FLT Proof: Simple analogy
James Harris
Fermat's Last Theorem.
In my proof I have an expression
F(x,y,z,v) = 0 (mod G(x,y,z,v)).
Here, for instance, F(x,y,z,v) just means an expression with those numbers.
(For example:
x^2 + 2xy + y^2 = T(x,y))
x^p + y^p = z^p, is given, and
v is just some integer that I can choose with the restriction that it not
have factors in common with x+y.
What F(x,y,z,v) = 0 (mod G(x,y,z,v))
actually means is simply that
F(x,y,z,v) has G(x,y,z,v) as a factor.
That's it.
Ok, the issue is that posters in replying to me, say that factoring
F(x,y,z,v) is "wrong".
Now, think again what the expression with the modulus is actually saying.
It's saying that
F(x,y,z,v) = G(x,y,z,v)*W(x,y,z,v)
where I don't care what W(x,y,z,v) is, which is why I'm using a modulus in
the first place.
It turns out that by inspection, you can see that
F(x,y,z,v) has the two factors
G(x,y,z,v) and W(x,y,z,v).
Now, I also have that
F(x,y,z,v) = F1(x,y,z,v)*F2(x,y,z,v),
where F1 and F2, are what I'll call a natural factorization.
(For instance, x^2 + 3x + 2 has the natural factorization of (x+2)(x+1), but
with my expressions, these factors are a bit more complicated as they
contain radicals.)
So, it's obviously true that
F1(x,y,z,v)F2(x,y,z,v) = G(x,y,z,v) W(x,y,z,v)
All disagreement with my proof depends on the assertion that
G(x,y,z,v) has the additional factors, g1 and g2, and W(x,y,z,v) has w1 and
w2, there's something like
F1(x,y,z,v) = f1 w1, while
F2(x,y,z,v) = f2 w2,
or
F1(x,y,z,v) = f1, while
F2(x,y,z,v) = f2 W(x,y,z,v),
or
etc.
Now, remember that
F1(x,y,z,v)F2(x,y,z,v) = G(x,y,z,v) W(x,y,z,v)
is not disputed.
So, why not accept that G(x,y,z,v) can simply be a factor of
F1(x,y,z,v) or F2(x,y,z,v)?
Good question.
So why do a lot of people keep trying to say I have to split up G(x,y,z,v)?
If I gave you a polynomial and told you that x+1 was a factor of it, would
you start looking for factors of x+1 to divide that polynomial with?
I don't think so.
Then, why do I think these posters have gotten away with convincing so many
of you that I'm wrong?
Here's what I think.
They keep emphasizing math that many of you aren't familar with.
Most people don't do much math with equations that have "mod" in them;
although, people use the concept all the time.
Our time keeping is modular, which is why we keep up with the hour in 12 or
24 hour increments.
Computer programmers get to use "mod" a bit as well for any number of things
in programs.
But, in math it is primarily number theorists who play around with these
type of equations.
So, the number theorists or people playing at being number theorists put up
examples that don't relate to my proof, and keep claiming it's wrong.
I bet you figure that they should know. After all, you don't know, it
sounds good and they're supposed to be the experts.
So, even those of you who are expert mathematicians in your own field simply
defer to the number theorists, and a lot of them seem to have forgotten
basic algebra because I don't think they're deliberately leading you astray.
I suggest you do this.
Where ever you see that "mod" in my proofs, say to yourself, it just means
that what's to the right of it is a factor.
Go back to your algebra, and you'll see how simple and obvious it all really
is.
--
James Harris
David Libert
"James Harris" (js...@mindspring.com) writes:
> This post is intended to answer the one continuing objection to my proof of
> Fermat's Last Theorem.
>
> In my proof I have an expression
>
> F(x,y,z,v) = 0 (mod G(x,y,z,v)).
>
> Here, for instance, F(x,y,z,v) just means an expression with those numbers.
>
> (For example:
>
> x^2 + 2xy + y^2 = T(x,y))
>
> x^p + y^p = z^p, is given, and
>
> v is just some integer that I can choose with the restriction that it not
> have factors in common with x+y.
>
> What F(x,y,z,v) = 0 (mod G(x,y,z,v))
>
> actually means is simply that
>
> F(x,y,z,v) has G(x,y,z,v) as a factor.
>
> That's it.
>
> Ok, the issue is that posters in replying to me, say that factoring
>
> F(x,y,z,v) is "wrong".
That has not been what I am saying. You have a congruence as above,
where F(x,y,z,v) can be rewritten as a quadratic in z^2. You pass
to a suffiently big ring R to have the quadratic formula solutions of
this quadratic. You factor your original quadratic:
F(x,y,z,v) = (z^2 - one quadratic formula solution) *
(z^2 - other quadratic formula solution) .
No problem there. As you do below, let us denote these right side
factors as F1(x,y,z,v) and F2(x,y,z,v) respectively.
So you mod out the last equation, obtaining the congruence:
(1) F(x,y,z,v) = F1(x,y,z,v)*F2(x,y,z,v) (mod G(x,y,z,v)).
No problem so far, I am agreeing with the validity of everything so
far, and with the correctness of (1).
I will repeat your last two lines, quoted above:
> Ok, the issue is that posters in replying to me, say that factoring
>
> F(x,y,z,v) is "wrong".
The factoring would be (1), and I am not saying it is wrong.
It is your next step I question. You use "+-" notation to refer to
the two solutions from the quadratic formula, and you claim disjunct of
two congruences, (expressed using +-) :
(2) z^2 = one or the other of the quadratic formula solutions
(mod G(x,y,z,v)).
I will paste the corresponding lines concluding this from your original
proof under discussion:
> Factoring with the quadratic formula gives
>
> z^2 -(5v^3 +/- sqr(5v^6 - 20v))xy/2(v^5 + 1)
>
> = 0(mod (x+y+vz)/h).
>
> (Here 2(v^5 + 1) is the denominator which might not be clear to all as
> presented.)
The F1 and F2 factors above were each of the form z^2 - a quadratic
formula solution, so (2) amounts to a disjunct of the congruences
(3) Fi(x,y,z,v) = 0 (mod G(x,y,z,v)) i = 1,2
I repeat again congruence (1) from above, that I do agree with
(renaming it as (4) now to keep numbers in sequence):
(4) F(x,y,z,v) = F1(x,y,z,v)*F2(x,y,z,v) (mod G(x,y,z,v)).
Your starting assumption was that F(x,y,z,v) is congruent to 0 in
that modulus, so from (4) you know that product on the right side is
congruent to 0. The question is how do you get from that to the (3)
conclusion that one factor must be congruent to 0. My recent numerical
examples have shown cases where this corresponding step fails.
It is misrepresenting my question to say I am questioning the
factoring. I have agreed with the (1):(4) factoring. I am
questioning the step to (3) based on that factoring.
My own numeric examples are of this corresponding form. They also
support the factoring. They refute the later step corresponding to
(3). It is the (3) step you claimed in your proof as I quoted above,
ie z^2 - quadratic formula expressions = 0 (mod (x+y+vz)/h) as quoted
above.
So I will repeat a third time your last two lines:
> Ok, the issue is that posters in replying to me, say that factoring
>
> F(x,y,z,v) is "wrong".
That is not what I have been saying. The factoring is right. A later
step is wrong.
> Now, think again what the expression with the modulus is actually saying.
>
> It's saying that
>
> F(x,y,z,v) = G(x,y,z,v)*W(x,y,z,v)
>
> where I don't care what W(x,y,z,v) is, which is why I'm using a modulus in
> the first place.
Right. Agreed.
> It turns out that by inspection, you can see that
>
> F(x,y,z,v) has the two factors
>
> G(x,y,z,v) and W(x,y,z,v).
>
> Now, I also have that
>
> F(x,y,z,v) = F1(x,y,z,v)*F2(x,y,z,v),
>
> where F1 and F2, are what I'll call a natural factorization.
I agree with all that just above. The F1 and F2 are as I wrote
earlier related to the quadratic formula.
> (For instance, x^2 + 3x + 2 has the natural factorization of (x+2)(x+1), but
> with my expressions, these factors are a bit more complicated as they
> contain radicals.)
>
> So, it's obviously true that
>
> F1(x,y,z,v)F2(x,y,z,v) = G(x,y,z,v) W(x,y,z,v)
Yes.
> All disagreement with my proof depends on the assertion that
>
> G(x,y,z,v) has the additional factors, g1 and g2, and W(x,y,z,v) has w1 and
> w2, there's something like
>
> F1(x,y,z,v) = f1 w1, while
>
> F2(x,y,z,v) = f2 w2,
>
> or
>
> F1(x,y,z,v) = f1, while
>
> F2(x,y,z,v) = f2 W(x,y,z,v),
>
> or
>
> etc.
Right, my questions about this point in the proof are about cases like
you just wrote.
> Now, remember that
>
> F1(x,y,z,v)F2(x,y,z,v) = G(x,y,z,v) W(x,y,z,v)
>
> is not disputed.
Right.
> So, why not accept that G(x,y,z,v) can simply be a factor of
>
> F1(x,y,z,v) or F2(x,y,z,v)?
Okay, if we do accept that, my question goes away. My whole question
was why we should accept that.
I am not disputing that the above is possible, that it is possible
that G(x,y,z,v) is a factor of F1(x,y,z,v) or F2(x,y,z,v) . As far
as I can see that might be true or it might be false.
But a proof of FLT is supposed to handle all conceivable cases. So it
is not a complete analysis if it does not cover all possibilities. And
until we have a proof in place it remains a conceivable case that
G(x,y,z,v) factors neither of F1(x,y,z,v) or F2(x,y,z,v) , so it
should be covered in a proof.
> Good question.
>
> So why do a lot of people keep trying to say I have to split up G(x,y,z,v)?
Because it is one of the cases to handle. Do you want to have an
incomplete proof?
> If I gave you a polynomial and told you that x+1 was a factor of it, would
> you start looking for factors of x+1 to divide that polynomial with?
>
> I don't think so.
Right, your example with a polynomial has no nontrivial
factorization. So I will discuss how the context in your proof differs
from this polynomial case.
You begin your proof by contradiction assuming x,y,z is a pairwise
coprime Fermat counterexample. So these are assumed to be integers.
They are unknown integers, in fact by the end of the proof the intent is
to show they didn't even exist after all. But for the early part of the
proof the assumptions in force have them existing. They are fixed
integers, fixed by the opening existence assumption. They are fixed but
unknown. This is common in proofs by contradiction.
It is true that in the late stages of your proof you will be
discussing variant readings of the proof in which x,y,z are no longer
necessarily all integers. But for the opening section there is at least
one reading of the proof in force, inherited from the opening
assumption, that x,y,z are fixed but unknown integers. Ie, by unknown
I mean that the proof does not pin down x, y or z as equalling any
specific numerals spelled out in the proof.
You work with these variables x,y,z , making polynomial or rational
expressions in them. Eventually in the proof you discussing factoring.
The important point is the factoring notion in force is factoring in
integers, not factoring in polynomials. These can diverge. If your
proof had involved x=3 (I know it didn't, but follow this
hypothetically for a moment), you could have 2 as a factor in integers
of x+1. But considered as polynomials, 2 is not a factor of the
polynomial x+1.
Of course, in your real proof things are more complicated, because
there are no specific assumptions like x=3, ie x,y,z are fixed but
*unknown*. So I can't come up with a specific expression like 2 in the
last version, which I can definitively claim must be a factor of
(x+y+vz)/h .
But the job of a complete proof is more than just covering all the
cases that you can explicitly describe. To make sure all cases are
covered, you handle cases that are just assumed to exist. Could
(x+y+vz)/h have a nontrivial factorization? Maybe, who knows. So a
complete FLT must cover these possibilities, and should do so by showing
how to handle them from just supposing they exist.
If a proof only handles the specific cases that the mathematician can
think of during the proof writing, who is to say that after the proof is
completed someone may not think of a new case?
The issue is not can I think of an explicit factor for (x+y+vz)/h ,
or even can I prove such nontrivial factors exist. The issue is can you
prove that they don't exist. Or if you can't, can you show how your
final conclusion is still correct even if they do exist.
Why is the burden on you like this? Because you are the one who
choose to work on the hard problem. You are the one who is trying to do
a general proof. The burden of proof on you here is no more than all
the rest of us face when we try to do our own proofs of other theorems.
Anyway, your proof is about analysing the fixed but unknown x,y,z .
So it is conceivable that for a case of this to be handled (x+y+vz)/h
could have some factorings that don't correspond to any general
polynomial factorings. Just as in the easier case of x=3, x+1 had the
2 factor which is not a factor of the general polynomial x+1.
> Then, why do I think these posters have gotten away with convincing so many
> of you that I'm wrong?
Of course this question carries background assumptions. "Gotten
away". If you get to pick all the questions you can spin the
discussion. "By what sneaky strategems did that faker David Libert pull
the wool over your eyes? And worse: just what is he after anyway??
..." Etc.
> Here's what I think.
>
> They keep emphasizing math that many of you aren't familar with.
>
> Most people don't do much math with equations that have "mod" in them;
> although, people use the concept all the time.
Well, how did I get to talking about mod? Because it was in your
proof. Anyway, your proof did an unusual manoever, using mod beyond
integers, unfamilar even from the normal mod. This was new to me, and I
had to discover how this worked to understand your proof. So it is
inevitable that discussions of your proofs get into these areas.
> Our time keeping is modular, which is why we keep up with the hour in 12 or
> 24 hour increments.
>
> Computer programmers get to use "mod" a bit as well for any number of things
> in programs.
>
> But, in math it is primarily number theorists who play around with these
> type of equations.
>
> So, the number theorists or people playing at being number theorists put up
> examples that don't relate to my proof, and keep claiming it's wrong.
Again, if you want to just describe my examples as examples that don't
relate to my proof, that is closing off the discussion right there. I
have laid out my examples in detail for people to judge for themselves
as to whether they relate.
> I bet you figure that they should know. After all, you don't know, it
> sounds good and they're supposed to be the experts.
I am not hoping for people to just accept my writing from authority.
(Not to even imply that I am an authority). I hope people will consider
the details of what I write, and accept or reject them accordingly based
on the intrinsic content. I have tried to give complete explanations so
people can check.
Ok, I realize there have been a few exceptions to that. Occasionally
I have made claims that I was too lazy to justify in detail. These have
been the exception. And I will go into more detail on these later if
they come up in discussion.
My hope also is that James himself will make a direct evaluation of
those details.
> So, even those of you who are expert mathematicians in your own field simply
> defer to the number theorists, and a lot of them seem to have forgotten
> basic algebra because I don't think they're deliberately leading you astray.
I don't want people to defer to my writing. I want them to consider
my details, and confirm them or refute them.
> I suggest you do this.
>
> Where ever you see that "mod" in my proofs, say to yourself, it just means
> that what's to the right of it is a factor.
>
> Go back to your algebra, and you'll see how simple and obvious it all really
> is.
I also hope for such clarity.
As I have said before, I am not commited to my present questioning. I
said before and repeat now, I would consider it a successful resolution
of this round of discussion if the proof steps I am asking about are
filled in, and we can continue with later parts of the proof.
> James Harris
--
David Libert (ah...@freenet.carleton.ca)
1. I used to be conceited but now I am perfect.
2. "So self-quoting doesn't seem so bad." -- David Libert
3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig
James Harris
"David Libert" wrote in message ...
>
> "James Harris" writes:
> >
> > Ok, the issue is that posters in replying to me, say that factoring
> >
> > F(x,y,z,v) is "wrong".
>
> That has not been what I am saying.
You're not all posters.
Still, what you've been saying is still off.
A good way to explain it is to keep with the algebra because I really do
think the "mod" is throwing people off.
What we have with p=5, is
(z^2 - k1 xy)(z^2 - k2 xy) = (x+y+vz)W/h,
where W is an expression in terms of x, y, and z, like I had before.
You are essentially claiming that
(x+y+vz) factors because otherwise we could divide by, say, z^2 - k2 xy, to
get
z^2 - k1 xy = (x+y+vz)W/h(z^2 - k2 xy).
Now any objections to the proof center around claiming that z^2 - k2 xy is
not a factor of W.
Those objections must mean that (x+y+vz) factors further.
There is no other choice.
More comments below.
You hang onto those examples no matter how many times I remind you that they
look nothing like the actual argument in the proof.
The argument is algebraic.
You must explain why (x+y+vz) must factor.
Not how 12 can.
> It is misrepresenting my question to say I am questioning the
> factoring. I have agreed with the (1):(4) factoring. I am
> questioning the step to (3) based on that factoring.
>
I did not misrepresent your question. I noted that your arguments require a
factorization of (x+y+vz) by using an analogy where I represented (x+y+vz)/h
by G(x,y,z,v), and said arguments like yours required that it have
additional factors g1 and g2.
Which is what I'm saying more explicitly now.
> My own numeric examples are of this >corresponding form.
Again, I think you're confused by your numeric examples.
>They also
> support the factoring.
So what?
You use a modulus of 12 in one of your examples.
x+y = 12, does not prove that x+y has an algebraic factoring just because
12 does.
>They refute the later step corresponding to
> (3).
Your own examples "prove" that x+y = 12 must factor.
Now, I can go ahead and factor it into
(x^{1/3} + y^{1/3})(x^{2/3} - (xy)^{1/3} + y^{2/3}),
and that shows the type of thing you are claiming must be true for
(x+y+vz)/h.
>It is the (3) step you claimed in your proof as I quoted above,
> ie z^2 - quadratic formula expressions = 0 (mod (x+y+vz)/h) as quoted
> above.
>
No matter how you dance around it, your objection depends on factoring
(x+y+vz).
> So I will repeat a third time your last two lines:
>
> > Ok, the issue is that posters in replying to me, say that factoring
> >
> > F(x,y,z,v) is "wrong".
>
> That is not what I have been saying. The factoring is right. A later
> step is wrong.
>
You can't say the step is wrong. Unless you *prove* that x+y+vz factors.
Handwaving arguments using a given integer number modulus mean nothing.
And like I said before that means you are requiring that (x+y+vz) factors.
There is no basis for that requirement in the math.
Let me give you a numerical example.
x^2 + y^2 = 65,
now based on your arguments,
because 65 = 5*13, x^2 + y^2 should factor.
But it *doesn't*, unless you want to go to fractional exponents like I
showed before.
x = 33, and y = 56, work just fine anyway.
Yeung Kit Ming
James Harris wrote:
>
>
> You must explain why (x+y+vz) must factor.
>
> Not how 12 can.
>
>
Hello David, by now you should see he does not understand what constitutes a
proof. It would be a waste of time to continue the discussion since Wiles and
Taylor had shown that no such x, y, z exist and therefore you have no way to
provide a counter example to what ever statement James or any body makes about
x, y, z.
--
Prof. Yeung Kit Ming
Department of Mathematics
The Chinese University of Hong Kong
Shatin, N. T.
HONG KONG
David Libert
"James Harris" (js...@mindspring.com) writes:
> "David Libert" wrote in message ...
>>
>> "James Harris" writes:
>> >
>> > Ok, the issue is that posters in replying to me, say that factoring
>> >
>> > F(x,y,z,v) is "wrong".
>>
>> That has not been what I am saying.
>
> You're not all posters.
>
> Still, what you've been saying is still off.
>
> A good way to explain it is to keep with the algebra because I really do
> think the "mod" is throwing people off.
>
> What we have with p=5, is
>
> (z^2 - k1 xy)(z^2 - k2 xy) = (x+y+vz)W/h,
>
> where W is an expression in terms of x, y, and z, like I had before.
>
> You are essentially claiming that
>
> (x+y+vz) factors because otherwise we could divide by, say, z^2 - k2 xy, to
> get
>
> z^2 - k1 xy = (x+y+vz)W/h(z^2 - k2 xy).
Yes, the problem cases I am raising all involve (x+y+vz) factoring
nontrivially. I am not claiming these results. I am just raising them
as a case to consider among others. That is different from claiming it,
but it still requires your proof to deal with it.
> Now any objections to the proof center around claiming that z^2 - k2 xy is
> not a factor of W.
>
> Those objections must mean that (x+y+vz) factors further.
>
> There is no other choice.
Yes, all the problem cases I have raised are like this.
See my other recent article about numeric examples in general as they
relate to your proof.
> You must explain why (x+y+vz) must factor.
No, this is a distorted version of what I am talking about. I am
talking about the hypothetical cases that need to be handled or
eliminated. Explaining why (x+y+vz) must factor would be about it
factoring in all cases. I am just saying we must consider some cases in
which it does. I am not even saying such cases must definitely arise.
I am just saying a proof must consider the possibility. For example, if
a proof showed the case does not arise, then the proof would be
actively considering the case to dismiss it. So even though in the end
the case is found not to arise the possibility is still addressed in the
proof.
These points were discussed at length in the closing section of the
article you are replying to. You do not include those sections in your
quote. And above you are overlooking the distinction spelled out in
that section you deleted from the quote.
> Not how 12 can.
12 is one example that can illustrate the possibilities I am raising.
A picture is worth a thousand words.
>> It is misrepresenting my question to say I am questioning the
>> factoring. I have agreed with the (1):(4) factoring. I am
>> questioning the step to (3) based on that factoring.
>>
>
> I did not misrepresent your question. I noted that your arguments require a
> factorization of (x+y+vz) by using an analogy where I represented (x+y+vz)/h
> by G(x,y,z,v), and said arguments like yours required that it have
> additional factors g1 and g2.
Here is what you wrote in the base article:
>> > Ok, the issue is that posters in replying to me, say that factoring
>> >
>> > F(x,y,z,v) is "wrong".
You often in other articles refer to me as "a poster" instead of by
name, and other recent exchanges you have been answering me saying that
my recent objections were about something not factoring. So I took that
quote as referring to me, though above you seemed to back off from that.
Anyway, that quote is about whether F(x,y,z,v) factors. It was not
about whether (x+y+vz) factors. And in your quote, the "other posters"
are the ones saying the factoring was "wrong". Ie in the quote, James
is on the factor side, and the others are on the non-factor side.
The point you are raising now above, is you are saying that I, David,
"requiring" that (x+y+vz) factors. So in this new version, it is the
other side, the other than you James, who is on the factor side.
So your previous article was about you saying F(x,y,z,v) factors,
and your new point is you questioning that (x+y+vz) factors. So your
original quote was about a different expression, and about a different
answer re factors/question-factors, than the point you raise now.
I was responding in the article you quote to your earlier post, about
your original comments about F(x,y,z,v) factoring. I noted that I had
always agreed that F(x,y,z,v) factors, and your attributing my dispute
as being about whether F(x,y,z,v) factors was a misrepresentation of
my position.
You now respond, saying there is no misrepresentation, becuase indeed
we are in disagreement whether something should be considered as
factoring. But in saying that you have switched the topic from
F(x,y,z,v) to (x+y+vz).
My comment still stands, that
>> > Ok, the issue is that posters in replying to me, say that factoring
>> >
>> > F(x,y,z,v) is "wrong".
would be a misrepresentation of my position.
Of course, now you are waffling about whether "posters" refers to me.
Who does it refer to then?
Anyway, your response to my quoted comments earlier about us not
disagreeing whether F(x,y,z,v) factors, and you say that there is no
misrepresentation because we are disagreeing about (x+y+vz) factoring,
that is itself a new layer of misrepresentation, since my comments that
you quoted weren't concerning (x+y+vz).
> Which is what I'm saying more explicitly now.
>
>> My own numeric examples are of this >corresponding form.
>
> Again, I think you're confused by your numeric examples.
My examples provide a graphic illustration of possibilities to
consider in other cases. As such, they are illuminating, not confusing.
>>They also
>> support the factoring.
>
> So what?
>
> You use a modulus of 12 in one of your examples.
>
> x+y = 12, does not prove that x+y has an algebraic factoring just because
> 12 does.
Right. And as the closing section of my last article (that you
deleted from quotation) discusses, what is relevant to the proof is
integer factoring, not polynomial algebraic factoring.
>>They refute the later step corresponding to
>> (3).
>
> Your own examples "prove" that x+y = 12 must factor.
Are you saying that 12 stops factoring just because you set it = to
x+y ?
> Now, I can go ahead and factor it into
>
> (x^{1/3} + y^{1/3})(x^{2/3} - (xy)^{1/3} + y^{2/3}),
>
> and that shows the type of thing you are claiming must be true for
>
> (x+y+vz)/h.
>
>>It is the (3) step you claimed in your proof as I quoted above,
>> ie z^2 - quadratic formula expressions = 0 (mod (x+y+vz)/h) as quoted
>> above.
>>
>
> No matter how you dance around it, your objection depends on factoring
> (x+y+vz).
I don't dance around it, I affirm it freely.
>> So I will repeat a third time your last two lines:
>>
>> > Ok, the issue is that posters in replying to me, say that factoring
>> >
>> > F(x,y,z,v) is "wrong".
>>
>> That is not what I have been saying. The factoring is right. A later
>> step is wrong.
>>
>
> You can't say the step is wrong. Unless you *prove* that x+y+vz factors.
>
> Handwaving arguments using a given integer number modulus mean nothing.
Ok, wrong is maybe an overloaded word here. Better is incomplete. A
proof must consider all cases. If a case cannot be resolved as to
whether it actually arises or not, the proof must allow for the worst
scenario and cover such cases anyway. The issue is not to prove x+y+vz
factors, the issue is to either prove it doesn't factor, or if that
cannot be done, to handle that case and show that if it should arise it
doesn't mess up the final conclusion.
This was all covered in the closing section you deleted.
Again the same point. Not can I prove it factors. It is can you
prove it doesn't factor. And if you can't, can you prove your final
conclusion even if it does factor.
> Let me give you a numerical example.
>
> x^2 + y^2 = 65,
>
> now based on your arguments,
>
> because 65 = 5*13, x^2 + y^2 should factor.
>
> But it *doesn't*, unless you want to go to fractional exponents like I
> showed before.
This is backwards. x,y,z,v in your proof are fixed but unknown
integers. So x+y+vz is a fixed but unknown integer. The property is
whether this integer factors in the sense of integers, not whether the
corresponding polynomial factors as a polynomial. The integer doesn't
lose its factoring just because it can be set to a polynomial expression
which as a polynomial doesn't factor.
> x = 33, and y = 56, work just fine anyway.
I am going to use the methods you suggest to prove a revolutionary
result in number theory. Even though this application is mine, I am
going to name the theorem in your honour, since the underlying methods
are yours, as suggested by your comments here.
Harris Theorem: every integer is prime
Proof: an arbitrary integer can be written in the form x+1.
But the polynomial x+1 has no non-trivial factorization.
QED
oooF
James Harris
"David Libert" wrote in message ...
>
> "James Harris" writes:
> >
> > You are essentially claiming that
> >
> > (x+y+vz) factors because otherwise we could divide by, say, z^2 - k2
xy, to
> > get
> >
> > z^2 - k1 xy = (x+y+vz)W/h(z^2 - k2 xy).
>
> Yes, the problem cases I am raising all involve (x+y+vz) factoring
> nontrivially. I am not claiming these results. I am just raising them
> as a case to consider among others. That is different from claiming it,
> but it still requires your proof to deal with it.
>
I'm glad you admitted that and I hate to tell you this David, but that
objection is just plain dumb.
Let me explain.
Remember the simple example for n=2?
(v^2 - 1)z^2 - 2xy = (x+y+vz)(-x-y+vz), when
x^2 + y^2 = z^2.
Now, really pay attention to it.
All my arguments are about expression like the above.
There is no more reason to believe that
x+y+vz factors than if, given
(x^2 + ax + b)(x^2 + cx + d) = (x+e)F,
that x+e factors here.
It's *basic* algebra.
You wouldn't argue with me about whether or not x+e is a factor of
x^2 + ax+ b or x^2 + cx + d, would you?
But now, here you are arguing *exactly* the same thing.
Think about it.
Erik Max Francis
> It's *basic* algebra.
hah! Tell us another one, Mr. Integers Are Irrational.
--
Erik Max Francis / m...@alcyone.com / http://www.alcyone.com/max/
__ San Jose, CA, US / 37 20 N 121 53 W / ICQ16063900 / &tSftDotIotE
/ \ Of all the perversions, chastity is the strangest.
\__/ Anatole France
blackgirl international / http://www.blackgirl.org/
The Internet resource for black women.
James Harris
fix a few typos.
"James Harris" wrote in message ...
>
> "David Libert" wrote in message ...
> >
> > "James Harris" writes:
> > >
> > > You are essentially claiming that
> > >
> > > (x+y+vz) factors because otherwise we could divide by, say, z^2 - k2
> xy, to
> > > get
> > >
> > > z^2 - k1 xy = (x+y+vz)W/h(z^2 - k2 xy).
> >
> > Yes, the problem cases I am raising all involve (x+y+vz) factoring
> > nontrivially. I am not claiming these results. I am just raising them
> > as a case to consider among others. That is different from claiming it,
> > but it still requires your proof to deal with it.
> >
>
> I'm glad you admitted that and I hate to tell you this David, but that
> objection is just plain dumb.
>
I debated about putting it that way, but I ultimately decided to go with the
hard line.
I'm not saying that David is dumb. I just think he got caught with a dumb
argument and defended it rather than just dropping it when it obviously
didn't apply.
> Let me explain.
>
> Remember the simple example for n=2?
>
> (v^2 - 1)z^2 - 2xy = (x+y+vz)(-x-y+vz), when
>
> x^2 + y^2 = z^2.
>
> Now, really pay attention to it.
>
> All my arguments are about expression like the above.
>
Of course that should be expressions.
What I'm trying to point out is that the expressions I give factor in the
same way.
It just turns out they get progressively more complicated as you go to
higher primes p, which shouldn't surprise anyone.
And, who out there thinks that (x+y+vz) *might* factor here?
Ok, it's too easy since you only have
z^2 - 2xy on the left, but what if
z^2 - 2xy = f1 f2, where neither
f1 or f2 equal x+y+vz.
David's argument would imply that you could factor x+y+vz to get this case.
I doubt any reasonable person out there believes I need to prove that this
possibility is not a decent objection.
> There is no more reason to believe that
>
> x+y+vz factors than if, given
>
> (x^2 + ax + b)(x^2 + cx + d) = (x+e)F,
>
> that x+e factors here.
>
> It's *basic* algebra.
>
I just want to emphasize that the expression we've been talking about with
the proof for p=5, is exactly like the above, but just like with n=2, it's
not obvious just by looking at it.
That should not be a problem for mathematicians.
Just because my expressions don't quite look like some other form, but still
belong to the same type, mathematicians shouldn't be unable to see the
underlying structure.
> You wouldn't argue with me about whether or not x+e is a factor of
>
> x^2 + ax+ b or x^2 + cx + d, would you?
>
> But now, here you are arguing *exactly* the same thing.
>
> Think about it.
>
I like that example because it's simple enough for any of you who got past
algebra in high school to understand what's going on.
And it's simple enough to help you understand why I've been wondering if the
people posting that I'm wrong, or the people who are quietly sitting back
while claims that my proof is false get put out, aren't doing so because
they're hoping that no one will ever see the truth.
And now you might understand why I've said more than once that if there
weren't a really simple proof that could be taught to high school students,
I wouldn't want to look for it and find it, because mathematicians could
make sure that no one ever believed it was true.
afraid of psychosis
David Libert
I have realized I should back off from a statement I agreed to in the
post below. I do not presently see that the statement is actually
false, but I also now realize my reasons for believing it true were
incorrect. So I presently hold off judgement on the statement.
David Libert (ah...@FreeNet.Carleton.CA) writes:
> "James Harris" (js...@mindspring.com) writes:
>> "David Libert" wrote in message ...
>>>
>>
>> Still, what you've been saying is still off.
>>
>> A good way to explain it is to keep with the algebra because I really do
>> think the "mod" is throwing people off.
>>
>> What we have with p=5, is
>>
>> (z^2 - k1 xy)(z^2 - k2 xy) = (x+y+vz)W/h,
>>
>> where W is an expression in terms of x, y, and z, like I had before.
>>
>> You are essentially claiming that
>>
>> (x+y+vz) factors because otherwise we could divide by, say, z^2 - k2 xy, to
>> get
>>
>> z^2 - k1 xy = (x+y+vz)W/h(z^2 - k2 xy).
>
> Yes, the problem cases I am raising all involve (x+y+vz) factoring
> nontrivially. I am not claiming these results. I am just raising them
> as a case to consider among others. That is different from claiming it,
> but it still requires your proof to deal with it.
>
>
>> Now any objections to the proof center around claiming that z^2 - k2 xy is
>> not a factor of W.
>>
>> Those objections must mean that (x+y+vz) factors further.
>>
>> There is no other choice.
>
> Yes, all the problem cases I have raised are like this.
So the above is concerning my questioning of a step in James' proof,
namely against James' step I have raised the possibility that a product
is a multiple of James' modulus (x+y+vz)/h while neither factor
multiplying to that product is a multiple of (x+y+vz)/h .
I had presented three parallel examples to these, each using a
different modulus replacing James' (x+y+vz)/h, and using different
factors.
James has raised the issue of such examples involving the modulus
factoring non-trivially in the ring. It turns out that this does happen
in each of my three examples.
My examples were presented as possible parallels to the cases arising
in James' proof. James' was suggesting that all the possibilities I
suggest for his actual cases also nust involve the modulus, (x+y+vz)/h
in Janes' cases, factoring nontrivially.
I considered this, and above posted my agreement on that point. I was
thinking how can the modulus divide the profuct but not the individual
factors. I thought the factors of the modulus must split between the
two factors on the product, meaning in particular the modulus factors
nontrivially.
I realized later such reasoning is implicitly assuming we are in a
unique factorization domain (UFD). These factoring of modulus questions
occur in the ring extension R, so there is the possibility that R is not
a UFD. So my previous reasons for agreeing to the nontrivial factoring
of modulus claim might lose their foundation.
I will define terminology to clarify the issue under discussion here,
and state what I do presently know related to it.
We will be working now in commutative rings with unit element 1.
These definitions come from Hungerford _Alegebra_ .
Def: An element of such a ring is defined to be a unit iff it has a
multiplicative inverse (ie something that multiplies by it to give 1).
Def: An element a is defined to be irreducible iff
a != 0 & a is not a unit &
for all b,c if a = b*c then b is a unit or c is a unit
Def: for a,b elements a|b ("a divides b") iff
a != 0 & there exists c s.t. b = a*c
Def: An element a is defined to be prime iff
a != 0 & a is not a unit &
for all b,c a|(b*c) -> a|b or a|c
In Z, the ring of integers, an integer is prime <-> it is
irreducible.
In elementary number theory, the common definition of the word "prime"
is closest to the ring theory definition of irreducible. But this is
not too significant in view of the last equivalence.
The Z equivalence of irreducible to prime breaks down in other rings.
I cite examples from Hungerford.
In Z/(6) (Z mod 6) 2 (or more accurately [2]), is prime but
not irreducible. (Ie: not irreducible: 2 = 2*4 and neither of these
are units).
The next example is the important one for the point of this article
about James' proof.
This comes from an exercise in Hungerford. Consider R = Z[sqrt(10)].
Hungerford's excercise includes that in R 2, 3, 4+sqrt(10), 4-sqrt(10)
are each irreducible. On the other hand
2*3 = 6 = (4+sqrt(10)) * (4-sqrt(10)) and this suffices for showing
each of 2, 3, 4+sqrt(10), 4-sqrt(10) nonprime.
(Ie, for example 2| ((4+sqrt(10)) * (4-sqrt(10))) and then also verify
2 doesn't divide either of these factors. Similarly for showing the
other three nonprime.)
So this last example shows that it is possible to have an extension of
Z by the solution of a quadratic equation (ie Z[sqrt(10)]) with
nonprime irreducible elements. This last possibility is relevant for
the point under discussion about James' proof.
So with that background I return to the topic above, on the relation
of my possible gap in James' proof to the modulus factoring
nontrivially. This discussion will involve the ring R from James'
proof, a proper extension of the integers, so for this discussion I will
use the general ring theoretic definitions of irreducible and prime, not
the "prime" definition from elementary number theory.
So I have raised the question of the modulus in James' proof dividing
a product without dividing either factor, this possibility conflicting
with James' step. This property is exactly a failure of primeness as
defined ring theoretically. So the question I raise for James' proof is
the possibility in the extension ring R that the modulus (x+y+vz)/h
has a specific failure of primeness.
James suggested that these cases all involve a nontrivial factoring of
the modulus, ie the modulus being reducible.
My possibility was about nonprimeness, and the new point James raised
was my nonprimeness implying reducibility. Or phrasing this as
contrapositives, does irreducubility -> primeness ? Hungerford's second
example was exactly a failure of this implication. And note this second
example occured in an extension of the integers by a solution of the
quadratic equation, exactly the type of ring extension appearing in
James' proof.
So this at least suggests the possibility of nonprime cases along the
lines I raise which are nontheless irreducible. In my previous article
I had agreed with James suggestion, making the classic mistake of being
misled by the UFD case, but considering these other cases now I retract
that. I do not know whether the cases I have suggested for
consideration involve the modulus factoring nontrivially.
This whole point I am raising now is similar to the first round of
discussion about my examples as related to James basic proof. Now the
same issue comes up with the new example from this article and James'
claim that my nonprime cases must be also reducible.
The common point is James' full assumptions in the background include
a Fermat counterexample, so I can never produce a complete
counterexample to any inference James suggests. My examples are instead
weaker ones, examples in cases parallel but not identical to his. Such
examples at least show that any elementary proof supporting the
questioned inferences in James' cases must at least be a detailed enough
argument to exploit differences between James' cases and my examples.
David Libert
>> (renaming it as (4) now to keep numbers in sequence):
>>
>> (4) F(x,y,z,v) = F1(x,y,z,v)*F2(x,y,z,v) (mod G(x,y,z,v)).
>>
>> Your starting assumption was that F(x,y,z,v) is congruent to 0 in
>> that modulus, so from (4) you know that product on the right side is
>> congruent to 0. The question is how do you get from that to the (3)
>> conclusion that one factor must be congruent to 0. My recent numerical
>> examples have shown cases where this corresponding step fails.
>>
>
> You hang onto those examples no matter how many times I remind you that they
> look nothing like the actual argument in the proof.
>
> The argument is algebraic.
>
> You must explain why (x+y+vz) must factor.
>
> Not how 12 can.
Since my earlier writing I have found a new proof related to this last
point. I cannot prove outright that (x+y+vz)/h must factor. But I
can come very close to this. Recall your proof starts with the
assumption for contradiction that x,y,z is a pairwise coprime Fermat
counterexample. You then define h from the Barlow Abel relations.
This choice of h is completely determined by the original choice of
x,y,z.
Then you define restrictions on v, and let v be an arbitrary
integer meeting these restrictions. These choices of x,y,z,v are all
of the choices made in the proof.
I will now present a new proof I have recently found that for each
choice as above of x,y,z there are infinitely many choices of v
meeting all your restrictions and producing (x+y+vz)/h that factors
nontrivially in Z, the integers.
In fact my statement on how many v's I can produce like this is
stronger than merely saying there are infinitely many v's : using more
refined measures of "how many" involving asymptotic density, we get
statements strictly stronger than merely being infinite. These
statements will be obvious from the discusion of the proof below.
There was an earlier round of discussion like this, in "questions
about JSH FLT". I noted a problem for some v's as James was then
allowing. He replied to just select those v's handling my problem, in
effect adding to the list of restrictions on the choice of v.
Such an answer may again be possible now. I have always found the
closing section of James' proof obscure. So I can't tell if that part
of the proof would be affected by further restricting the choice of v to
avoid the cases I will be presenting here.
Even if this response is possible and is done, the fact remains that
my proof below refutes James' recent claims that his assumptions to date
are enough to conclude (x+y+vz)/h can't factor.
Toward my proof, I will paste a review of the restrictions on v from
my article
[1] Aug 9, '00 "Re: on JSH FLT"
http://forum.swarthmore.edu/epigone/sci.math/toulanpi/8mqt0m$pvc$1...@freenet9.carleton.ca
> We assume v is an integer coprime to h, and v != 1, v != -1. In
>the recent thread "questions about JSH FLT" James has recently added the
>assumption on v that xy is coprime to (x+y+vz)/h. This was to fix
>some problems with the proof discussed there. As discussed there, there
>are still many v's satisfying this extra assumption.
My base article starting the thread "questions about JSH FLT" asked
about a problem for the first version of the proof. In that thread
James suggested putting extra restrictions on v to fix that problem,
as just mentioned above in the quote. Below is my article expanding on
James suggestion.
[2] Aug 6 00 "Re: questions about JSH FLT"
http://forum.swarthmore.edu/epigone/sci.math/yesmolwor/8ml3o7$14i$1...@freenet9.carleton.ca
I quote a key passage from [2] :
> Ok this fixes the point I was questioning in my first post and the
>second related point from my followup post.
>
> Your original writeup wanted to have an integer v coprime to h.
>
> So we will add to this condition the new requirement that xy be
>coprime to (x+y+vz)/h . This condition can be realized.
>
> Suppose we have a starting v, an integer corpime to h . I will
>define the new integer v1 = xyv. I claim this v1 can be used to meet
>the new version: ie v1 is still coprime to h, and as required
>(x+y+ v1*z)/h is coprime to xy.
[2] continues with a proof of this last claim.
I will use these properties on the restrictions on v to prove there
are many v 's meeting all these restrictions and producing
(x+y+vz)/h which factors nontrivially.
So by those closing comments quoted above, xy*(anything coprime to h)
will give a value meeting the two important clauses on v.
The only other clauses on v are v != +-1.
So consider the arithmetic sequence: for n>0 a_n = xy*(1 + n*h).
x and y are assumed coprime, so neither = +-1, so xy != +-1, so for
all n>0 a_n != +-1.
For all n>0, 1+nh is coprime to h, so by the comments above a_n
meets the other two requirements on v.
We conclude a_n (n>0) is an arithmetic sequence of integers each
meeting all of the restrictions on v.
We return now to the expression (x+y+vz)/h . [1] proved that for the
present assumptions of x,y,z a pairwise coprime Fermat counterexample,
and for h from the Barlow Abel relations for x,y,z , we have in
integers h|(x+y) and h|z.
So (x+y+vz)/h = (x+y)/h + v*(z/h) and viewing the right
side as a linear expression in v, the constant coefficient (x+y)/h
and the linear coefficient z/h are both integers.
Now define for n>0 b_n = (x+y)/h + a_n*(z/h) .
b_n is a linear expression in a_n which was itself an arithmetic
progression, so b_n is itself an arithmetic progression with integer
coefficients as n varies: n>0.
A consequence of the Prime Number Theorem which gives a formula for
the asymptotic density of primes in positive integers is that the limit
of the number of primes below m divided by m tends to 0. In
particular this means that within any arithmetic progression the
fraction of primes enumerated tends to 0, otherwise the fixed density of
the progression within all integers * the limsup density of primes in
the progression would give a positive lower bound on limsup the density
of primes in all integers, contradicting the density of primes in all
integers being defined and = 0.
So there are infinitely many nonprime b_n 's, in fact asymptotically
the fraction of nonprimes tends to 1.
These b_n 's are the respective (x+y+a_n*z)/h for a_n 's, each of
which met all restrictions on v.
So these are not primes. So they are either powers of pure primes, or
products of mixed prime powers. Either way they factor nontrivially.
David Libert
In my recent articles I have identified a gap in James' proof, related
to the possibility of James' modulus (x+y+vz)/h being nonprime in the
ring theoretic sense in the ring R appearing late in James' proof.
James has been dismissing this as a possible case that must be
considered, by questioning the possibility of his modulus factoring
nontrivially, ie in ring theoretic language questioning the possibility
of his modulus being reducible in R.
I have just posted some further articles on this, but now I want to
summarize all the weak points in James' answer that these recent
articles have raised.
So my gap concerns the possibility that (x+y+vz)/h has a specific
instance of nonprimeness in R: a particular pair of factors that the
modulus does not divide individually but divides their product.
Nonprimeness in general says this happens for some pair of factors. My
question was having this happen for two specific factors.
So James' reply saying his proof need not consider this case was that
(x+y+vz)/h doesn't factor nontrivially, ie (x+y+vz)/h is
irreducible, therefore according to James my suggested instance of
nonprimeness can't occur.
This argument of James to dismiss the case I raise passes sequentially
through three distinct gaps, any one of which individually would leave
the question unsettled.
Gap 1: So (x+y+vz)/h is irreducible in Z? But I just posted that
with the current restrictions on v there are infinitely many v for
which (x+y+vz)/h is reducible. In fact there is asymptotic density 1
ratio of such v within an arithmetic progression of acceptable v 's.
Oops, infinitely many counterexamples to the opening claim.
Gap 2: So what about the left over v's that got by the first gap (if
any, I don't actually have a proof that there are any such v's).
The significant ring for the proof by the end is not Z, it is R a
ring extension of Z. The gap I raised is about nonprimeness of the
modulus in R, not in Z.
So factoring of the modulus in R is closer to the nonprimeness
property in R finally desired.
So for v's getting past the first gap and irreducible in Z, what might
happen? As usual in all this, I can't make complete counterexamples
when James assumes a Fermat counterexample to start, but in terms of
my parallel examples, one of my three parallel problem examples was
modulus 7 and another was modulus 2. These are both moduli
irredicible in Z, and so getting by that corresponding to gap 1.
But in those two examples 7 and 2 each factored nontrivially in the
respective R extension rings: 7 = (3 + sqrt(2)) * (3 - sqrt(2))
2 = (sqrt(3) + 1) * (sqrt(3) - 1) .
So just because a case of (x+y+vz)/h avoided the first gap and was
irreducible in Z, there is still the general possibility of
(x+y+vz)/h factoring in R, which is the important context for
considering primeness in R.
Gap 3: Suppose we have v's passing both of the first two gaps. So
these v's yield a (x+y+vz)/h actually irreducible in R.
Must these (x+y+vz)/h be prime in R, ruling out my suggested gap of
an instance of nonprimeness?
My other recent post cited an example of an extension of Z by a
quadratic root (the same general form arising in James' p=5 proof)
having elements nonprime and irreducible. So even if we get past the
first two hurdles and get (x+y+vz)/h irreducible in R, there remain
cases parallel to James' in which primeness fails.
As usual, all these efforts must rely on examples distinct from James'
actual cases which are based on false premises of a Fermat
counterexample. But these examples raising possibilities at least
suggest when things are not trivial or obvious. Originally James
skipped over the cases I raised as if they need not be mentioned. Now
responding to my articles he dismisses these cases by reference to the
modulus not factoring. But that is as much as he says, there is no
detail as to why the modulus doesn't factor, nor why that nonfactoring
fact if actually true answers my question. So the internals of these
questions are so far completely glossed over in James' discussions just
as his previous rounds of discussion glossed over the very case I
raised.
My examples above show that these points James skips over now have
their own content. If there is an answer as to why the three gaps I
note above for parallel cases don't apply to James' cases it depends on
more than just skipping over the whole topic in discussion.
Tom Osborn
"trivial", or "any real mathematician could see". [Tradition leads one to
expect such "exercises left to the reader" on good faith].
Thanks to Andrew Wiles, and with reasonably well devised partial
arguments by JSH, we can't expect to find falsifications for _all_ Harris's
gaps. Some of his chains of reasoning will die because of false
assumptions, but the truth or falsity of others won't be decideable
(and who knows which)! Some will end up proveable. Ie, apart
from flawed assumptions and application, JSH's achievements (less
gaps) cannot be inconsistent with what Wiles has already established.
[He's trying to explain which horse will win a race, and why, but after
the race has been decided. The "why" can contain as much sophistry
as he can type. Alternative proof? ...].
What has been offensive is the continual bluffs by JSH about the gaps.
What has been left out, _isn't_ there. Nothing more can be assumed.
T.
--
Dr Tom Osborn
Director of Modelling
The NTF Group
Level 7, 1 York Street
SYDNEY NSW 2000