learning years. I have.
>
> Clearly only Pythagoras triples have to do with rectangular triangles
> (n=2). Generalized: orthogonality <--> inner vector product=0. No news..
>
> But you overlook the fact that this proves *nothing* about the very
> existence of such integers {x,y,z} for n>2 (to be disproved for FLT).
There we go again, implying that FLT must be proven.
> --
> Ciao, Nico Benschop
> bens...@iae.nl -- http://www.iae.nl/users/benschop
Frank Manus
Jim Ferry
>
> Yes, but ther x,y,z roots here are 3,2,5 and they are not valid roots, they
> are not pythagorean. All you show here is an equation which is the sum of two
> integers and you are right that it holds. But you are not comparing apples
> with apples. The starting point is the valid equation when n=2. Any other
> value of n (always using constant valid roots) cause the equation to become
> invalid. You are using the equation when n=1 alone, not in the set of all
> equations, and, in your case, the equation is only x+y=z, not x^1+y^1=z^1.
> Yours is taken out of context
I haven't read this thread thoroughly. Could you (Frank) summarize
it by answering two quick multiple choice questions?
1) The statement of Fermat's Last Theorem most resembles
a) "For any integer n >= 3, and any non-zero integers x, y
and z, x^n + y^n != z^n."
b) "For any integer n >= 3, and any non-zero integers x, y
and z such that x^2 + y^2 = z^2, x^n + y^n != z^n."
2) The statement you have proven most resembles
a) "For any integer n >= 3, and any non-zero integers x, y
and z, x^n + y^n != z^n."
b) "For any integer n >= 3, and any non-zero integers x, y
and z such that x^2 + y^2 = z^2, x^n + y^n != z^n."
Here "!=" means "not equal to."
| Jim Ferry | Center for Simulation |
+------------------------------------+ of Advanced Rockets |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@expunge_this_field.uiuc.edu | University of Illinois |
Keith Ellul
> will only form scalene triangles with no 90 degree angle. Does this not show
> that the n=2 form of the equation is totally different from the n<>2 form? I
YES! It is totally different! No one is denying that!
It is impossible to form a right triangle with sides a^n, b^n, and c^n
if n is not 2. Fine. Great. This is NOT what Fermat's last theorem
states. Fermat's last theorem makes no mention of triangles, right
or otherwise.
Fermat's last theorem states that a^n + b^n = c^n has no
positive integer solutions when n > 2. You proving that this
is totally different from some other equation is irrelevant.
It is totally different from just about any random equation
that I happen to mention.
What you have done is shown that the two equations are totally different.
What you have not done is shown that Fermat's equation has no solutions.
The fact that two equations are "totally different" is irrelevant.
They might both have solutions. Different solutions.
curmu...@aol.com
> them. You must use a constant set of roots (pythagorean) in all forms of the
> x^n+y^n=z^n equation from n=1 to infinity. Then, with these constant roots,
(I know I said I wouldn't respond any more... so, I lied...)
See, the problem is that there is no restriction that when investigating the
equation
x^n+y^n=z^n
that we must stick to the roots of the equation when n=2. In fact, the
point of most posts to yours has been that when n=1, the values for {x,y,z}
are totally different for n=1 than n=2. This should be a clue to you that
your reasoning does not lend any new or useful information to the resolution
of FLT or the claiming of the Wolfskehl prize.
Further, the solutions to the equation above do not behave the same for
different values of n (as you've so oft noted). That's no big deal. In fact,
that's to be expected. Regardless, whether or not the equation has integer
solutions when n=1 and n=2 has absolutely no bearing on whether the equation
has solutions when n>2. Whether or not right triangles are formed also has no
bearing. The points that your missing is this...
You are assuming that because when n<>2, right triangles are not formed that
there is some significance. There is not, at least as far as FLT is concerned
(or the Wolfskehl prize, for that matter). Your argument that FLT is
Pythagorean in nature (i.e., that all roots must behave the n=2 case) is
flawed. There is no such restriction.
Joe
bens...@iae.nl
fjm...@thegrid.net wrote:
> In article <36B05F...@iae.nl>,
> Nico Benschop <bens...@iae.nl> wrote:
>
> > I must admit I did not in detail follow your geometric reasoning, since
> > your whole line of 'proof' concentrates only on showing {x,y,z} triples
> > for n=2 and n<>2 are distinct [no problem there, easier full-proof by
> > algebraic function f(x,y,z;n) as I showed you] -- and then "stopping*!
>
> But I don't need to go any further for there is nothing to solve.
> I think these thoughts suffice to show that no nth power (except 2)
Well, Frank, *you* said it: your aim (despite your denials;-) *is*
proving FLT: the sum of two n-th powers is not an n-th power for n>2.
vvv: yes he did (in a very indirect way)
> Even Wiles did not show this and that was all Fermat said.
> I wish you would follow all the details of my reasoning and look
> past trying to solve FLT as I have done. I don't think anyone has
> ever put in figures, turned the crank, and enjoyed seeing what came out like
> we all have done during our learning years. I have.
Figures are nice to get idea's, but *ot* to prove anything beyond doubt.
For that your better use algebra & arithmetic, so called 'formal' proof.
Your trails show how imcomplete figures can be (in fact loosing sight
of the main line;-(
> > Clearly only Pythagoras triples have to do with rectangular triangles
> > (n=2). Generalized: orthogonality <--> inner vector product=0. No news..
> >
> > But you overlook the fact that this proves *nothing* about the very
> > existence of such integers {x,y,z} for n>2 (to be disproved for FLT).
>
> There we go again, implying that FLT must be proven. -- Frank Manus
Indeed: as you yourself wrote two paragraphs ago! (see above).
Is your memory *that* short?
I better call it quits now;-)
Ciao, Nico Benschop
bens...@iae.nl -- http://www.iae.nl/users/benschop/ferm.htm
Clive Tooth
> Clive Tooth wrote:
> >
> > [ I hope nobody responds with an opinion of the truth or falsity of CLT.
> > ]
> > Clive Tooth
> >
>
> 3^3 + 4^3 + 5^3 = 6^3
>
> End of story for CLT.
>
> --Chuck Giffen
<sigh>
fjm...@thegrid.net
Don't <sigh>, Clive. All is not lost. Charles H. Gillian lucked out finding
that series of numbers but I can't prove it yet. The proof will parallel my
thoughts on FLT since these bricks are related to triangles; sorta cousins.I
have not answered your post to me yet because I must do more thinking to give
it the answer it deserves.
But back to Gillians brick: His space diagonal of 6 is totally wrong (it
should be 7.07, but the cube of 7.07 will not equal the sum of his other
three cubes where as 6^3 will. I can't visualize the proof yet but it will be
similiar to my FLT thoughts because, if you'll notice, his space diagonal has
decreased from 7.07 (which it should be) to 6 (which it is in his
example--which is right) just as one side of my FLT triangles decrease an *n*
increases. Draw the brick and you will see that his inner space triangle has
sides of 5 and 5. The hypotenuse (space diagonal) of these sides is 7.07, not
6 as his is. Something is wrong!! I'll start working on it, along with my
other stuff, of course.
One last thing in closing to make you feel better. Consider that you have a
brick 3 x 4 x 12 and his is 3 x 4 x 5. Is he not working with half a brick??
ROWL (rolling over with laughter).
Frank
fjm...@thegrid.net
(curmudgeon) are starting to make me question my faith in my FLT thoughts.
Don't worry, I am not a martyr. I have nine more replies to contemplate this
weekend. They may cause my self-demise to occur (thereby following JSH's
action) or I may accidentally reveal some fact that will make someone
comprehind what I myself may not know how to explain. I just don't like to
give up on something that may be plausible just because I can't explain what
I can see properly (even if I may be the only one seeing it); but, I also
don't like being alone in these sightings (no, not UFO sightings!). I can
take it but I don't like it. And I don't have any mathematically inclined
friends to discuss this with so I do very much appreciate this newsgroups
help and responses. And I will not abuse this help and responsiveness.
Earle Jones
wrote:
> In article <36B2D406...@pisquaredoversix.force9.co.uk>,
> Clive Tooth <cl...@pisquaredoversix.force9.co.uk> wrote:
> > "Charles H. Giffen" wrote:
> >
> > > Clive Tooth wrote:
> > > >
> > > > [ I hope nobody responds with an opinion of the truth or falsity of CLT.
> > > > ]
> > > > Clive Tooth
> > > >
> > >
> > > 3^3 + 4^3 + 5^3 = 6^3
--
By the way, what is the next member of this series:
1. 3^2 + 4^2 = 5^2
2. 3^3 + 4^3 + 5^3 = 6^3
3. ??
earle
--
Jim Ferry
> By the way, what is the next member of this series:
>
> 1. 3^2 + 4^2 = 5^2
>
> 2. 3^3 + 4^3 + 5^3 = 6^3
>
> 3. ??
I don't know what it is, but I know what it isn't: it's not
of the form
(a-1)^4 + a^4 + (a+1)^4 + (a+2)^4 = n^4, since the R.H.S. is
2 (2a(a+1)(a^2+a+8) + 9), which, being divisible by 2 for all a
but never by 4, is not n^k for any integers k>1 and n.
Clive Tooth
> Earle Jones wrote:
>
> > By the way, what is the next member of this series:
> >
> > 1. 3^2 + 4^2 = 5^2
> >
> > 2. 3^3 + 4^3 + 5^3 = 6^3
> >
> > 3. ??
>
> I don't know what it is, but I know what it isn't: it's not
> of the form
>
> (a-1)^4 + a^4 + (a+1)^4 + (a+2)^4 = n^4, since the R.H.S. is
>
> 2 (2a(a+1)(a^2+a+8) + 9), which, being divisible by 2 for all a
> but never by 4, is not n^k for any integers k>1 and n.
Well, why we are talking about this let us remember that:
95800^4+217519^4+414560^4 = 422481^4
and
2682440^4+15365639^4+18796760^4 = 20615673^4
And, of course,
1439965710648954492268506771833175267850201426615300442218292336336633^4+
4417264698994538496943597489754952845854672497179047898864124209346920^4+
9033964577482532388059482429398457291004947925005743028147465732645880^4
=
9161781830035436847832452398267266038227002962257243662070370888722169^4
fjm...@thegrid.net
Clive Tooth <cl...@pisquaredoversix.force9.co.uk> wrote:
>
> Frank,
>
> I am probably going to hate myself for doing this. But I am going to
> have one last try at convincing you of the error of your ways. I will
> try a different approach.
>
> I will state (one formulation of) Fermat's Last Theorem:
>
> "When n is an integer, n>2, the equation x^n+y^n=z^n has no solution in
> positive integers x, y, z."
>
> Frank: Do you accept this is a valid formulation of FLT?
Yes I do but I prefer the one about no cube being the sum of two other cubes.
> As you have rightly pointed out: in a right angled triangle with sides
> x, y and hypotenuse z we always have x^2+y^2=z^2.
Ok so far.
> Imagine a cuboid (otherwise called a brick, or a rectangular
> parallelepiped) with sides x, y and z. Let the length of its space
> diagonal (from one corner to the opposite corner) be d. It is a fact
> that x^2+y^2+z^2=d^2. I do not know if you knew this.
No, I never knew this. Never played with bricks before, just triangles. But I
am sure you'll agree that your bricks are to FLT what spherical geometry is to
geometry and my thoughts will apply the same way since triangles are involved.
>
> So, for example, a 3x4x12 brick has a diagonal of 13, because
> 3^2+4^2+12^2 = 9+16+144 = 169 = 13^2.
>
> I will now formulate Clive's Last Theorem (CLT):
>
> "When n is an integer, n>2, the equation x^n+y^n+z^n=d^n has no solution
> in positive integers x, y, z, d."
It can't because the equation becomes invalid, just like the pythagorean
equation that Fermat used. I bet the space diagonal formed in your brick will
start shrinking away from one of its corner as n increases to infinity and at
infinity will equal the length of the longer side. Your brick will have turned
in on itself!!
> My question to you, Frank, is this:
>
> Do you think that the truth or falsity of CLT has anything to do with
> the fact that the formula for the length of the space diagonal of a
> cuboid is x^2+y^2+z^2=d^2 ?
No, the truth or falsity of CLT has no effect on this formula. But (and I am
sorry) the formula will only *work* with powers of 2. Even Gillians' 3^3 + 4^3
+ 5^3 = 6^3 example didn't work (altho it blew a hole in CLT). The equation
becomes invalid when n>2. The space diagonal will begin shrinking and approach
the length of the larger side as n increases. Bricks or triangles, the results
are the same. The Pythagorean theorem only works with squares. And that's what
FLT is (or should be); showing that the Pythagorean theorem only works with
squares.
Frank
> [ I hope nobody responds with an opinion of the truth or falsity of CLT.
> ]
>
> --
>
> Clive Tooth
> http://www.pisquaredoversix.force9.co.uk/
> End of document
>
-----------== Posted via Deja News, The Discussion Network ==----------
fjm...@thegrid.net
> |> In article <78o065$1d5$1...@sun500.nas.nasa.gov>,
> |> ho...@nas.nasa.gov wrote:
> Except that you seem to fundamentally confused about just exactly
> what Fermat's Last Theorem says *and* what sorts of things constitute
> a mathematical proof.
I am not confused about these items. I am just going about it in a way no one
seems to have done in the past 360 years. And to come up with a simple way to
*prove* the statements' validity.
> |> ^^^^^^^^^^^^^^^^^^^^^^^^ Not only this but that it is an invalid
equation.
> What do you mean by the statement that "it is an invalid equation" ??
The equation is invalid when z^n will not equal the sum of x^n and y^n and
this only happens when n<>2. When n=2 x^n+y^n will equal z^n. Fermat said
that no cube is the sum of two cubes and this shows he was right by the
invalidity of when n<>2
> There's nothing wrong with it _qua_ equation -- it just so happens that
> it has no solutions in positive integers x,y,z for n > 2. But, if you
> look for solutions in _real_ numbers (for example), you'll find that this
> equation has _lots_ of solutions for whatever value of n ...
No, it doesn't have many solutions, unless you assign values to x and y and
solve for z. Then if you use these root values and try to make right triangles
with them, you won't be able to.
> No, by "trivial" I mean that's it's a result that's very easy to prove
> and one which doesn't shed much light on anything. Let me try to write
> out an actual proof of what I understand to be your major claim (or, in
> fact, a generalization of it):
I did not understand any of your *actual proof* so I snipped it. Sorry.
> So, what you say is _true_ ... it just doesn't have _anything_ to do
> with Fermat's Last Theorem, which is:
>
> Fermat's Last Theorem [Wiles _et_al._]: The equation x^n + y^n = z^n
> has no solutions with x, y, z positive integers, when n > 2.
>
> The fact that an alleged solution to Fermat's equation could not _also_
> be a solution of the Pythagorean version does NOT demonstrate the truth
> of Fermat's Last Theorem ... because there's absolutely no reason to
> suppose that the two equations have to have the same solutions. You
> appear to believe that, but no one else does. And that's what everyone
> has been trying to tell you.
I do not believe that and am not saying that. I am saying that when n=2 there
are solutions, when n<>2 there are no solutions, and hence the invalidity of
the equation.
>
> |> > Since you claim that n = 1 is ruled out for exactly the same reason
> |> > as n > 2, I'd think that the following (selected pretty much at random)
> |> > would give you pause:
> |> ^^^^^^^^^^^^
> |> > 1 + 2 = 3
> |> > 42 + 69 = 111
> |> > 2 + 7 = 9
> |> >
> |> > And so forth ... Each of those lines exhibits a solution in positive
> |> > integers x, y, z of the equation x + y = z. If you agree with that
> |> > assertion _and_ you agree that your argument holds for n = 1 to just
> |> > exactly the same extent that it holds for n > 2, then you're pretty
> |> > much forced to acknowledge that your position is (not to be
uncharitable)
> |> > flawed ...
> |>
> |> No pause at all!! The three sets of integers you give above are supposed to
> |> be x,y,z roots useable in the equation to compare apples with apples. They
> |> are not valid roots, simply the sum of two integers. For example, in your
>
> But, you see, that was the _whole_point_ -- your argument applies to the
> case n = 1 just as it does to n > 2. So you are claiming that the equation
Not with respect to the original equation when n=2 and again, x+y=z is the
sum of two numbers, not the sum of the roots of two powers. It takes this
form only because of n equalling 1.
> x + y = z can have _no_ solutions in positive integers x, y, z. Hence, the
> randomly-chosen counterexamples to your claim that I displayed up above
> _should_ put an end to the discussion. Your argument "proves" something
> that is patently false - _that's_ why the above should give you pause ...
It did not and for the same reasons. Please re-read the paragraph. There is no
1,2,3 triangle, nor a 42,69,111 triangle, nor 2,7,9. I do no use randomly
chosen numbers and the roots I use are valid and constant throughout the range
of all n's.
>
> |> first line, 1+2 does = 3 but 1^2 + 2^2 does not = 3^2, nor does 1^3 + 2^3
=
> |> 3^3, etc.,etc. You must start with valid roots for when n=2 and then when
> |> using them when n<>2 these valid roots cause the equation to become
invalid.
> |> What you have done is solve for the 3rd root by adding the other two
instead
> |> of getting the square root of the sum of the others' squares. Using any of
> |> your examples when n=2 and solving for the 3rd root will give valid roots
> |> (one that will form right triangles). But not when n<>2. Then these roots
> |> will only form scalene triangles with no 90 degree angle. Does this not
show
> |> that the n=2 form of the equation is totally different from the n<>2 form?
I
>
> Yes, it (sort of) shows that the nontrivial solutions, if there are any,
> of the Fermat equation have to be different for different exponents. I say
> "sort of", because nothing you've said really amounts to a proof, as that
> term is usually understood hereabouts.
>
> Again, though, this observation is a far cry from a proof of Fermat's
> Last Theorem -- if you wanted to turn it into one of those, you'd have
But I don't want to turn my *proof* into one of those. That's what I am
rebelling against.
Frank
> to prove (in addition) that any solution of the Fermat equation with
> n > 2 is _necessarily_ a solution of the Pythagorean equation as well.
> I don't think that you're likely to be able to do that, though :-(
> Ed Hook | Copula eam, se non posit
> MRJ Technology Solutions, Inc. | acceptera jocularum.
> NAS, NASA Ames Research Center | I can barely speak for myself, much
> Internet: ho...@nas.nasa.gov | less for my employer
>
>
-----------== Posted via Deja News, The Discussion Network ==----------
fjm...@thegrid.net
Jim Ferry <jferry@delete_this_field.uiuc.edu> wrote:
> fjm...@thegrid.net wrote:
> >
> > Yes, but ther x,y,z roots here are 3,2,5 and they are not valid roots, they
two
> > integers and you are right that it holds. But you are not comparing apples
> > with apples. The starting point is the valid equation when n=2. Any other
become
> > invalid. You are using the equation when n=1 alone, not in the set of all
> > equations, and, in your case, the equation is only x+y=z, not x^1+y^1=z^1.
> > Yours is taken out of context
>
> I haven't read this thread thoroughly. Could you (Frank) summarize
> it by answering two quick multiple choice questions?
>
> 1) The statement of Fermat's Last Theorem most resembles
>
> a) "For any integer n >= 3, and any non-zero integers x, y
> and z, x^n + y^n != z^n."
>
> b) "For any integer n >= 3, and any non-zero integers x, y
> and z such that x^2 + y^2 = z^2, x^n + y^n != z^n."
>
> 2) The statement you have proven most resembles
>
> a) "For any integer n >= 3, and any non-zero integers x, y
> and z, x^n + y^n != z^n."
>
> b) "For any integer n >= 3, and any non-zero integers x, y
> and z such that x^2 + y^2 = z^2, x^n + y^n != z^n."
>
> Here "!=" means "not equal to."
>
> +------------------------------------+ of Advanced Rockets |
> | http://www.uiuc.edu/ph/www/jferry/ +------------------------+
> | jferry@expunge_this_field.uiuc.edu | University of Illinois |
a and b are the same for both 1 and 2. In both b lines, I don't know why
x^2+y^2=y^2 as been added in. 1a is the only true statement. I have not proven
2a for I am not offering any proof. I am only showing that x^n+y^n=z^n is
invalid when n<>2 unless this constitutes a proof.
fjm...@thegrid.net
> > that the n=2 form of the equation is totally different from the n<>2 form? I
>
>
> It is impossible to form a right triangle with sides a^n, b^n, and c^n
> if n is not 2. Fine. Great. This is NOT what Fermat's last theorem
> states. Fermat's last theorem makes no mention of triangles, right
> or otherwise.
But by using triangles you can directly see what the equation is doing for
different values of n and see when the equation will not *work*. His theorem
did not *mention it* but it is obvious that it refers to to right triangles.
So why not use triangles and their associated roots to show the validity/non-
validity of the equation instead of trying to prove the equation will not
work with other values of n. It does the same thing but easier.
> Fermat's last theorem states that a^n + b^n = c^n has no
> positive integer solutions when n > 2. You proving that this
> is totally different from some other equation is irrelevant.
> It is totally different from just about any random equation
> that I happen to mention.
But we can't use random equations. I am using the same equation with all
values of n and showing it is only valid when n=2. And I am only using the
same roots throughout the range to keep things consistant.
> What you have done is shown that the two equations are totally different.
Besides showing this, I have shown which form of the equation *works* and how
to see this.
> What you have not done is shown that Fermat's equation has no solutions.
But I show that the equation only *works* when n=2 and how to actually see
this or atleast the results of an increasing n when constant roots are used.
Why show there are no solutions when n>2 when the equation is then invalid?
> The fact that two equations are "totally different" is irrelevant.
If this is irrelevant, why then are people trying to show that the equation
will not work when n=3 but will work when n=2. If this were irrelevant the
equation would work when n=3, but it doesn't.
> They might both have solutions. Different solutions.
Not true. N=2 has integer solutions. When n=3, to have a solution you must
get the cube root of the sum of the other two cubes and it will not be an
integer. And the new roots change the inherent shape of a triangle formed by
these roots when n was 2.
Frank
> -Keith!
>
> ---------------------------------------------------------------
> Keith Ellul 4th Year Pure Math / Computer Science
> kbe...@golden.net University of Waterloo
> ---------------------------------------------------------------
>
>
-----------== Posted via Deja News, The Discussion Network ==----------
fjm...@thegrid.net
bens...@iae.nl wrote:
> In article <78tdu0$d72$1...@nnrp1.dejanews.com>,
> fjm...@thegrid.net wrote:
> > In article <36B05F...@iae.nl>,
> > Nico Benschop <bens...@iae.nl> wrote:
> >
> > > I must admit I did not in detail follow your geometric reasoning, since
> > > your whole line of 'proof' concentrates only on showing {x,y,z} triples
> > > for n=2 and n<>2 are distinct [no problem there, easier full-proof by
> > > algebraic function f(x,y,z;n) as I showed you] -- and then "stopping*!
> >
> > But I don't need to go any further for there is nothing to solve.
> > I think these thoughts suffice to show that no nth power (except 2)
> > is the sum of two nth powers (except 2).
>
> Well, Frank, *you* said it: your aim (despite your denials;-) *is*
> proving FLT: the sum of two n-th powers is not an n-th power for n>2.
Then does this mean that I have since I can show that only one form of the
equations' roots produces right triangles as it should and the other forms
don't? You are giving me hope, Nico. I may have indirectly *solved* it like
Wiles did, and in less than 2 months!
> vvv: yes he did (in a very indirect way)
Yes, extremely indirect. Mine, if it is a proof, is extremely simple. But,
yet, so far, difficult to envision, just like his.
> > Even Wiles did not show this and that was all Fermat said.
> > I wish you would follow all the details of my reasoning and look
> > past trying to solve FLT as I have done. I don't think anyone has
> > ever put in figures, turned the crank, and enjoyed seeing what came out like
> > we all have done during our learning years. I have.
>
> Figures are nice to get idea's, but *ot* to prove anything beyond doubt.
> For that your better use algebra & arithmetic, so called 'formal' proof.
> Your trails show how imcomplete figures can be (in fact loosing sight
> of the main line;-(
I can, using figures, do this but it can't be shown on a computer because
shapes have to be drawn to show the effects of different values of n. And I
have done it with different sets of roots and have seen the verifications via
the Law of Cosines and the Sin^ + Cos^=1 identity. I can visually show the
results of an increasing n value and visually show the resultant invalidity of
n<>2 equations. I can do this using arithmetic; no problem. I can show
everything I have said to be true. Perhaps I should make some kind of formal
proof for I never have completely.
>
> > > Clearly only Pythagoras triples have to do with rectangular triangles
> > > (n=2). Generalized: orthogonality <--> inner vector product=0. No news..
> > >
> > > But you overlook the fact that this proves *nothing* about the very
> > > existence of such integers {x,y,z} for n>2 (to be disproved for FLT).
> >
> > There we go again, implying that FLT must be proven. -- Frank Manus
>
> Indeed: as you yourself wrote two paragraphs ago! (see above).
> Is your memory *that* short?
Yah! My short term memory has gotten short; part of aging I guess. But then I
have had these thoughts for a couple years now and my long-term memory is
good.
> I better call it quits now;-)
Hope you doesn't mean you are giving up.
Frank
>
> Ciao, Nico Benschop
> bens...@iae.nl -- http://www.iae.nl/users/benschop/ferm.htm
>
fjm...@thegrid.net
Planar <Damien....@inria.fr> wrote:
> >From: fjm...@thegrid.net
>
> >No pause at all!! The three sets of integers you give above are supposed to
> >be x,y,z roots useable in the equation to compare apples with apples. They
> >are not valid roots, simply the sum of two integers. For example, in your
> >3^3, etc.,etc. You must start with valid roots for when n=2 and then when
>
> OK, I think I get it. You are trying (with success) to prove:
>
> There does not exist x, y, z in N such that for all n in N,
>
>
> There does not exist x, y, z in N such that there exists n in N
> such that n > 2 and x^n + y^n = z^n
>
> Do you understand the difference between the two
No I don't.
Frank
>
> --
> Planar
fjm...@thegrid.net
Planar <Damien....@inria.fr> wrote:
> >From: fjm...@my-dejanews.com
>
> >No, it falls in my category of invalid forms of the equation when n<>2.
> >Please tell me what the problem is.
>
The equation is invalid when z^n will not equal the sum of x^n+y^n and this
only happens when n<>2. When n=2, x^n+y^n will equal z^n. Fermat said that no
cube is the sum of two cubes and this shows he was right by the invalidity of
when n<>2.
> > Using my sample roots of 3,4,5 in the
> >above equation does 3^1 + 4^1 = 5^1? I don't think it does! Is the form when
> >n=1 not invalid? And this applies also when n>2 but not when n=2. Frank
>
> So you have proved that 3^n + 4^n = 5^n only has solutions when n=2.
> Unfortunately, FLT is a lot more general than that.
Not 3^n+4^n=5^n but x^n+y^n=z^n. I have said many times that 3,4,5 values for
x,y,z respectively are only examples. Any valid roots, integer or irrational,
will work. And, in case you are preparing to ask, valid roots are roots which
will form right triangles since this equation, contrary to some people, deals
with right triangles.
fjm...@thegrid.net
curmu...@aol.com wrote:
> In article <78t1v8$23g$1...@nnrp1.dejanews.com>,
> > them. You must use a constant set of roots (pythagorean) in all forms of the
> > x^n+y^n=z^n equation from n=1 to infinity. Then, with these constant roots,
>
> (I know I said I wouldn't respond any more... so, I lied...)
You never *really* said this altho you did imply it. Welcome back.
> See, the problem is that there is no restriction that when investigating the
> equation
>
> x^n+y^n=z^n
>
> that we must stick to the roots of the equation when n=2. In fact, the
I only stick to the same roots to show the differing effects of n on the shape
of the triangle that the equations' roots generate. These are not restrictions
but since I am only comparing results, I must maintain a constant base of
comparison. Any root sets will work as long as they are used always and only n
is changed, which is what is required--- different powers only.
> point of most posts to yours has been that when n=1, the values for {x,y,z}
> are totally different for n=1 than n=2. This should be a clue to you that
And by using a constant root and varying n you will see the changes it causes
to the trianglar shape. But no triangular shape occurs when n=2. The values
for (x,y,z) are not tatollt different since I am using a constant root. All
that is showing is that the sum of the powers of the two roots does not equal
the third power except when n=2. Isn't that what it is supposed to do? I am
just giving a visual proof of this.
> your reasoning does not lend any new or useful information to the resolution
> of FLT or the claiming of the Wolfskehl prize.
But no one has ever came up with a visual proof that shows the resolution.
> Further, the solutions to the equation above do not behave the same for
> different values of n (as you've so oft noted). That's no big deal. In fact,
But it is a big deal. You agree they don't behave the same but you can't agree
that only when n=2 will the equation function properly, namely that the sum of
the two powers equals a third power.
> that's to be expected. Regardless, whether or not the equation has integer
> solutions when n=1 and n=2 has absolutely no bearing on whether the equation
There are no integer solutions when n=1, only when n=2.
> has solutions when n>2. Whether or not right triangles are formed also has no
> bearing. The points that your missing is this...
True is has no bearing, it is only used to prove (no, show) what is happening.
> You are assuming that because when n<>2, right triangles are not formed that
> there is some significance. There is not, at least as far as FLT is concerned
Probably not where FLT is concerned. But when used to show that the equations
are different from when n=2, the only way to do so is to show the effect on
the right triangle the roots cause when n changes. That is the only
significance to the right triangles.
> (or the Wolfskehl prize, for that matter). Your argument that FLT is
> Pythagorean in nature (i.e., that all roots must behave the n=2 case) is
> flawed. There is no such restriction.
I am not saying that all roots must behave as when n=2 and I have put no
restrictions on them. I am saying that roots in the n=2 case change when n
changes, thereby causing a recognizable change in the original triangle. I am
not saying all roots must behave like the n=2 case. They start there but they
don't remain there.
Frank
>
> Joe
Keith Ellul
> > "When n is an integer, n>2, the equation x^n+y^n=z^n has no solution in
> > positive integers x, y, z."
> >
> > Frank: Do you accept this is a valid formulation of FLT?
>
> Yes I do but I prefer the one about no cube being the sum of two other cubes.
[...]
> are the same. The Pythagorean theorem only works with squares. And that's what
> FLT is (or should be); showing that the Pythagorean theorem only works with
> squares.
Ahhh.. here's your mistake. See, the one that you "prefer" is not
Fermat's Last Theorem. See, FLT states exactly what Clive wrote above.
Your statement does not.. it is NOT EQUIVALENT to FLT. So, if you
want to talk about the fact that "the Pythagorean Theorem only works
with squares", then you can talk about that all you want (it's true and
trivial to show.. you can do it algebraically in about 3 lines) but don't
call it Fermat's Last Theorem, because you will confuse people... FLT
refers to the statement which Clive made above.
Keith Ellul
> > YES! It is totally different! No one is denying that!
> >
> > It is impossible to form a right triangle with sides a^n, b^n, and c^n
> > if n is not 2. Fine. Great. This is NOT what Fermat's last theorem
> > states. Fermat's last theorem makes no mention of triangles, right
> > or otherwise.
>
> But by using triangles you can directly see what the equation is doing for
> different values of n and see when the equation will not *work*. His theorem
> did not *mention it* but it is obvious that it refers to to right triangles.
> So why not use triangles and their associated roots to show the validity/non-
> validity of the equation instead of trying to prove the equation will not
> work with other values of n. It does the same thing but easier.
No. It ISN'T the same thing. See... this is the difference. Try to
pay attention here. Read this very, very carefully before responding.
You are saying:
If x, y, and z, are the lengths of the sides of a right triangle, then
x^n + y^n = z^n is false if x is not 2.
And what FLT is saying:
For arbitrary integers x, y, and z, x^n + y^n = z^n is false if n is not 2.
It is really important that you see the difference between these two
statements. There is a HUGE difference. In particular, your statement
doesn't say anything about x^n + y^n = z^n if x, y, and z aren't the
sides of a right triangle.
> > They might both have solutions. Different solutions.
>
> Not true. N=2 has integer solutions. When n=3, to have a solution you must
> get the cube root of the sum of the other two cubes and it will not be an
> integer. And the new roots change the inherent shape of a triangle formed by
> these roots when n was 2.
But... what if for some particular x, y, and z, we have
(1) x^2 + y^2 IS NOT EQUAL TO z^2
but, (2) x^2001 + y^2001 = z^2001.
You can't make statemnts about any triangle... because of (1), we know that
there is no right triangle with sides x, y, and z. But, to prove FLT, you
still have to show that (2) is correct. To try to make this even simpler...
(A) x^2 + y^2 = z^2
(B) x^n + y^n = z^n for n > 3.
You have shown that if (A) is true, then (B) is false.
FLT states that (B) is always false.
That is, (B) is false for ANY x, y, and z, regardless of whether or not
they are roots of (A)
For example, consider these two statements:
(I) x < 5
(II) x > 10
Now, I can show that if (I) is true, then (II) is false. But this does
not mean that (II) is always false.... we only know that (II) is false if
(I) is true.
Similarly, above, you showed that if (A) is true, then (B) is false. But
this is not the same as saying that (B) is always false... and this is
precisely what FLT states: (B) is always false.
It is not enough to show that A implies (not B) You must show (not
B)... REGARDLESS OF A. In particular, you cannot assume that A is true.
I really don't see what else I can say. People keep trying to explain
your logical error (it really is quite glaring) and you just don't seem
to get it. I can't explain it any better. If you don't understand this,
I can only suggest that you go read some basic math books.
Planar
>> There does not exist x, y, z in N such that for all n in N,
>> x^n + y^n = z^n
>>
>> But this is not FLT. FLT is the following.
>>
>> There does not exist x, y, z in N such that there exists n in N
>> such that n > 2 and x^n + y^n = z^n
>>
>> Do you understand the difference between the two
>From: fjm...@thegrid.net
>No I don't.
Then you really need to learn some math.
>The equation is invalid when z^n will not equal the sum of x^n+y^n and this
>only happens when n<>2.
You are saying that "the equation is invalid" means "the equation has
no solution".
> When n=2, x^n+y^n will equal z^n.
No, this is wrong. For example, 2^2 + 3^2 is not equal to z^2 for any
integer z.
> Fermat said that no
>cube is the sum of two cubes and this shows he was right by the invalidity of
>when n<>2.
The "invalidity" of [the equation] when n<>2 is FALSE. The
"invalidity" of [the equation] when n>2 is what you need to prove.
>Not 3^n+4^n=5^n but x^n+y^n=z^n. I have said many times that 3,4,5 values for
>x,y,z respectively are only examples. Any valid roots, integer or irrational,
>will work. And, in case you are preparing to ask, valid roots are roots which
>will form right triangles since this equation, contrary to some people, deals
>with right triangles.
You are saying that "valid" roots are roots which form right
triangles. You are proving that there are no "valid" solutions of
x^n+y^n=z^n when n<>2. This is not very interesting.
--
Planar
Nico Benschop
>
>In article <78uf7e$7vs$1...@nnrp1.dejanews.com>,
> bens...@iae.nl wrote:
>> In article <78tdu0$d72$1...@nnrp1.dejanews.com>,
>> fjm...@thegrid.net wrote:
>>> In article <36B05F...@iae.nl>,
>>> Nico Benschop <bens...@iae.nl> wrote:
>>>
>>>> your whole 'proof' concentrates only on showing {x,y,z} triples
>>>> for n=2 and n<>2 are distinct [no problem there, easier proof by
>>>> algebraic function f(x,y,z;n) as I showed you] and then "stopping*!
>>>
>>> But I don't need to go any further for there is nothing to solve.
>>> I think these thoughts suffice to show that no nth power (except 2)
>>> is the sum of two nth powers (except 2).
>>
>> Well, Frank, *you* said it: your aim (despite your denials;-) *is*
>> proving FLT: the sum of two n-th powers is not an n-th power for n>2.
>
> of the equations' roots produces right triangles as it should and the
> other forms don't? You are giving me hope, Nico. I may have indirectly
> *solved* it like Wiles did, and in less than 2 months!
>
>
> Yes, extremely indirect. Mine, if it is a proof, is extremely simple.
> But, yet, so far, difficult to envision, just like his.
>
>>> I wish you would follow all the details of my reasoning and look
>>> past trying to solve FLT as I have done. I don't think anyone has
>>> ever put in figures, turned the crank, and enjoyed seeing what came
>>> out like we all have done during our learning years. I have.
>>
>> doubt. For that your better use algebra & arithmetic, so called
>
> I can, using figures, do this but it can't be shown on a computer
> because shapes have to be drawn to show the effects of different
> values of n. And I have done it with different sets of roots and have
> identity. I can visually show the results of an increasing n value and
> visually show the resultant invalidity of n<>2 equations. I can do
> this using arithmetic; no problem.
> I can show everything I have said to be true.
> for I never have completely.
>
>>
>>>>
>>>> But you overlook the fact that this proves *nothing* about the very
>>>> existence of such integers {x,y,z} for n>2 (to be disproved for
>>>> FLT).
>>>
>>> There we go again, implying that FLT must be proven. -- Frank Manus
>>
>> Indeed: as you yourself wrote two paragraphs ago! (see above).
>
> Yah! My short term memory has gotten short; part of aging I guess.
> But then I have had these thoughts for a couple years now and my
> long-term memory is good.
>
>
> Hope you doesn't mean you are giving up. -- Frank
don't
Yes Frank, I'm afraid so ;-( The amount of advice you got, from me
and many others, is so overwhelming, that I fear it is impossible to
progress further. Unless you forget about your years-old (& incomplete)
ideas. As it were: start from scratch! I can definitely advise you
to look (with open mind;-) at Keith Ellul's recent post: a simpler
explanation than *that* is hardly possible... And then try make,
as you said, a "formal proof" ... not depending (only) on figures...,
but on complete logic sequences of statements, leaving "no hole";-)
{the hole in your 'proof' is so huge, as Keith & others showed, that
I really do hope you'll see it, eventually -- otherwise: lost case}
As intro, check out my .../ferm.htm (semi-historical;-) text:
>> Ciao, Nico Benschop
>> bens...@iae.nl -- http://www.iae.nl/users/benschop/ferm.htm
>>
PS:
Indeed - Wiles' FLT proof is very indirect (going "via the Northpole").
In fact proving something much more interesting [about the modularity
of cubic equations in two variables] linking two dissimilar operations
(+) and (^), as in x^p+y^p=z^p, via modular forms f(x) = (ax+b)/(cx+d).
For intro, see http://www.iae.nl/users/benschop/sgrp-flt.htm
and http://www.iae.nl/users/benschop/selmer.htm -- NB.
Charles H. Giffen
>
[snip]
> that series of numbers but I can't prove it yet. The proof will parallel my
> thoughts on FLT since these bricks are related to triangles; sorta cousins.I
> have not answered your post to me yet because I must do more thinking to give
> it the answer it deserves.
>
> But back to Gillians brick: His space diagonal of 6 is totally wrong (it
> should be 7.07, but the cube of 7.07 will not equal the sum of his other
> three cubes where as 6^3 will. I can't visualize the proof yet but it will be
> similiar to my FLT thoughts because, if you'll notice, his space diagonal has
> decreased from 7.07 (which it should be) to 6 (which it is in his
> example--which is right) just as one side of my FLT triangles decrease an *n*
> increases. Draw the brick and you will see that his inner space triangle has
> sides of 5 and 5. The hypotenuse (space diagonal) of these sides is 7.07, not
> 6 as his is. Something is wrong!! I'll start working on it, along with my
> other stuff, of course.
>
> One last thing in closing to make you feel better. Consider that you have a
> brick 3 x 4 x 12 and his is 3 x 4 x 5. Is he not working with half a brick??
> ROWL (rolling over with laughter).
>
> Frank
>
> http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Are you ever confused! *Cubes* of numbers have nothing to do with
diagonals of rectangular boxes. It is a matter of "Euclidean luck"
that the square of the diagonal c of a rectangle with sides a, b
is a^2 + b^2 = c^2 and that this just happens to coincide with
a "Fermat equation" a^n + b^n = c^n (when n = 2). There is no
such "length of diagonal" meaning to the Fermat equation when
n > 2.
And, finally, although I have a daughter whose name is Anne, she
is not Anne Gillian -- but rather, Anne Giffen. By the way, my
name is not Gillian, either.
--Chuck Giffen
Charles H. Giffen
>
[snip]
> > My question to you, Frank, is this:
> >
> > Do you think that the truth or falsity of CLT has anything to do with
> > the fact that the formula for the length of the space diagonal of a
> > cuboid is x^2+y^2+z^2=d^2 ?
>
> No, the truth or falsity of CLT has no effect on this formula. But (and I am
> sorry) the formula will only *work* with powers of 2. Even Gillians' 3^3 + 4^3
> + 5^3 = 6^3 example didn't work (altho it blew a hole in CLT). The equation
> becomes invalid when n>2. The space diagonal will begin shrinking and approach
> the length of the larger side as n increases. Bricks or triangles, the results
> FLT is (or should be); showing that the Pythagorean theorem only works with
> squares.
>
>
> > [ I hope nobody responds with an opinion of the truth or falsity of CLT.
> > ]
> >
> > --
> >
> > Clive Tooth
> > http://www.pisquaredoversix.force9.co.uk/
> > End of document
> >
>
> http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
You may be trying to "visualize" FLT -- but you are in no way
giving anything close to an indication as to *why* FLT is true
-- psychobabble juwt doesn't cut it.
And, I'm sorry that I spilled the beans and revealed the falsity of
CLT and thereby deprived you of the fun (anguish?) of trying to see
whether it was true or false (for Clive indeed was aware of the
counterexample I gave).
If my daughter *were* Ms. Gillian -- she would probably tell me that,
with all this eerie nonsense, I just might be dealing with an X-file.
As it is, I'm wondering if this thread ought to be directed to ones
Killfile, instead.
--Chuck GiFFen
Victor Eijkhout
> See, the problem is that there is no restriction that when investigating the
> equation
>
> x^n+y^n=z^n
>
> that we must stick to the roots of the equation when n=2.
You and others have said this before, and he didn't get it before.
I'm a bit amazed at how much time, beyond what's reasonable, people
are investing here in trying to educate/convince some obvious
blockheads.
--
Victor Eijkhout
"[A] man was excluded from the jury after asking the judge to tell him the
date of birth of the defendant so that he could draw up an astrological
chart to help him to decide on guilt or innocence." [Times 1998/12/29]
Ed Hook
|>
|>
|> > |> In article <78o065$1d5$1...@sun500.nas.nasa.gov>,
|> > |> ho...@nas.nasa.gov wrote:
|>
|> > Except that you seem to fundamentally confused about just exactly
|> > what Fermat's Last Theorem says *and* what sorts of things constitute
|> > a mathematical proof.
|>
|> I am not confused about these items. I am just going about it in a way no one
|> seems to have done in the past 360 years. And to come up with a simple way to
|> *prove* the statements' validity.
Actually, to the extent that you claim to be talking about Fermat's
Last Theorem, you _are_ confused. I notice that a number of others
have posted attempts to explain to you your various misapprehensions.
In particular, I thought that Keith did a very good job of this --
perhaps you'll want to concentrate on understanding his explanation ??
In any case, I sense that you and I are _not_ communicating, so I
suspect that this will be _my_ last attempt at setting you straight ...
|>
|> > |> ^^^^^^^^^^^^^^^^^^^^^^^^ Not only this but that it is an invalid
|> equation.
|>
|> > What do you mean by the statement that "it is an invalid equation" ??
|>
|> The equation is invalid when z^n will not equal the sum of x^n and y^n and
|> this only happens when n<>2. When n=2 x^n+y^n will equal z^n. Fermat said
|> that no cube is the sum of two cubes and this shows he was right by the
|> invalidity of when n<>2
You'd probably do well to stop referring to "invalid equation", if
what you mean is that the equation has no solutions in positive
integers, since *all* of the equations that you're discussing _do_
have solutions in (say) real numbers. For this reason, your choice of
terminology confuses your listeners needlessly. And (for a slightly
different reason) you should try to understand _why_ the fact that
your argument [suitably interpreted] works for all values of n different
from 2 means that it's simply wrong. One last time: the "Fermat" equation
in the case n = 1 is x + y = z and this equation has *lots*and*lots*
of solutions. Hence, an argument that claims to show that this equation
has no solutions at all is _totally_bogus_.
And, just to get it on the record, Fermat's claim was that the equation
x^n + y^n = z^n has no solutions in positive integers x, y, z when n > 2.
His marginal note (the one mentioning the marvelous proof that was, alas,
just a bit too long to fit in that margin) nowhere mentions right
triangles. That's because right triangles have exactly *nothing* to do
with the result. Also, in posts that came later in time than the one that
I'm responding to, it has become clear that you believe that Fermat's
Last Theorem amounts to the statement that there is no triple of positive
integers x, y, z that satisfy *all* of the equations x^n + y^n = z^n for
*all* positive integers n _simultaneously_. That's true (it's more-or-less
the result that I called "Frank's Theorem", which you snipped down below),
but (again) it's got precious little to do with Fermat's actual claim,
which concerns solutions of the Fermat equation for a *fixed*but*arbitrary*
n > 2.
|>
|> > There's nothing wrong with it _qua_ equation -- it just so happens that
|> > it has no solutions in positive integers x,y,z for n > 2. But, if you
|> > look for solutions in _real_ numbers (for example), you'll find that this
|> > equation has _lots_ of solutions for whatever value of n ...
|>
|> No, it doesn't have many solutions, unless you assign values to x and y and
|> solve for z. Then if you use these root values and try to make right triangles
|> with them, you won't be able to.
|>
See, it's clear even to you that the equation _has_ solutions -- there's
no requirement that these solutions correspond to right triangles and, in
fact, they don't. If you try to build any sort of triangle using a solution
to x + y = z, you end up with a degenerate triangle (indistinguishable
from a line segment).
|> > No, by "trivial" I mean that's it's a result that's very easy to prove
|> > and one which doesn't shed much light on anything. Let me try to write
|> > out an actual proof of what I understand to be your major claim (or, in
|> > fact, a generalization of it):
|>
|> I did not understand any of your *actual proof* so I snipped it. Sorry.
|>
That's OK. Hopefully, though, you at least understood the statement --
no two "Fermat equations" with differing exponents can share a nontrivial
solution. That really is the _positive_ content that can be extracted
from an attempt to understand your various claims. As noted previously,
this is a true statement, but not one which has anything much to do with
Fermat's Last Theorem.
|> > So, what you say is _true_ ... it just doesn't have _anything_ to do
|> > with Fermat's Last Theorem, which is:
|> >
|> > Fermat's Last Theorem [Wiles _et_al._]: The equation x^n + y^n = z^n
|> > has no solutions with x, y, z positive integers, when n > 2.
|> >
|> > The fact that an alleged solution to Fermat's equation could not _also_
|> > be a solution of the Pythagorean version does NOT demonstrate the truth
|> > of Fermat's Last Theorem ... because there's absolutely no reason to
|> > suppose that the two equations have to have the same solutions. You
|> > appear to believe that, but no one else does. And that's what everyone
|> > has been trying to tell you.
|>
|> I do not believe that and am not saying that. I am saying that when n=2 there
|> are solutions, when n<>2 there are no solutions, and hence the invalidity of
|> the equation.
Again, later posts by you contradict your claim that you don't believe
what I said you believe. In addition, after correcting for your "n = 1
_faux_pas_", "when n > 2 there are no solutions" is Fermat's Last Theorem;
it's _really_ not sufficient to wave your hands and airily assert the
result, in a thread where you claim to be proving Fermat's assertion.
(I *know* that you sometimes say that you're _not_ proving anything, but
then again you claim [equally often] that you _are_ -- the above is in
answer to the various posts of the latter sort ...)
|> >
|> > |> > Since you claim that n = 1 is ruled out for exactly the same reason
|> > |> > as n > 2, I'd think that the following (selected pretty much at random)
|> > |> > would give you pause:
|> > |> ^^^^^^^^^^^^
|> > |> > 1 + 2 = 3
|> > |> > 42 + 69 = 111
|> > |> > 2 + 7 = 9
|> > |> >
|> > |> > And so forth ... Each of those lines exhibits a solution in positive
|> > |> > integers x, y, z of the equation x + y = z. If you agree with that
|> > |> > assertion _and_ you agree that your argument holds for n = 1 to just
|> > |> > exactly the same extent that it holds for n > 2, then you're pretty
|> > |> > much forced to acknowledge that your position is (not to be
|> uncharitable)
|> > |> > flawed ...
|> > |>
|> > |> No pause at all!! The three sets of integers you give above are supposed to
|> > |> be x,y,z roots useable in the equation to compare apples with apples. They
|> > |> are not valid roots, simply the sum of two integers. For example, in your
|> >
|> > But, you see, that was the _whole_point_ -- your argument applies to the
|> > case n = 1 just as it does to n > 2. So you are claiming that the equation
|>
|> Not with respect to the original equation when n=2 and again, x+y=z is the
|> sum of two numbers, not the sum of the roots of two powers. It takes this
|> form only because of n equalling 1.
|>
Just to belabor the point, you're the one who keeps saying "n<>2", which
means that you're allowing n = 1. And, in that case, x^n + y^n = z^n is
precisely the equation x + y = z. That equation has lots of solutions,
which you seem to regard as "nonexistent". Again, I urge you to think
about the implications for your argument of this pesky fact ...
|> > x + y = z can have _no_ solutions in positive integers x, y, z. Hence, the
|> > randomly-chosen counterexamples to your claim that I displayed up above
|> > _should_ put an end to the discussion. Your argument "proves" something
|> > that is patently false - _that's_ why the above should give you pause ...
|>
|> It did not and for the same reasons. Please re-read the paragraph. There is no
|> 1,2,3 triangle, nor a 42,69,111 triangle, nor 2,7,9. I do no use randomly
|> chosen numbers and the roots I use are valid and constant throughout the range
|> of all n's.
See, you just admitted that you are only considering solutions that solve
the Fermat equation for _all_ n _simultaneously_. We (that is, you and I ...
and probably everyone else in sci.math [excepting, possibly, the resident
batch of idiots the likes of <whoever> Deeth and Edward Meisner]) are in
violent agreement that there are _*NO*_ such triples of positive integers --
if that's all that you want to claim, then there's no point in continuing,
since you're right. But, then, _please_ stop saying that you're talking
about Fermat's Last Theorem -- 'cause you're not. Each instance of _his_
claim concerns one particular version of the equation, for a fixed value
of n > 2. And _his_ is a LOT harder to prove (which is, no doubt, the
reason that it took mathematicians more than 350 years of sustained effort
to verify the truth of the assertion -- and these were persons who would
have had no difficulty whatsoever in coming up with any _valid_ argument
that proved the result and was as simple as yours).
|> >
|> > |> first line, 1+2 does = 3 but 1^2 + 2^2 does not = 3^2, nor does 1^3 + 2^3
|> =
|> > |> 3^3, etc.,etc. You must start with valid roots for when n=2 and then when
|> > |> using them when n<>2 these valid roots cause the equation to become
|> invalid.
|> > |> What you have done is solve for the 3rd root by adding the other two
|> instead
|> > |> of getting the square root of the sum of the others' squares. Using any of
|> > |> your examples when n=2 and solving for the 3rd root will give valid roots
|> > |> (one that will form right triangles). But not when n<>2. Then these roots
|> > |> will only form scalene triangles with no 90 degree angle. Does this not
|> show
|> > |> that the n=2 form of the equation is totally different from the n<>2 form?
|> I
|> >
|> > Yes, it (sort of) shows that the nontrivial solutions, if there are any,
|> > of the Fermat equation have to be different for different exponents. I say
|> > "sort of", because nothing you've said really amounts to a proof, as that
|> > term is usually understood hereabouts.
|> >
|> > Again, though, this observation is a far cry from a proof of Fermat's
|> > Last Theorem -- if you wanted to turn it into one of those, you'd have
|>
|> But I don't want to turn my *proof* into one of those. That's what I am
|> rebelling against.
|>
So what _do_ you want to turn your proof into ?? Elsewhere you've
claimed that what you've got here _is_ a proof of Fermat's Last
Theorem. What I'm saying is that it's not (it's mostly confused and
erroneous, when examined in that context) and that you could only
turn it into a proof by taking my proffered suggestion ...
|> Frank
|>
|> > to prove (in addition) that any solution of the Fermat equation with
|> > n > 2 is _necessarily_ a solution of the Pythagorean equation as well.
|> > I don't think that you're likely to be able to do that, though :-(
--
fjm...@thegrid.net
Nico Benschop <bens...@iae.nl> wrote:
> Yes Frank, I'm afraid so ;-( The amount of advice you got, from me
> and many others, is so overwhelming, that I fear it is impossible to
> progress further. Unless you forget about your years-old (& incomplete)
> ideas. As it were: start from scratch! I can definitely advise you
> to look (with open mind;-) at Keith Ellul's recent post: a simpler
> explanation than *that* is hardly possible... And then try make,
> as you said, a "formal proof" ... not depending (only) on figures...,
> but on complete logic sequences of statements, leaving "no hole";-)
> {the hole in your 'proof' is so huge, as Keith & others showed, that
> I really do hope you'll see it, eventually -- otherwise: lost case}
>
> As intro, check out my .../ferm.htm (semi-historical;-) text:
>
> >> Ciao, Nico Benschop
> >> bens...@iae.nl -- http://www.iae.nl/users/benschop/ferm.htm
Thank you, Nico,for the above advice. I am studying Keith Ellul's last post
now very carefully, and Planar's too. My mind is and has always been very
open to remarks about my FLT thoughts and I intently study all replies to
learn from them. I do hate going round and round and sounding like a broken
record about my ideas on this and I hope I am not upsetting anyone.
Apparently I am not proving FLT in the traditional, excepted way but am just
trying to show that what Fermat said was true. I believe this is what people
must be trying to tell me and I am starting to realize that. And I now can
see that this is two different items apparently. That must be the "hole" you
refer to above and it is like a black hole; information is going in but not
coming out!
Later, in silence, I shall work on that "formal proof" which I have always
wanted to do. Thanks again.
Frank
fjm...@thegrid.net
Victor Eijkhout <eijk...@nala.cs.utk.edu> wrote:
> curmu...@aol.com writes:
>
> > See, the problem is that there is no restriction that when investigating the
> > equation
> >
> > x^n+y^n=z^n
> >
> > that we must stick to the roots of the equation when n=2.
>
> You and others have said this before, and he didn't get it before.
>
> I'm a bit amazed at how much time, beyond what's reasonable, people
> are investing here in trying to educate/convince some obvious
> blockheads.
Apparently you can't realize that they are patient, sensitive people whose
love of mathematics cause them to give of themselves and their knowledge in
the hopes of educating, without trying to convince, others (who you call
blockheads) who are sincerely interested in learning from them.
> --
> Victor Eijkhout
> "[A] man was excluded from the jury after asking the judge to tell him the
> date of birth of the defendant so that he could draw up an astrological
> chart to help him to decide on guilt or innocence." [Times 1998/12/29]
You probably consider the above person a blockhead, don't you. Others would
think he was pretty smart in what he done. In some states you can get jail
time for not serving on a jury.
Best regards,
Frank Manus
fjm...@thegrid.net
> > > positive integers x, y, z."
> > >
> > > Frank: Do you accept this is a valid formulation of FLT?
> >
> > Yes I do but I prefer the one about no cube being the sum of two other
cubes.
>
> [...]
>
what
> > FLT is (or should be); showing that the Pythagorean theorem only works with
> > squares.
>
> Fermat's Last Theorem. See, FLT states exactly what Clive wrote above.
> Your statement does not.. it is NOT EQUIVALENT to FLT. So, if you
> want to talk about the fact that "the Pythagorean Theorem only works
> with squares", then you can talk about that all you want (it's true and
> trivial to show.. you can do it algebraically in about 3 lines) but don't
> call it Fermat's Last Theorem, because you will confuse people... FLT
> refers to the statement which Clive made above.
>
>
> ---------------------------------------------------------------
> Keith Ellul 4th Year Pure Math / Computer Science
> kbe...@golden.net University of Waterloo
> ---------------------------------------------------------------
>
statement Fermat is said to have written: "Cubum autem in duos cubos, aut
quadratoquadratum in duos quadratoquadratos". Now it doesn't take much
knowledge of latin to figure out what "Cubum" and "duos cubos" must mean and
the same for the bigger words that must mean 4th powers. The statement at the
top line above that you say every one uses for FLT has caused a constant
mindset among mathematicians to solve it. Then, apparently, I am doing
something different (wrong?) by wanting just to show that Fermat was right and
proving that rightfulness instead of trying to solve an unsolveable problem.
And did Mr. Wolfskehl ask to prove the problem or show that Fermat was right?
Frank
bensc...@my-dejanews.com
You're welcome. Just two more hints, FLT: x^p+y^p=z^p has no pos.integer
solution x,y,z for p>2, then: 1. It suffices to show this for prime exponents
only: (^) distributes over (.) 2. FLT has two cases. Case1: p not divides
x,y,z. Case2: p divides one of x,y,z.
Case1 is quite easy to prove, by expanding Farmat's Small Thm: n^p=n mod p,
for prime p, and all n. Namely consider arithmetic mod p^2 (2-digits base p),
then there are (p-1)p residues coprime to p (case1 !), and in general mod
p^k: there are p^k - p^{k-1} = (p-1)p^{k-1}. Any product of two nrs coprime
to p is also coprime to p (closure). They form a "group" G, and it is known
that this group is cyclic --> generated by the powers of just one nr, say G =
{g^i} for i=1..|G| (the order (p-1)p of the group / cycle). So mod p^2 the
p-th powers, that is: each p-th iteration of g, form a cycle of length p-1
---> Just like FST: n^p=n mod p means namely: a p-1 cycle, which Fermat
discovered around 1637, just when he made his marginal note (!). Only this
p-1 cycle mod p^2 is for 2-digit residues, and x^p=x, y^p=y, z^p=z mod p^2
for any x,y,z in this "core" cycle. Clearly, for any solution "in core":
(x+y)^p = x+y = x^p + y^p mod p^2....[2]
Notice that *every* solution of p-th power integers is "in core" mod p^2,
and that in [2] the Exponent p Distributes over a Sum (EDS property, as
trivially holds for FST mod p) --> Hence such solution in Core mod p^2
cannot be extended to integer p-th powers, of which the 2 lsd's necessarily
solve [2],because of inequality (X+Y)^p > X^p + Y^p for the corresponding
integer p-th powers < p^{2p}, where X,Y,Z are 2-digit integers solving [2]
mod p^2. ...QED (FLTcase1).
For details http://www.iae.nl/users/benschop/nfb0.dvi (full text)
http://www.iae.nl/users/benschop/scimat98.htm (intro)
http://www.iae.nl/users/benschop/sgrp-flt.htm (functions>arithm)
http://www.iae.nl/users/benschop/selmer.htm (Hensel lift escape)
http://www.iae.nl/users/benschop/campaign.htm (Hensel lift escape)
http://www.ams.org/preprints/11/199711/0index.html (abstract)
and let me know what you tink of _that_ "simple" solution (for case1).
I bet you: simpler cannot (via direct extension of Fermat's own Small
Theorem;-)
Ciao xxxxxxxxxxxxxxxxxxxxxxxx1.1xxxxxxxxxxxxxxxxxxxxxxxx Nico
FLT--------- AHA: One is Always Halfway Anyway -----------EDS
Victor Eijkhout
> > I'm a bit amazed at how much time, beyond what's reasonable, people
> > are investing here in trying to educate/convince some obvious
> > blockheads.
>
> Apparently you can't realize that they are patient, sensitive people whose
> love of mathematics cause them to give of themselves and their knowledge in
> the hopes of educating, without trying to convince, others (who you call
> blockheads) who are sincerely interested in learning from them.
"In the hopes of educating". My amazement is for the repeated explanations
given to those who have proved themselves beyond such hopes.
O&O.
Keith Ellul
> > Ahhh.. here's your mistake. See, the one that you "prefer" is not
> > Fermat's Last Theorem. See, FLT states exactly what Clive wrote above.
> > Your statement does not.. it is NOT EQUIVALENT to FLT. So, if you
> > want to talk about the fact that "the Pythagorean Theorem only works
> > with squares", then you can talk about that all you want (it's true and
> > trivial to show.. you can do it algebraically in about 3 lines) but don't
> > call it Fermat's Last Theorem, because you will confuse people... FLT
> > refers to the statement which Clive made above.
> >
> Keith, I do not know latin at all, but here are the first 10 words of the
> statement Fermat is said to have written: "Cubum autem in duos cubos, aut
> quadratoquadratum in duos quadratoquadratos". Now it doesn't take much
> knowledge of latin to figure out what "Cubum" and "duos cubos" must mean and
> the same for the bigger words that must mean 4th powers. The statement at the
> top line above that you say every one uses for FLT has caused a constant
> mindset among mathematicians to solve it. Then, apparently, I am doing
> something different (wrong?) by wanting just to show that Fermat was right and
> proving that rightfulness instead of trying to solve an unsolveable problem.
> And did Mr. Wolfskehl ask to prove the problem or show that Fermat was right?
Ok, I don't know a word of latin either, but that means "A cube is not
the sum of two cubes, a 4th power is not the sum of 4th powers...."
Then he goes on to say that no nth power is the sum of two nth powers.
(n > 2)
Or, in simple terms, z^n cannot be written as x^n + y^n, (n > 2)
Or, in even simpler terms, x^n + y^n = z^n has no solutions. (n > 2)
Ok, so, we want to show that this equation has no solutions. You have
shown that (3, 4, 5) is not a solution to the equation. You have
shown that ANY puthagorean triple is not a solution to the equation.
But what about (12, 20, 60)? You have not shown that (12, 20, 60) is
not a solution to x^n + y^n = z^n for some n. You have only shown it
for pythagorean triples... you need to show it for arbitrary triples.
BTW -- Who is Mr Wolfskehl? And proving that this equation has no
solutions is equivalent to showuing that Fermat was right.
David Kastrup
> fjm...@thegrid.net writes:
>
> > > I'm a bit amazed at how much time, beyond what's reasonable, people
> > > are investing here in trying to educate/convince some obvious
> > > blockheads.
> >
> > Apparently you can't realize that they are patient, sensitive people whose
> > love of mathematics cause them to give of themselves and their knowledge in
> > the hopes of educating, without trying to convince, others (who you call
> > blockheads) who are sincerely interested in learning from them.
>
> "In the hopes of educating". My amazement is for the repeated explanations
> given to those who have proved themselves beyond such hopes.
A knife will never be sharp enough to cut a whetstone, yet it gets
sharper by applying it to one. A whetstone is dull and constantly
going in circles.
I have found that my violin skills improve quite a bit when I practice
on pieces that are definitely way beyond what I could hope to master
satisfactorily.
Explaining math to someone who accepts it already will not make you
hone your pedagogic skills.
--
David Kastrup Phone: +49-234-700-5570
Email: d...@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany
Lee Rudolph
>A knife will never be sharp enough to cut a whetstone, yet it gets
>sharper by applying it to one. A whetstone is dull and constantly
>going in circles.
>
>I have found that my violin skills improve quite a bit when I practice
>on pieces that are definitely way beyond what I could hope to master
>satisfactorily.
>
>Explaining math to someone who accepts it already will not make you
>hone your pedagogic skills.
Well, it's said that a wise person can learn from anyone. On
the other hand, it's also said that when you've gotten the message,
you should hang up the phone.
Lee Rudolph
fjm...@thegrid.net
I can do this but my posting has come to an end for some time. Incidently,
(12,20,60) are impossible roots for they can never form a triangle unless a
curved hypotenuse is allowable.
>
> BTW -- Who is Mr Wolfskehl? And proving that this equation has no
> solutions is equivalent to showuing that Fermat was right.
>
> -Keith!
Mr Wolfskehl was the person who posted a large prize in Germany in 1908 to
anyone who could prove FLT, post the proof and have it be excepted by his
peers. But I think he wanted it done in the usual way that people try instead
of just to show that Fermat was right (as I have tried).
Frank
> ---------------------------------------------------------------
> Keith Ellul 4th Year Pure Math / Computer Science
> kbe...@golden.net University of Waterloo
> ---------------------------------------------------------------
>
>
-----------== Posted via Deja News, The Discussion Network ==----------
fjm...@thegrid.net
> >In article <78re00$8...@src-news.pa.dec.com>,
> > Planar <Damien....@inria.fr> wrote:
>
> >> There does not exist x, y, z in N such that for all n in N,
> >>
> >>
> >> There does not exist x, y, z in N such that there exists n in N
> >> such that n > 2 and x^n + y^n = z^n
> >>
> >> Do you understand the difference between the two
>
> >From: fjm...@thegrid.net
>
> >No I don't.
>
> Then you really need to learn some math.
Not really, triangles are simple and they are my friends. Its all those N's
that I don't understand.
>
> >The equation is invalid when z^n will not equal the sum of x^n+y^n and this
> >only happens when n<>2.
>
> You are saying that "the equation is invalid" means "the equation has
> no solution".
No, just that the equations zth root reduces in order to *have* a solution as
n increases. There always are solutions and they need not be integers. But
now I understand that this is where I have been wrong. They are required to
be integers to qualify for FLT proofs.
>
> > When n=2, x^n+y^n will equal z^n.
>
> No, this is wrong. For example, 2^2 + 3^2 is not equal to z^2 for any
> integer z.
No, that was right. I never said z^n is integer. There is no (2,3,?) triangle
(? being an integer root) but there is a (2,3,3.606) triangle. And these roots
are valid with which to start the series of equations required.
Frank
> Planar
fjm...@thegrid.net
"Charles H. Giffen" <ch...@virginia.edu> wrote:
> You may be trying to "visualize" FLT -- but you are in no way
> giving anything close to an indication as to *why* FLT is true
> -- psychobabble juwt doesn't cut it.
I just lack the ability to express my thoughts properly, that's all, but it
will come. But in this case, I now see that my thoughts don't follow the norm.
>
> And, I'm sorry that I spilled the beans and revealed the falsity of
> CLT and thereby deprived you of the fun (anguish?) of trying to see
> whether it was true or false (for Clive indeed was aware of the
> counterexample I gave).
But there would have been no anguish and you spilled no beans. I knew that
the truth or falsity of CLT has no effect on the diagonal formula as I said
before and that was all that Clive asked. But then you came up with that
unique set of cubes that blew his CLT up even though the value of its roots
can't be found in any brick. Invalid because of being a cube? Who knows!
Thanks Mr. Giffen.
Frank
>
> If my daughter *were* Ms. Gillian -- she would probably tell me that,
> with all this eerie nonsense, I just might be dealing with an X-file.
> As it is, I'm wondering if this thread ought to be directed to ones
> Killfile, instead.
>
> --Chuck GiFFen
>
-----------== Posted via Deja News, The Discussion Network ==----------
fjm...@thegrid.net
ho...@nas.nasa.gov wrote: A very lengthy reply:
Ed, that sure was a long reply and I am truly grateful for it and the time you
put into it on my behalf. I don't think you will mind my not replying to it
since I am on my way out thanks to your help and others (See my *No more FLT
posts* message). I promise that I shall study all you wrote carefully and to
learn from it. Thank you.
Frank
fjm...@thegrid.net
> diagonals of rectangular boxes. It is a matter of "Euclidean luck"
> that the square of the diagonal c of a rectangle with sides a, b
> is a^2 + b^2 = c^2 and that this just happens to coincide with
> a "Fermat equation" a^n + b^n = c^n (when n = 2). There is no
> such "length of diagonal" meaning to the Fermat equation when
> n > 2.
This is not "Euclidian luck". Euclid may have started it but Pythagorus made
it noticeable, Fermat made a problem of it, and I am trying to tie it all
together. And it is not luck about the diagonal either because it is derived
from similar triangles within the brick; simple Pythagorean triangles.
>
> And, finally, although I have a daughter whose name is Anne, she
> is not Anne Gillian -- but rather, Anne Giffen. By the way, my
> name is not Gillian, either.
>
> --Chuck Giffen
Incidently, I figured out how I spelled your name wrong, for which I
apologized to you via email. In my notes of my reply to Clive, I had it
mispelled and forgot to look it up before I sent the post out. Again I am
sorry.
Best regards,
Gus Gassmann
>
> Mr Wolfskehl was the person who posted a large prize in Germany in 1908 to
> anyone who could prove FLT, post the proof and have it be excepted by his
> peers. But I think he wanted it done in the usual way that people try instead
> of just to show that Fermat was right (as I have tried).
>
> Frank
Is that why you think you qualify???
gus gassmann