JSH: Negative Pell's Equation, consequences
JSH
simple result about the negative Pell's Equation and its connection to
Pell's Equation, where posters claimed the result was previously
widely known.
That result is that given j^2 - Dk^2 = -1, the negative Pell's
Equation, x, for x^2 - Dy^2 = 1, is given by x = 2j^2 + 1.
I've noted that is not stated in mainstream mathematical literature.
Posters have argued with me--often insultingly--claiming it is.
In response to a request for citations giving the result, they gave,
if anything, citations to equations that could LEAD to that result, if
you knew where to look!!! All math is derived from prior equations,
except the basic axioms.
Now here are some obvious consequences of that result to further test
your credibility in believing it is already part of mainstream
literature.
Given non-zero integer solutions to x^2 - Dy^2 = 1:
1. x must be odd, when a solution to the negative Pell's Equation
exists.
2. x = -1 mod D is required when a solution to the negative Pell's
Equation exists.
The issue is whether that is mainstream knowledge in MODERN number
theory, so whether or not you think you know number theory and know
these trivial results is an issue because I am trying to teach here.
The people arguing with me are, in my opinion, hoodlums disrupting a
class discussion.
Knowledge just is. But when some people hate knowledge because they
think it hurts their class positions or their personal views of
themselves, they can behave very badly in their attempts to hide
knowledge.
I think Pell's Equation is kind of fun. There are interesting little
results around it easily available to people who enjoy mathematics.
It's not my fault if these results are not taught by mainstream
mathematicians.
And I should not be attacked for giving them now.
James Harris
juandiego
> A telling discussion erupted when I challenged these newsgroups on a
> simple result about the negative Pell's Equation and its connection to
> Pell's Equation, where posters claimed the result was previously
> widely known.
No you were claiming that the result was not noted in the modern
mathematical literature.
I gave you two references Williams "Solving the Pell Equation"
2009 and Ed Barbeau's book 2003 where your bogus claim to
orihinality is treated as an EXCERCISE as it is in
the Carmichael reference 1915 which I also gave you.
These facts are very easy to understand.
Maybe you are so simple minded that you think if
authors are using different symbols for the
variables in Pell's equation they are deriving
different results or that a slightly different order
of derivation means the results are different.
It is impossible to tell what you are actually posting about.
I doubt if you know youself.
juandiego
> And I should not be attacked for giving them now.
No-one is attacking you for giving the result.
People are merely pointing out your cretinous
inability to see that a trivial result has been
derived by many thousands of people many thousands
of times over a period of almost 1500 years.
In your profound ignorance you seem to think that
you have received some sort of divine revelation
of sublime truth - rather like a child who, having
realized for the first time that 2+2 =4, goes
around telling everyome that 2+2 = 4, oblivious
to the fact that everyone else knows this already.
Obviously one expects a child to behave in this way.
An adult, and I suppose you are nominally an adult,
doing the same thing should expect a different reaction.
Jens Stueckelberger
should indeed not be attacked because of that - but you are too much fun.
JSH
> On 17 May, 15:00, JSH <jst...@gmail.com> wrote:
>
> > A telling discussion erupted when I challenged these newsgroups on a
> > simple result about the negative Pell's Equation and its connection to
> > Pell's Equation, where posters claimed the result was previously
> > widely known.
>
> No you were claiming that the result was not noted in the modern
> mathematical literature.
I suggest readers do a web search on "negative Pell's Equation".
> I gave you two references Williams "Solving the Pell Equation"
> 2009 and Ed Barbeau's book 2003 where your bogus claim to
> orihinality is treated as an EXCERCISE as it is in
> the Carmichael reference 1915 which I also gave you.
That's remarkable.
You say that the exercise states that EVERY solution to the negative
Pell's Equation gives a solution to Pell's Equation?
Then why bother solving for one if you have the other?
Isn't that stupid?
If what you say is true, then anyone in modern times who ever solved
for Pell's Equation after having the solution to the negative Pell's
Equation or solved for the negative Pell's Equation when they had a
solution for Pell's Equation is an idiot, correct?
James Harris
juandiego
You almost has this straight in your head a few posts
ago.
How are you going to find a solution to x^2 -Dy^2 = -1
in the first place and how do you know it is the smallest
solution when you have found it. If it is not the smallest
solution you will miss out solutions to x^2 -Dy^2 = 1.
The equation x^2 -Dy^2 = -1 is not solvable for
every non-square D and you don't know in advance for
which D, x^2 -Dy^2 = -1 is solvable
When do you stop looking for solutions to x^2 -Dy^2 = -1
if it is more than likely you will never find one ?
x^2 -Dy^2 = 1 is always solvable by, say,
continued fractions and you never need to calculate the
actual valoe of x^2 -Dy^2 at any stage of the algorithm.
Why slow down the calculation by evaluating x^2 -Dy^2
for every convergent to see if the value is -1 when
all you need to know is whether the period of the CF is
odd or even ?
The only idiot is you.
JSH
Pell's Equation gives you a solution.
My point, which you clearly ignored--and yes you are like a hoodlum
disrupting a class discussion--is that you NEED ONLY ONE and you have
the other.
> solution when you have found it. If it is not the smallest
> solution you will miss out solutions to x^2 -Dy^2 = 1.
>
> The equation x^2 -Dy^2 = -1 is not solvable for
> every non-square D and you don't know in advance for
> which D, x^2 -Dy^2 = -1 is solvable
That's not true and this exchange is beyond tedious.
The negative Pell's Equation ALWAYS has a solution if for every prime
factor p of D, p = 1 mod 4.
And your position is that you can evaluate for the world, as my
readership is worldwide--my math blog has hits from 100 countries/
territories as Google Analytics puts it, already this year--whether or
not a result is known to the mainstream world or not and you don't
know the simplest things about the negative Pell's Equation?
You are a hoodlum. It's that simple.
A hooligan on Usenet. Nothing more.
> When do you stop looking for solutions to x^2 -Dy^2 = -1
> if it is more than likely you will never find one ?
>
> x^2 -Dy^2 = 1 is always solvable by, say,
> continued fractions and you never need to calculate the
> actual valoe of x^2 -Dy^2 at any stage of the algorithm.
> Why slow down the calculation by evaluating x^2 -Dy^2
> for every convergent to see if the value is -1 when
> all you need to know is whether the period of the CF is
> odd or even ?
The result is that you need ONLY ONE. With one, you have the other.
With a solution to the negative Pell's Equation you have solved Pell's
Equation. With a solution to Pell's Equation, when it exists--and
I've given a way to know when it does exist--you have a solution to
the negative Pell's Equation.
> The only idiot is you.
Only for continuing to discuss serious mathematics with a person who
shows not only contempt for it, but for rational discourse, and for
every person who cares about mathematical knowledge with an interest--
all over the world--who reads these posts. You hate. It's that
simple.
But your hatred for humanity is not an excuse to be such a rude
person.
What manners did your parents teach you?
None?
James Harris
José Carlos Santos
Fascinating, coming from the same guy who wrote:
--------------------------------------------------------------------
You stupid shit head!!! What the fuck is wrong with you Ullrich?
No matter how many fucking times I tell you to fuck off, you keep
replying to me!!!
What the fuck is your problem you shithead?
You Ullrich are a stupid piece of dumb shit who refuses to get the
message when someone does NOT want to talk to you, you stupid fucking
shitty asshole.
You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass
stupid self somewhere to GET A FUCKING CLUE and QUIT FUCKING REPLYING
TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!
FUCK OFF!!!!
Can't you get it through your stupid head?
FUCK OFF!!!!!!!!!!!!!!!!!
--------------------------------------------------------------------
right here:
http://mathforum.org/kb/plaintext.jspa?messageID=540148
Therefore, we can safely deduce that your parents taught you no manners,
right? :-)
Best regards,
Jose Carlos Santos
Don Redmond
> A telling discussion erupted when I challenged these newsgroups on a
> simple result about the negative Pell's Equation and its connection to
> Pell's Equation, where posters claimed the result was previously
> widely known.
>
> That result is that given j^2 - Dk^2 = -1, the negative Pell's
> Equation, x, for x^2 - Dy^2 = 1, is given by x = 2j^2 + 1.
>
> I've noted that is not stated in mainstream mathematical literature.
> Posters have argued with me--often insultingly--claiming it is.
>
> In response to a request for citations giving the result, they gave,
> if anything, citations to equations that could LEAD to that result, if
> you knew where to look!!! All math is derived from prior equations,
> except the basic axioms.
>
>
Katz's
History of Mathematics, 3rd ed., the exercises on the history of
Indian mathematics
chapter have problems that give you explicitly solutions to x^2 - Dy^2
= 1 given
solutions to u^ - Dv^2 = -1, u^2 - Dv^2 = +-2 and u^2 - Dv^ = +-4.
These are the
results of Brahmagupta that everyone has been telling you about.
Don
Rotwang
> simple result about the negative Pell's Equation and its connection to
> Pell's Equation, where posters claimed the result was previously
> widely known.
>
> That result is that given j^2 - Dk^2 = -1, the negative Pell's
> Equation, x, for x^2 - Dy^2 = 1, is given by x = 2j^2 + 1.
>
> I've noted that is not stated in mainstream mathematical literature.
> Posters have argued with me--often insultingly--claiming it is.
>
> In response to a request for citations giving the result, they gave,
> if anything, citations to equations that could LEAD to that result, if
> you knew where to look!!! All math is derived from prior equations,
> except the basic axioms.
>
> Now here are some obvious consequences of that result to further test
> your credibility in believing it is already part of mainstream
> literature.
>
> Given non-zero integer solutions to x^2 - Dy^2 = 1:
>
> 1. x must be odd, when a solution to the negative Pell's Equation
> exists.
>
> 2. x = -1 mod D is required when a solution to the negative Pell's
> Equation exists.
Counterexample: D = 5, x = 161, y = 72.
Rotwang
On 17 May, 18:40, Rotwang <sg...@hotmail.co.uk> wrote:
> On 17 May, 15:00, JSH <jst...@gmail.com> wrote:
>
> [...]
>
> > Now here are some obvious consequences of that result to further test
> > your credibility in believing it is already part of mainstream
> > literature.
>
> > Given non-zero integer solutions to x^2 - Dy^2 = 1:
>
> > 1. x must be odd, when a solution to the negative Pell's Equation
> > exists.
Assuming you made the same mistake as with the second claim below
(i.e. assuming that every solution to Pell's equation is given by x =
2j^2 + 1 for some j such that j^2 - Dk^2 = -1, whenever the latter
equation can be solved), you haven't proved this. But after thinking
about it for a bit, here is what I think is an actual proof: suppose
that x^2 - Dy^2 = 1, j^2 - Dk^2 = -1 and that x is even. Note that D
must be odd, and that x^2 = 0 mod 4. Suppose D = 1 mod 4. Then
0 - 1*y^2 = 1 mod 4,
so y^2 = -1 mod 4, which is impossible. Therefore we must have D = 3 =
-1 mod 4. Then
j^2 + k^2 = -1 mod 4,
but this is easily shown to be impossible by exhaustion, so again we
have a contradiction. QED. Assuming the above is correct, note that I
was able to come up with a proof despite knowing nothing about this
area. It's therefore safe to assume that this fact is well known.
juandiego
> On May 17, 9:07 am, juandiego <sttscitr...@tesco.net> wrote:
>
>
>
>
>
> > On 17 May, 16:38, JSH <jst...@gmail.com> wrote:
>
> > > On May 17, 7:49 am, juandiego <sttscitr...@tesco.net> wrote:
>
> > > > On 17 May, 15:00, JSH <jst...@gmail.com> wrote:
> > How are you going to find a solution to x^2 -Dy^2 = -1
> > in the first place and how do you know it is the smallest
>
> Pell's Equation gives you a solution.
Are you a complete idiot ?
Supposedly, you are using x^2 -Dy^2 = -1 to solve x^2 -Dy^2 = 1
and not the other way round.
You don't seem to know what you are solving at any
stage of your deranged outpourings.
> My point, which you clearly ignored--and yes you are like a hoodlum
> disrupting a class discussion--is that you NEED ONLY ONE and you have
> the other.
Yes, but x^2 -Dy^2 = -1 is not ALWAYS solvable.
Are you too thick to grasp that ?
> > solution when you have found it. If it is not the smallest
> > solution you will miss out solutions to x^2 -Dy^2 = 1.
>
> > The equation x^2 -Dy^2 = -1 is not solvable for
> > every non-square D and you don't know in advance for
> > which D, x^2 -Dy^2 = -1 is solvable
>
> That's not true and this exchange is beyond tedious.
What is the solution of x^2 -35y^2 = -1 ?
> The negative Pell's Equation ALWAYS has a solution if for every prime
> factor p of D, p = 1 mod 4.
13 == 1 mod 4
17 == 1 mod 4
D= 13*17 = 221
What is the solution of x^2- 221y^2 = -1 ?
As usual your mathematical eructations are nonsense.
You seem incapable of understanding plain English.
What I said was x^2 -Dy^2 = -1 is not always solvable.
You are obviously too mentally confused to hold more
than one thought at a time.
> > When do you stop looking for solutions to x^2 -Dy^2 = -1
> > if it is more than likely you will never find one ?
>
> > x^2 -Dy^2 = 1 is always solvable by, say,
> > continued fractions and you never need to calculate the
> > actual valoe of x^2 -Dy^2 at any stage of the algorithm.
> > Why slow down the calculation by evaluating x^2 -Dy^2
> > for every convergent to see if the value is -1 when
> > all you need to know is whether the period of the CF is
> > odd or even ?
>
> The result is that you need ONLY ONE. With one, you have the other.
Yes, but you can't always find a solution to x^2 -Dy^2 = -1
What is the solution to x^2 -35y^2 = -1 ?
> With a solution to the negative Pell's Equation you have solved Pell's
> Equation. With a solution to Pell's Equation, when it exists--and
> I've given a way to know when it does exist--you have a solution to
> the negative Pell's Equation.
Buffoon. x^2 -Dy^2 = 1 always has a solution
x^2 -Dy^2 = -1 doesn't.
6^2 -35*1^2 =1 what is the solution of x^2 -35y^2 = -1 ?
marcus_...@yahoo.com
> On May 17, 7:49 am, juandiego <sttscitr...@tesco.net> wrote:
>
> > On 17 May, 15:00, JSH <jst...@gmail.com> wrote:
>
> > > A telling discussion erupted when I challenged these newsgroups on a
> > > simple result about the negative Pell's Equation and its connection to
> > > Pell's Equation, where posters claimed the result was previously
> > > widely known.
>
> > No you were claiming that the result was not noted in the modern
> > mathematical literature.
>
> I suggest readers do a web search on "negative Pell's Equation".
>
> > I gave you two references Williams "Solving the Pell Equation"
> > 2009 and Ed Barbeau's book 2003 where your bogus claim to
> > orihinality is treated as an EXCERCISE as it is in
> > the Carmichael reference 1915 which I also gave you.
>
> That's remarkable.
>
> You say that the exercise states that EVERY solution to the negative
> Pell's Equation gives a solution to Pell's Equation?
>
> Then why bother solving for one if you have the other?
>
> Isn't that stupid?
>
Hmm, let's see. Solution to negative Pell equation
gives a solution to positive Pell equation. Yes, that
works.
But what about the reverse? Does a solution to the
positive Pell equation give a solution to the negative
Pell equation?
No. The negative Pell equation, for some values of D,
does not have a solution.
This is pretty low-level logic, isn't it? You seem to
think that
A implies B
is the same as
B implies A.
So for example, "All men are mortal" implies "All mortals
are men."
You might want to check on that.
> If what you say is true, then anyone in modern times who ever solved
> for Pell's Equation after having the solution to the negative Pell's
> Equation or solved for the negative Pell's Equation when they had a
> solution for Pell's Equation is an idiot, correct?
>
Would an idiot say that "A implies B" is equivalent to
"B implies A" ?
Marcus.
> James Harris
JSH
You are correct. The above requirements are true when x is the
minimal solution.
Notice then that using x = 2j^2 - 1, gives j = 9, and 9^2 - 5*4^2 = 1.
So with your example, Pell's Equation is its own alternate, which is
only possible if x is not the minimum.
Good catch!
James Harris
JSH
And I've noted that Fermat and Euler were probably aware of them as
well.
That's not the issue.
It's what you and people like you knew TODAY before I brought this
subject up.
> These are the
> results of Brahmagupta that everyone has been telling you about.
>
> Don
I've said repeatedly the results are fairly trivial. That's not an
issue of disagreement.
I've not claimed they are new either.
I'm simply noting that none of you knew them, that's all.
Your much vaunted educations are woefully incomplete.
You don't even know some of the basics about one of the most famous
equations in mathematics.
Having lost details known as you mention back in antiquity.
James Harris
JSH
But still that means you can just walk down. As now using the Pell's
Equation solution given from the previous one, you NOW get the
solution to the negative Pell's Equation from sqrt((9-1)/2) = 2. 2^2
- 5*1^2 = -1.
So the primary result remains: you only need one to have the other!!!
This mathematics is trivial. Euler, and Fermat would have known it.
It has been argued that Brahmagupta knew it.
If you didn't, if you've ever tried to solve one of the alternates to
Pell's Equation when you had a solution to Pell's Equation, you did
not know this result!
Because that's a move that requires basic ignorance of the
mathematics.
And remember my other issue is with posters attacking me for simply
noting what's true and talking about interesting mathematics!!!
That is not the behavior of people who value knowledge. It's the
behavior of people who hate it.
James Harris
Mark Murray
>> The equation x^2 -Dy^2 = -1 is not solvable for
>> every non-square D and you don't know in advance for
>> which D, x^2 -Dy^2 = -1 is solvable
>
> That's not true and this exchange is beyond tedious.
As you are now aware, counterexamples have been demonstrated.
> The negative Pell's Equation ALWAYS has a solution if for every prime
> factor p of D, p = 1 mod 4.
Counterexamples to this too.
> And your position is that you can evaluate for the world, as my
> readership is worldwide--my math blog has hits from 100 countries/
> territories as Google Analytics puts it, already this year--whether or
> not a result is known to the mainstream world or not and you don't
> know the simplest things about the negative Pell's Equation?
>
> You are a hoodlum. It's that simple.
>
> A hooligan on Usenet. Nothing more.
... and given the above, you now owe an apology.
Your own incompetence, and inability to check your naive assertions has
yet again dropped you in the crap. Can you see the pattern yet?
M
JSH
> JSH wrote:
> >> The equation x^2 -Dy^2 = -1 is not solvable for
> >> every non-square D and you don't know in advance for
> >> which D, x^2 -Dy^2 = -1 is solvable
>
> > That's not true and this exchange is beyond tedious.
>
> As you are now aware, counterexamples have been demonstrated.
>
> > The negative Pell's Equation ALWAYS has a solution if for every prime
> > factor p of D, p = 1 mod 4.
>
> Counterexamples to this too.
Give one then.
___JSH
Mark Murray
> This mathematics is trivial. Euler, and Fermat would have known it.
> It has been argued that Brahmagupta knew it.
Correct.
So why do you keep on yammering about it like it was some earth-shattering
discovery?
M
Mark Murray
D = 221
221 = 13*17
13 = 1 mod 4
17 = 1 mod 4
M
juandiego
But of course you are not noting what is true.
According to you x^2 -221y^2 = -1 must have a solution
but doesn't.
You may be talking about interesting mathematics but unforntuantely
what you are saying is vacuous nonsense.
It is obvious you are regurgiating results from the internet which
you have picked up without understanding them .
JSH
Sigh. I haven't. In fact I've repeatedly said it was probably known
to Fermat and Euler. I accept that it was probably known to
Brahmagupta.
Notice, nothing has changed. The result still remains the same: given
a solution to the negative Pell's Equation j^2 - Dk^2 = -1, you MUST
have a solution to Pell's Equation x^2 - Dy^2 = 1, from x = 2j^2 + 1.
It is a trivial result about Pell's Equation.
Further, knowing Pell's Equation, if the negative Pell's Equation has
a solution you can ALWAYS get that solution using a Pell's Equation
solution.
One reader noted that some solutions may not directly give it, but I
noted that then you can walk down to the solution.
It's all trivial, easy algebra.
James Harris
Mark Murray
>> So why do you keep on yammering about it like it was some earth-shattering
>> discovery?
>>
>> M
>
> Sigh. I haven't. In fact I've repeatedly said it was probably known
> to Fermat and Euler. I accept that it was probably known to
> Brahmagupta.
Yet you persist in bringing it up and writing articles about it.
> Notice, nothing has changed. The result still remains the same: given
> a solution to the negative Pell's Equation j^2 - Dk^2 = -1, you MUST
> have a solution to Pell's Equation x^2 - Dy^2 = 1, from x = 2j^2 + 1.
Repeated /ad nauseam/.
> It is a trivial result about Pell's Equation.
Yes. Trivial. Please explain why you constantly need to repeat it, then.
> Further, knowing Pell's Equation, if the negative Pell's Equation has
> a solution you can ALWAYS get that solution using a Pell's Equation
> solution.
Yes. Trivial.
> One reader noted that some solutions may not directly give it, but I
> noted that then you can walk down to the solution.
>
> It's all trivial, easy algebra.
Yes. No earth-shattering discoveries here.
M
Achava Nakhash, the Loving Snake
JSH does not seem to be claiming that every solution to Pell's
equation arises from a solution to the negative Pell equation. In
fact, he appears to be surprised by that fact in this proof. His
repeated claim is that the smallest solution to Pell's equation arises
from the smallest solution to the negative Pell equation. Both
statements are true.
Rotwang, your attempted proof of the fact that every solution of the
Pell equation for a given D comes from the solutions to the negative
Pell equation for the same D, when there are such solutions, leaves me
puzzled. You successfully prove that x can't be even. So what?
Also, your claim that D must be odd is odd itself in view of the fact
that (1^2) - 2*(1^2) = -1 and 3^2 - 2*(2^2) = 1.
The proof, at least the one that I see, comes from knowning the
structure of the solutions to the Pell equation. In all cases where
there are solutions to the negative Pell equation, the solutions to
the pair of equations, x^2 - Dy^2 = +/-4 arise as follows: If (x, y)
is a solution, consider the number
alpha = (x/2) + (y/2)*sqrt(D). We can assume without loss of
generality that x and y are positive, since we can always switch the
signs of x and y and we really don't care about the solution where y =
0. Choose the solution (x_0, y_0) that givens the smallest value,
alpha_0 of the above expression. It is called the fundamental unit of
the quadratic field Q(sqrt(D)). Then all solutions (a, b) are given
by (a/2) + (b/2)*sqrt(D) = (x_0 + y_0*sqrt(D))^n for some positive
integer n. The fundamental solution is always a solution of the
negative Pell-4 equation when such exists. Then the solutions of the
negative Pell-4 equation are given by the odd powers of the
fundamental unit and the solutions of the regular old Pell-4 equation
are given by the even powers of the fundamental unit. Now how does
this translate to solutions of the actual Pell and negative Pell
equations? These will just be the powers of alpha_0 for which a and
b both come out even. In that case, the solutions of the Pell /
negative Pell equation will just be given by a/2 and b/2. Now if we
suppose that (a, b) is a solution of the Pell equation, then (a/2) +
(b/2)*sqrt(D) must be an even power, say 2r of alpha_0. This is
because norms are multiplicative, and the fundamental unit has norm -1
when there are units with norm -1, which is to say, solutions of the
negative Pell equation. Then
(alph_0)^r is easily seen to be a solution of the negative Pell
equation, since if it had positive norm, by the minimality of our
solutionn to the Pell equation, it would have to be of the form
(u/2) + (v/2)*sqrt(D), and the square of this quantity does not
produce even rational and imaginary parts (since 2uv is not divisible
by 4). Hence our smallest Pell solution must come from a negative
Pell solution.
If D is 2 or 3 mod 4 we do not have the annoyance of having to
consider expressions of the form (x/2) + (y/2)*sqrt(D). We can stick
with x + y*sqrt(D) and so the units of the quadratic field are
identical to the solutions of the Pell equation. A similar analysis
handles the case when D is 2 mod 4. Of course D can't be 3 mod 4 if
there is a solution of the negative Pell equation since then it would
be divisible by a prime congruent to 3 mod 4 of which -1 would have to
be and can't be a quadratic residue. However the analysis of the
structure of the solutions to the Pell equation is essentially what I
have above in the more complicated case that occurs when D is 1 mod 4.
Regards,
Achava
Don Redmond
I probably shouldn't bother to answer you, but you're so charming that
I can't help myself.
I knew of these results at least back to the 80s or 90s when I first
taught out of Katz's first edition. As you sasy the negative Pell
result is trivial so I rediscovered it myself in high school in the
60s.
I'm sure I found out about the other results as well then since I read
both Dickson's History of the Theory of Numbers (which you might find
interesting) as well as the Carmichael book referenced elsewhere. Is
that
far enough before today?
Don
chum ley
"JSH" <jst...@gmail.com> wrote in message
news:d2a2a159-f5f7-45ef...@x31g2000prc.googlegroups.com...
>Give one then.
>___JSH
Ha, ha, ha, you silly fool, presenting old stuff by others as yours, silly
dorkus, monkey butt.
Achava Nakhash, the Loving Snake
> JSH wrote:
<snip>
> > Further, knowing Pell's Equation, if the negative Pell's Equation has
> > a solution you can ALWAYS get that solution using a Pell's Equation
> > solution.
>
> Yes. Trivial.
>
It is not clear to me that it is trivial that you get all of the
solutions to the Pell equation from the solutions to the negative Pell
equation for a fixed D. I have proved it, at least I think I have
proved it in a previous post in this thread, but, while easy, they
required some knowledge not evidently assimilated by JSH. Notice that
the implication is that if x^2 - Dy^2 = 1, and if there exists a
solution to a^2 - bD^2 = -1, then necessarily (x-1)/2 is a square. I
don't see how this comes from trivial algebra. Am I wrong? JSH has
now clearly made that assertion, and everyone is saying it is
trivial. What am I not seeing? Please show me the trivial proof.
<snip>
Regards,
Achava
Rotwang
Indeed he isn't. But he was tacitly assuming that this is so when he
gave the two statements at the start of this thread, and my
counterexample was intended to show that his conclusion was false.
> [...]
>
> Rotwang, your attempted proof of the fact that every solution of the
> Pell equation for a given D comes from the solutions to the negative
> Pell equation for the same D, when there are such solutions, leaves me
> puzzled.
No, you misunderstand - that's not what I was attempting to prove at
all. I proved by contradiction the first of James's claims in this
thread, namely that if x^2 - D*y^2 = 1 and there exists a solution to
the negative Pell's equation j^2 - D*k^2 = -1 then x must be odd. I
showed that this is true, despite the fact that James's probable
reason for believing it (i.e. the same assumption mentioned above) is
false.
> You successfully prove that x can't be even. So what?
"So what", indeed. But that's exactly what I was trying to prove.
> Also, your claim that D must be odd is odd itself in view of the fact
> that (1^2) - 2*(1^2) = -1 and 3^2 - 2*(2^2) = 1.
I meant that D must be odd /provided that x is even/.
Junoexpress
You know James, I was never totally sure if you were just very clever
and leading the math people on or just really stupid. But now, I know
the answer. Thanks for clearing that up,
M
JSH
> A telling discussion erupted when I challenged these newsgroups on a
> simple result about the negative Pell's Equation and its connection to
> Pell's Equation, where posters claimed the result was previously
> widely known.
>
> That result is that given j^2 - Dk^2 = -1, the negative Pell's
> Equation, x, for x^2 - Dy^2 = 1, is given by x = 2j^2 + 1.
>
> I've noted that is not stated in mainstream mathematical literature.
> Posters have argued with me--often insultingly--claiming it is.
>
> In response to a request for citations giving the result, they gave,
> if anything, citations to equations that could LEAD to that result, if
> you knew where to look!!! All math is derived from prior equations,
> except the basic axioms.
>
> your credibility in believing it is already part of mainstream
> literature.
>
> Given non-zero integer solutions to x^2 - Dy^2 = 1:
>
> 1. x must be odd, when a solution to the negative Pell's Equation
> exists.
>
> Equation exists.
It was noted further in this thread that that those requirements are
absolute for the minimum x solution to Pell's Equation.
James Harris
juandiego
> Achava- Hide quoted text -
>
> - Show quoted text -
juandiego
I don't quite see why you are intoducing half integers.
If you are dealing with integral bases etc. I can see where
the half-integers might come from but why can't you
simply define the fundamental solution of the Pell
equation to be least positive solution (u,v) of |x^2-Dy^2|=1 ?
In the negative and positive cases 0 < u+v*sqrt(D) <1.
Then it is easy to show by reductio ad absurdum that there are
no more units between consecutive powers of the fundamental unit.
Then every solution to the PPE is a power of the fundamental unit
The CF method always finds the fundamental unit.
Achava Nakhash, the Loving Snake
I am working this from the point of view of all the units in the
quadratic field of
Q(sqrt(D)). For these you need the half-integers. For example, if D
= 5, then the fundamental unit is (1 + sqrt(5)) / 2, and yet the
smallest solution of the Pell equation, if I remember correctly, is x
= 9, y = 4, which comes from the sixth power of the fundamental unit.
The cube of the fundamental unit thus generates the smallest solution
of the negative Pell equation, namely x = 2, y = 1. As I had never
worked through the relationship between the Pell equation and the
fundamental units in the case of D = 1 mod 4, I was given the excuse
and so I went for it. From previous posts, you seem knowledgeable, so
I pose the question of the index of the "solutions of the Pell
equation or negative Pell equation" in the unit group. Do you happen
to know, or is it actuallly trivial?
juandiego
Yes, I see what you are doing now.
You are looking at the integers of the form (a +bsqrt(D)/2
a==b==1 mod, D= 4m+1, for which |(a +bsqrt(D)/2| = -1
The equation of the negative unit is x^2 -ax -1 =0
so if s =(a +bsqrt(D)/2
s^2 -as -1 =0 and so s^2 == s+1 mod 2
s ==s mod 2
s^2 == s+1 mod 2
s^3 == s^2 +s == 2s+1 == 1 mod2
s^4 == s mod 2
s^5 == s^2 == s+1 mod 2
s^6 == 1 mod 2
For s^3 , |s^3| = -1 and s^3 = 2m( (a +bsqrt(D)/2|) + 2n+1 =
|(ma+2n+1) + mb*sqrt(D)| = -1
For s^6, |s^6| = 1 , |(pa+2q+1) + pb*sqrt(D)| = 1
So what you were saying seems to be correct.
|(3 + sqrt(13))/2| = -1
s^2 -3s -1 =0
s^2 = 3s+1
s^3 = 3s^2 +s = 9s+3 +s =10s+3
10s+3 = 5(3 +sqrt(13)) + 3 => |18 + 5sqrt(13)| = -1
s^4 = 10s^2 + 3s = 30s +10 +3s = 33s +10
s^5 = 33s^2 +10s = 99s+ 33 +10s = 109s +33
s^6 = 109s^2 + 33s = 327s +109 +33s = 360s +109
s^6 = 180((3 + sqrt(13)) +109 = 649 +180sqrt(13)
|649 +180sqrt(13)| = 1
Achava Nakhash, the Loving Snake
Looks good to me. So the index of the Pell solutions is going to be
either 2 or 6 and the negative Pell solutions 1 or 3, depending on
whether the fundamental unit requires the use of half-integers. Now
we have a great new research topic, eh, although I suspect this one
has already been done. Any takers?
Regards,
Achava
juandiego
If you want a another similar challenge.
given r^3 - D=0 is the equation of r a cubic irrational for what
values
of D is |e| =|(A +Br +Crr)/3| = 1 and for what n does e^n
give the fundamental solution of X^3 +DY^3 +DDZ^3 -3DXYZ = 1
It would be more interesting to know
the necessary and sufficient conditions for the existence of solutions
to
x^2 -Dy^2 = -1 when D==1 mod 4
Achava Nakhash, the Loving Snake
> On 18 May, 00:13, "Achava Nakhash, the Loving Snake"
>
> > equation arises from a solution to the negative Pell equation.
>
> Indeed he isn't. But he was tacitly assuming that this is so when he
> gave the two statements at the start of this thread, and my
> counterexample was intended to show that his conclusion was false.
He actually appears to make this claim later. I was really more
interested in working out a proof of this claim about all solutions of
the Pell equation coming from the negative Pell equation than in
criticizing you. A number of posters have either claimed or blindly
accepted that this is trivial. It is not difficult, but it is
probably difficult enough to be beyond James.
> > [...]
>
> > Rotwang, your attempted proof of the fact that every solution of the
> > Pell equation for a given D comes from the solutions to the negative
> > Pell equation for the same D, when there are such solutions, leaves me
> > puzzled.
>
> No, you misunderstand - that's not what I was attempting to prove at
> all. I proved by contradiction the first of James's claims in this
> thread, namely that if x^2 - D*y^2 = 1 and there exists a solution to
> the negative Pell's equation j^2 - D*k^2 = -1 then x must be odd. I
> showed that this is true, despite the fact that James's probable
> reason for believing it (i.e. the same assumption mentioned above) is
> false.
Apparently I misunderstood you. Sorry about that.
> > You successfully prove that x can't be even. So what?
>
> "So what", indeed. But that's exactly what I was trying to prove.
>
> > Also, your claim that D must be odd is odd itself in view of the fact
> > that (1^2) - 2*(1^2) = -1 and 3^2 - 2*(2^2) = 1.
>
> I meant that D must be odd /provided that x is even/>
While working on the first part, I almost talked myself into believing
the statement that I just debunked.
Regards,
Achava
Achava Nakhash, the Loving Snake
> On May 17, 7:49 am, juandiego <sttscitr...@tesco.net> wrote:
>
> > On 17 May, 15:00,JSH<jst...@gmail.com> wrote:
>
<snip>
> That's remarkable.
>
> You say that the exercise states that EVERY solution to the negative
> Pell's Equation gives a solution to Pell's Equation?
>
> Then why bother solving for one if you have the other?
>
> Isn't that stupid?
>
> If what you say is true, then anyone in modern times who ever solved
> for Pell's Equation after having the solution to the negative Pell's
> Equation or solved for the negative Pell's Equation when they had a
> solution for Pell's Equation is an idiot, correct?
>
> James Harris
James, that is not correct. The way you expess yourself here,
particularly the question that you ask above, "Then why bother
solving for one if you have the other?" as well as numerous other
comments sprinkled throughout your postings on this topic and others,
betrays a mindset very different from that of all of the
mathematicians that I have discussed such things with. I was kinda/
sorta a professional, "mainstream" mathematician for a little while
and certainly went to a graduate school that was loaded with such
people.
Mathematicians are not as a rule about solving single equations in the
way you appear to think of it, which as I see it, is simply to
generate some numbers that work when you plug them in. I find it
quite difficult to express the feel I get of what you are trying to
do. An example is this: if you are building some Rube Goldberg
device that shoots a projectile upwards at a given velocity and at
given initial angle, then in order to figure out how to work it so
that it falls onto a lever that activiates the next device in the
chain, you have to solve for v_0 and theta so that the proctile goes
exactly the right distancee and lands on you lever. Any solution will
do, but you might be concerned about some constraints that cause you
to look for
solutions with the angle in a specified range or the velocity not
above a certain amount or whatever. This kind of problem arises in
engineering quite routinely, so I called it the engineering approach
or some such in a previous thread.
Naturally mathematicians have to generate solutions to do much of
anything, if it is an equation we are dealing with. We are by no
means always dealing with equations. However, that being said, we are
more commonly interested in finding the patterns of the solutions,
whether they exist at all, describing the structure of the solution
set, and so on and so forth. Since good and quite remarkable patterns
have been found for the solutions of the Pell and negative Pell
equations, since the Pell equations have been known to always have
solutions at least from the time of Euler, and no doubt other reasons,
we aren't that interested in simply cranking out solutions to any
given Pell or negative Pell equation unless we are making a table of
solutions to study something about them that interests us. That is
why most of what you have written about Pell equations is of no
interest to mathematicians. I also believe that the difference in
mindset is why you have such a problem understainding our reactions
other than those that are sheer rudeness, for which I really don't see
a place here, and also why mathematicians find your writing so
peculiar - also among other reasons.
On the subject of whether or not your result about the solution of
Pell equations coming from the solution of the negative Pell equation
when there are solutions for that D of the negative Pell equation, and
if (x, y) is a solution of the negative Pell equation, then (2x^2 +
1, ???) is a solution of the Pell equation. Too many words to make a
good sense back there, but you wanted to know whether these were
generally known ot the modern mathematical community. You never
specify the y value here. The full identity is that (2x^2 + 1)^2 -
((2xy)^2)D = 1. In other words, the generated solution is (2x^2 + 1,
2xy) . To any one (meaning modern mathematician) who knows anything
about the Pell equation, the fact that a solution of the negative Pell
equation generates a solution (actually an infinite number of
solutions) is so blindingly obvious that it might or might not
actually be written down. Who doesn't know that (-1)^2 = 1? Also the
method for calculating the generated solution would be known. The
standard method yields a new x value of x^2 + Dy^2, but since (x, y)
is a solution of the negative Pell equation you can substitute out
Dy^2 = x^2 + 1 to obtain your result. It is interesting that only the
x value of the original solution is needed to get the x value of the
generated solution. This is probably in the literature somewhere, a
textbook or a problem set in a textbook or an offhand mention
somewhere. Nobody would dream of presenting it as other than a
throwaway line in a research paper, as it is completely trivial and of
no significance whatsoever. I expect all number theorists know all of
this, as it is part of our field and the Pell and negativie Pell
equations are the starting point of very interesting stuff about
algebraic number fields. I would be very surprised if all that many
mathematicians not involved with number theory would know much of
anything about the Pell equations. They can all correct me if I am
wrong.
Regards,
Achava
pumpledu...@gmail.com
> On 19 May, 01:02, "AchavaNakhash, the Loving Snake"> > > On 18 May, 17:50, "AchavaNakhash, the Loving Snake"
>
> > > <ach...@hotmail.com> wrote:
> > > > On May 18, 1:43 am, juandiego <sttscitr...@tesco.net> wrote:
>
Well, a couple of necessary and sufficient conditions for the
fundamental
unit in Q(sqrt(d)) to have norm -1 are
(a) the simple continued fraction to sqrt(d) has odd period length
and
(b) the ordinary and narrow class groups associated to the
quadratic field are isomorphic.
Both of these are nice enough, but not, in some sense, what one
would like to have (for computations, the latter is arguably a
better characterization than the former).
de P
juandiego
Yes, I mentioned that somewhere else as a way of telling whether
the convergents at the end of a period give 1 or -1.
> (b) the ordinary and narrow class groups associated to the
> quadratic field are isomorphic.
How would that work for D = 205 or 221 ?
It doesn't seem to have anything to do with the class number
whch is 2 for D=85 for example.
> Both of these are nice enough, but not, in some sense, what one
> would like to have (for computations, the latter is arguably a
> better characterization than the former).
Yes, the criterion is basically doing the compuation .
JSH
Yeah, you're right. Thanks for catching that error.
The correct existence conditions are, with odd D:
1. For each prime factor p of D, p = 1 mod 4.
2. (D-1)/4 must be a quadratic residue modulo D.
For instance with D=17, (D-1)/4 = 4, so it has solutions.
With D=13, (D-1)/4 = 3, and 3 is a quadratic residue modulo 13, so it
has solutions.
But, D = 221 = 13*17, (D-1)/4 = 55, so 55 must not be a quadratic
residue modulo 221.
Good catch. I'm still pondering the even D case.
James Harris
JSH
<ach...@hotmail.com> wrote:
> On May 17, 2:53 pm, Mark Murray <w.h.o...@example.com> wrote:
>
> > JSH wrote:
>
> <snip>
>
> > > Further, knowing Pell's Equation, if the negative Pell's Equation has
> > > a solution you can ALWAYS get that solution using a Pell's Equation
> > > solution.
>
> > Yes. Trivial.
>
> It is not clear to me that it is trivial that you get all of the
> solutions to the Pell equation from the solutions to the negative Pell
> equation for a fixed D. I have proved it, at least I think I have
If the negative Pell's Equation j^2 - kD^2 = -1, has non-zero integer
solutions, then Pell's Equation
x^2 - Dy^2 = 1
has the solution x = 2j^2 + 1.
It's a trivial result.
> proved it in a previous post in this thread, but, while easy, they
> required some knowledge not evidently assimilated by JSH. Notice that
> the implication is that if x^2 - Dy^2 = 1, and if there exists a
> solution to a^2 - bD^2 = -1, then necessarily (x-1)/2 is a square. I
> don't see how this comes from trivial algebra. Am I wrong? JSH has
(x-1)(x+1) = Dy^2
to help you assume D is prime, and note that x-1 or x+1 must have D as
a factor, right?
But then the OTHER must be a square. Note that x must be odd, and
you're done.
Interestingly, you can then prove that D must divide x+1. when x is a
positive integer, the negative Pell's Equation exists, and x is the
minimum solution to Pell's Equation.
> now clearly made that assertion, and everyone is saying it is
> trivial. What am I not seeing? Please show me the trivial proof.
The result is easily proven with basic algebra as I noted above, and
that's NOT the point.
The point is that it's not given in mainstream literature!
So people may naively try to solve the negative Pell's Equation when
they have Pell's Equation, or vice versa.
But you need only solve one, if the negative Pell's Equation exists.
Work through the other cases and you'll have the other alternates.
Then expand from D prime to D composite and figure out the final
alternate equation.
It's fun algebra. So easy it doesn't take much mental effort so it is
a way to pass the time without exerting yourself.
Kind of like chewing gum.
James Harris
I
pumpledu...@gmail.com
I didn't say it had anything (directly) to do with
the class number. I said it had to do with the relation
between the ordinary and narrow class groups. Feel free to
compute these for your favourite D to check....
de P
Martin Musatov
that, for all n ≥ 1, .... np(p + q) n-1. = np. T
This is why. --Martin Musatov (earth shatters here)
Mark Murray
> Yeah, you're right. Thanks for catching that error.
You're welcome, but it wasn't me that caught it; it was reported originally by someone else, to whom I have not given the credit, and to whom I apologise for not doing so.
> The correct existence conditions are, with odd D:
>
> 1. For each prime factor p of D, p = 1 mod 4.
>
> 2. (D-1)/4 must be a quadratic residue modulo D.
>
> For instance with D=17, (D-1)/4 = 4, so it has solutions.
> With D=13, (D-1)/4 = 3, and 3 is a quadratic residue modulo 13, so it
> has solutions.
>
> But, D = 221 = 13*17, (D-1)/4 = 55, so 55 must not be a quadratic
> residue modulo 221.
To what extent have you checked this?
M
Rotwang
> JSH wrote:
> > Yeah, you're right. Thanks for catching that error.
>
> You're welcome, but it wasn't me that caught it; it was reported
> originally by someone else, to whom I have not given the credit,
> and to whom I apologise for not doing so.
It was juandiego I believe.
> > The correct existence conditions are, with odd D:
>
> > 1. For each prime factor p of D, p = 1 mod 4.
>
> > 2. (D-1)/4 must be a quadratic residue modulo D.
>
> > For instance with D=17, (D-1)/4 = 4, so it has solutions.
> > With D=13, (D-1)/4 = 3, and 3 is a quadratic residue modulo 13, so it
> > has solutions.
>
> > But, D = 221 = 13*17, (D-1)/4 = 55, so 55 must not be a quadratic
> > residue modulo 221.
>
> To what extent have you checked this?
Like you have to ask ;)
In fact, 55 = 87^2 mod 221.
juandiego
Yes, the Moron of Morons has made a great strategic
mistake in choosing something as concrete as the Pell
equation for incessant exrectolocution. He is in his
element when dealing with proofs by contradiction
where he can prevaricate and obfuscate till the cows come
home. With Pell, he only produces trivia or is just
wrong all the time. Will he astound us all and clearly
formulate the necessary and sufficient conditions for the
solvability of x^2-Dy^2 = -1, D = 1 mod 4 before hell freezes over ?
With proof of course.
rossum
wrote:
1829572648 + 29857264396 = 31686837044 is not given in the mainstream
literature either. That is another trivial result derived from
already known principles, hence we do not expect to see it copied
everywhere. Your Pell result is the same, it is not significant
enough to get a wide coverage.
>
>So people may naively try to solve the negative Pell's Equation when
>they have Pell's Equation, or vice versa.
>
>But you need only solve one, if the negative Pell's Equation exists.
So how can we tell *in advance* whether to try for the negative Pell
equation rather then the positive Pell equation? How do we avoid
going on a wild goose chase looking for a solution of the negative
Pell that does not exist?
rossum
Achava Nakhash, the Loving Snake
> On May 17, 4:23 pm, "Achava Nakhash, the Loving Snake"
>
Trivial. - This is from James, I think.
>
> > It is not clear to me that it is trivial that you get all of the
> > solutions to the Pell equation from the solutions to the negative Pell
> > equation for a fixed D. I have proved it, at least I think I have
>
Thank you for responding.
> If the negative Pell's Equation j^2 - kD^2 = -1, has non-zero integer
> solutions, then Pell's Equation
>
> x^2 - Dy^2 = 1
>
> has the solution x = 2j^2 + 1.
>
> It's a trivial result.
>
Yes, it is trivial, but this comment is quite irrelevant to my
question which was about something like the converse of this fact,
namely that every solution of the Pell equation comes from a solution
to the negative Pell equation when there are such solutions.
> > proved it in a previous post in this thread, but, while easy, they
> > required some knowledge not evidently assimilated by JSH. Notice that
> > the implication is that if x^2 - Dy^2 = 1, and if there exists a
> > solution to a^2 - bD^2 = -1, then necessarily (x-1)/2 is a square. I
> > don't see how this comes from trivial algebra. Am I wrong? JSH has
>
> (x-1)(x+1) = Dy^2
>
> to help you assume D is prime, and note that x-1 or x+1 must have D as
> a factor, right?
>
I don't need this sort of help, actually. Let us also note that
proving a theorem for a special case is not the same as proving the
theorem. Therefore it should not be just left at special cases which
are perhaps easier than the general theorem. I am writing as I read,
so perhaps you address the case later. We shall see. And yes, if D
is a prime, it ust divide eithe x - 1 or x + 1.
> But then the OTHER must be a square. Note that x must be odd, and
> you're done.
Only if (x-1) and (x+1) are relatively prime. If x is odd, they have
a common factor of 2, so this conclusion does not follow. Why must x
be odd?
OK, if D is even, then obviously x is odd. Otherwise, if x is even
then
-Dy^2 = 1 mod 4, but D = 1 mod 4 so -1 is a square mod 4 which is
impossible. All right x is odd, but you still have a problem.
What am I done with? You have tried unsuccessfully to prove that x is
a square plus or minus 1. How do you get a solution of the negative
Pell equation from this? That, after all, is what you are supposedly
trying to prove, namely that all solutions of the Pell equation come
as 2j^2 + 1 where j is an x value of the solution of some negative
Pell equation. You don't even have that x is of the form twice a
square plus one at this point!
> Interestingly, you can then prove that D must divide x+1. when x is a
> positive integer, the negative Pell's Equation exists, and x is the
> minimum solution to Pell's Equation.
I am sure that this is interesting, but if you can prove it then
please do so. Bare assertions at this level are really not at all
interesting. I wanted to know why the result I talked about is
trivial and I haven't got anything close to a reply.
> > now clearly made that assertion, and everyone is saying it is
> > trivial. What am I not seeing? Please show me the trivial proof.
> The result is easily proven with basic algebra as I noted above, and
> that's NOT the point.
I would like to see this trivial proof. I haven't seen it yet. I
believe the result, because I can prove it, and I did in another of
your threads, eaning the result I asked about, but the result is not
trivial unless you know something far from trivial about the structure
of the solutions to the Pell equation.
> The point is that it's not given in mainstream literature!
It's maybe true and maybe not true that your result about 2j^2 + 1 is
not in the mainstream literature. It probably is, since there is a
lot of mainstream literature that I haven't seen relating to attracive
results that are on a level that doesn't recquire an advanced degree
to understand. However, even if you are right that it is not given in
the mainstream literature, the reason is that it is trivial. Nobody
makes a big point of presenting trivial results. I take that back.
Almost nobody makes a point of presenting trivial results. This is
surprising?
> So people may naively try to solve the negative Pell's Equation when
> they have Pell's Equation, or vice versa.
>
> But you need only solve one, if the negative Pell's Equation exists.
>
> Work through the other cases and you'll have the other alternates.
> Then expand from D prime to D composite and figure out the final
> alternate equation.
Read my post in this thread about your "engineering mindset."
Mainstream mathematicians are not out there hunting to solutions for
the Pell equation or negative Pell equation and waiting with bated
breath for the latest algorithm breakthrough that gives you a
particular solution for a particular value of D. That is just not
what we generally do in these circumstances.
> It's fun algebra. So easy it doesn't take much mental effort so it is
> a way to pass the time without exerting yourself.
>
> Kind of like chewing gum.
I just had tooth out this afternoon and I am still in a lot of pain in
spite of pain meds. Chewing gum sounds like a hideous prospect to
me. I am glad you enjoy the algebra. I always hope that you will
learn enough about elementary number theory and the spirit of real
mathematics that you will find it even more fun when you understand
why you have such a minimal concept of what a proof is, and you have
some real fun when you can actually understand when your work is valid
and when it isn't. It is indeed a wonderful joy to get all the parts
to work together and really prove something that interests you, even
if it interests no one else.
> James Harris
>
Regards,
Achava
JSH
But it is.
Seems I'm at a loss on what the rules are for when the negative Pell's
Equation must have non-zero integer solutions!
That throws a lot of things up in the air, like my claims of having
generally solved binary quadratic Diophantine equations.
Oh well.
James Harris
JSH
Ok, got 'em.
1. For every prime factor p of D, p = 1 mod 4.
2. If D-1 has 8 as a factor it must have 16 as a factor as well.
3. D cannot equal 1 mod 5.
Those are the first set of conditions required for the existence of
the negative Pell's Equation.
There may be more, but probably not.
James Harris
Mark Murray
> 2. If D-1 has 8 as a factor it must have 16 as a factor as well.
What does this mean? That
a) D-1 = 8 times 16 times SomethingElse or
b) D-1 = 8 times SomeThing AND D-1 = 16 times SomethingElse
?
If a) this means that D-1 is a multiple of 128
If b) this means that D-1 is a multiple of 16
M
JSH
Crap.
> Those are the first set of conditions required for the existence of
> the negative Pell's Equation.
Wrong.
James Harris
juandiego
It's not clear whether you mean
A 1) and 2) and 3) imply the existence of the NPE
B 1) or 2) or 3) imply the existence of the NPE
C) the existence of the NPE implies 1) and 2) and 3)
D) the existence of the NPE implies 1) or 2) or 3)
You have already seen that x^2 -221y^2 =-1 has no solutions
and also x^2 -26y^=-1 has the solution (5,1), 2 divides 26 but
2 <> 1 mod 4
I'm not sure what 2) means
In any case x^2 -9425y^2 = -1 has no solutions
and D-1 = 9425 -1 = 9424 = 16*589
D = 1 mod 5 implies that x^2-Dy^2 =-1 has no solutions
is false as x^2 -26y^=-1 and 26 = 1 mod 5
It seems that a necessary condition for the existence of
a solution to the NPE is that D = a^2 +b^2, GCD (a,b) =1
D not a square.
This is not sufficient as D=221 = 5^2 +14^2 shows.
The period length of the CF of sqrt(221) is 6 and so is even.
But if D is not equal to the sum of two squares
you can conclude that x^2-Dy^2 =-1 has no solutions.
But a genius of your magnitude should be able to find
necessary and sufficient conditions.
Mark Murray
> Seems I'm at a loss on what the rules are for when the negative Pell's
> Equation must have non-zero integer solutions!
>
> That throws a lot of things up in the air, like my claims of having
> generally solved binary quadratic Diophantine equations.
>
> Oh well.
Not surprising at all.
Other things also not surprising:
* Silent removal of grandiose claims on your "My Math" blog.
* Total lack of apology to those you insulted when they pointed
out your errors.
But your ability to accept faults is improving. This is a good thing.
Time to revisit your paper and your book, and you are done.
M
--
Martin Musatov
Mark Murray wrote:
> JSH wrote:
> > On May 17, 1:50 pm, Mark Murray <w.h.o...@example.com> wrote:
> >> JSH wrote:
> >>>> The equation x^2 -Dy^2 = -1 is not solvable for
> >>>> every non-square D and you don't know in advance for
> >>>> which D, x^2 -Dy^2 = -1 is solvable
> >>> That's not true and this exchange is beyond tedious.
> >> As you are now aware, counterexamples have been demonstrated.
> >>
> >>> The negative Pell's Equation ALWAYS has a solution if for every prime
> >>> factor p of D, p = 1 mod 4.
> >> Counterexamples to this too.
> >
> > Give one then.
>
> D = 221
>
> 221 = 13*17
>
> 13 = 1 mod 4
> 17 = 1 mod 4
>
> M
Lagrange Proof. Proof ξ. 1 n is a positive solution by Prop. 1.
So .... NP∩co-NP. Regulator. Bigger picture. * Hallgren's can be
expanded to solve other ... 2s 2 Z b/c p > 2 in denominator of s ) p.
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M
QED
Martin Musatov
Junoexpress
> > generally solved binary quadratic Diophantine equations.
> > Oh well.
> Not surprising at all.
>
> Other things also not surprising:
>
> * Silent removal of grandiose claims on your "My Math" blog.
>
> * Total lack of apology to those you insulted when they pointed
> out your errors.
>
> But your ability to accept faults is improving. This is a good thing.
> Time to revisit your paper and your book, and you are done.
>
> M
> --
Oh, naive one! While you are quite correct that this result is
surprising to nobody except JSH ;>), you are quite in error when you
assume that James has accepted his "faults".
In Jame's mind, he made a mistake. This is not a "fault" but rather
part of the discovery process called "brain-storming" and is to be
accepted. And in fact, it is just these mistakes that will lead him to
his next brilliant observation, as you will see when he begins his
next post.
The Catch-22 of course is that reality is in your corner, and so like
Sisyphus, JSH will keep performing his brain-storming process, success
always just seemingly a tantalizing distance from his grasp, without
ever being able to achieve it.
No, you can't change JSH, but you can enjoy him!
M
Martin Musatov
Let the universe be the judge of it!
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--
Martin Musatov
Los Angeles, CA