Re: [Fractint] Extended Fermat's Last Theorem - Slightly Off-topic
IIRC, a pair of Japanise (?) students proposed that all elliptic curves are also modular forms. Someone else pointed out that, assuming Fermat was wrong, his equation x^n + y^n = z^n would be an elliptic equation, and the corresponding modular form is decidedly weird. Yet ANOTHER mathematition later proved that this modular form is in fact too weird to actually exist. In other words, assuming that EVERY elliptic curve REALLY IS a modular form, Fermat's Last Theorum MUST be true. Andrew Wiles proved that elliptic curves ARE modular, and as a side effect proved Fermat right. Andrew. PS. I've read the book. And I'm kinda bored today...
From: "Morgan L. Owens" <packrat@nznet.gen.nz> Reply-To: fractint@mailman.xmission.com To: fractint@mailman.xmission.com, fractint@mailman.xmission.com Subject: Re: [Fractint] Extended Fermat's Last Theorem - Slightly Off-topic Date: Mon, 19 Aug 2002 13:43:43 +1200
At 01:09 19/08/2002, dirving@box.net.au wrote:
"Ricardo M. Forno" wrote:
Does someone in the list know if the following conjecture, that
includes as
a particular case Fermat's Last Theorem, has ever been posed, and proved false or true? x[1]^p + x[2]^p + ... x[n]^p = z^p, for x[i] > 0 and 1 < n < p, has no integer solutions. TIA.
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I would conjecture ... no.
Reasonable: a proof of this extended conjecture would in turn provide an alternatve proof of FLT as a special case.
So if there has been a proof, it has been found more recently than (and probably develops the ideas of) Wiles' tour de force.
Morgan L. Owens "Yes, Andrew; a corollary :-)"
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On Wed, 21 Aug 2002 13:51:10 +0000, Andrew Coppin wrote:
IIRC, a pair of Japanise (?) students proposed that all elliptic curves are also modular forms. Someone else pointed out that, assuming Fermat was wrong, his equation x^n + y^n = z^n would be an elliptic equation, and the corresponding modular form is decidedly weird. Yet ANOTHER mathematition later proved that this modular form is in fact too weird to actually exist. In other words, assuming that EVERY elliptic curve REALLY IS a modular form, Fermat's Last Theorum MUST be true. Andrew Wiles proved that elliptic curves ARE modular, and as a side effect proved Fermat right.
What does "modular form" actually mean (to the layman - I feel sometimes that I am in too deep with fractals!) John -- John Lewis, jlewis@clara.net on 08/21/2002
At 01:51 22/08/2002, Andrew Coppin wrote:
IIRC, a pair of Japanise (?) students proposed that all elliptic curves are also modular forms. Someone else pointed out that, assuming Fermat was wrong, his equation x^n + y^n = z^n would be an elliptic equation, and the corresponding modular form is decidedly weird. Yet ANOTHER mathematition later proved that this modular form is in fact too weird to actually exist. In other words, assuming that EVERY elliptic curve REALLY IS a modular form, Fermat's Last Theorum MUST be true. Andrew Wiles proved that elliptic curves ARE modular, and as a side effect proved Fermat right.
An elliptic curve is basically a set of solutions to the equation y^2=Ax^3+Bx^2+Cx+D - where x and y are elements in some field (originally real numbers under addition and multiplication, but why restrict ourselves?) A modular function is one of the form f( (az + b)/(cz + d) ) = f(z) for complex a,b,c,d (it derives from linear fractional transforms (also known as bilinear) of the complex plane - those are the affine transforms without translation - and hence draws a lot on group theory.) The two theorems run: If FLT is false, then y^2=x(x-z^n)(x+y^n) is semistable, but not modular. (Conjectured in 1982 by Gerhard Frey; proven in 1986 by Ken Ribet via Jean-Pierre Serre). and All semistable elliptic curves are modular. (Conjectured by Yutaka Taniyama in 1955, later generalised by Goro Shimura and Andre Weil by dropping the "semistable" bit; proven in 1994 by Andrew Wiles). Most of that material can be found on this page on the subject, which leads up to the proof and covers things like elliptic curves and modular forms along the way: http://www.mbay.net/~cgd/flt/fltmain.htm As has already been noted, the given extended version of FLT is false, since counterexamples have been found. So the question has become: for _which_ pairs 1 < n < p do solutions exist? Morgan L. Owens "All a mathematician needs a computer for is emacs, TeX, and mail."
At 13:29 22/08/02 +1200, you wrote:
At 01:51 22/08/2002, Andrew Coppin wrote: translation - and hence draws a lot on group theory.)
The two theorems run: If FLT is false, then y^2=x(x-z^n)(x+y^n) is semistable, but not modular. (Conjectured in 1982 by Gerhard Frey; proven in 1986 by Ken Ribet via Jean-Pierre Serre).
and
All semistable elliptic curves are modular. (Conjectured by Yutaka Taniyama in 1955, later generalised by Goro Shimura and Andre Weil by dropping the "semistable" bit; proven in 1994 by Andrew Wiles).
put these strange things into a nice little fractint formula.. so we will be happy.. 8-))) cheers, guy
At 07:27 23/08/2002, Guy Marson wrote:
put these strange things into a nice little fractint formula.. so we will be happy.. 8-)))
cheers,
guy
comment{ This finds a solution to the elliptic equation y^2 = x^3 + Ax + B (In truth, I solve f(x,y) == x^3 + Ax + B - y^2 = 0 g(x,y) == x - y = 0 i.e., the (three) points on the elliptic curve that intersect the main diagonal. ) via the Newton method for various A and B, which get their values from pixel; starting point is provided in p1. In other words, this is a dual to the normal way of using Newton's method: instead of having one equation to solve and starting from various points, we're using a single starting point and solving for a whole family of equations. Does anyone get the feeling that it's all semantics? You should - all I'm doing is swapping the labels "parameter" and "variable". I just felt like doing that. To keep people satisfied, I offer the more traditional relabeling afterwards. } newton_elliptics{ x=real(p1) y=imag(p1) A=real(pixel) B=imag(pixel) z=x+flip(y) : oz=z x=real(z) y=imag(z) d=2*y-3*sqr(x)-A f=(sqr(x)+A)*x+B-sqr(y) g=x-y nx=x+(-f-2*y*g)/d ny=y+(-f+(3*sqr(x)+A)*g)/d z=nx+flip(ny) |z-oz|>0.0001 } newton_elliptics_2{ x=real(pixel) y=imag(pixel) A=real(p1) B=imag(p1) z=x+flip(y) : oz=z x=real(z) y=imag(z) d=2*y-3*sqr(x)-A f=(sqr(x)+A)*x+B-sqr(y) g=x-y nx=x+(-f-2*y*g)/d ny=y+(-f+(3*sqr(x)+A)*g)/d z=nx+flip(ny) |z-oz|>0.0001 } comment { Why stick with x-y=0? Now we're solving for g(x,y) == tan(phi)*x+t-y=0, where phi and t are given by p2. } newton_elliptics_3{ x=real(p1) y=imag(p1) A=real(pixel) B=imag(pixel) phi=tan(real(p2)) t=imag(p2) z=x+flip(y) : oz=z x=real(z) y=imag(z) d=2*y*phi-(3*sqr(x)+A) f=(sqr(x)+A)*x+B-sqr(y) g=phi*x+t-y nx=x-f+2*y*g ny=y-phi*f+(3*sqr(x)+A)*g z=nx+flip(ny) |z-oz|>0.0001 } newton_elliptics_4{ x=real(pixel) y=imag(pixel) A=real(p1) B=imag(p1) phi=tan(real(p2)) t=imag(p2) z=x+flip(y) : oz=z x=real(z) y=imag(z) d=2*y*phi-(3*sqr(x)+A) f=(sqr(x)+A)*x+B-sqr(y) g=phi*x+t-y nx=x-f+2*y*g ny=y-phi*f+(3*sqr(x)+A)*g z=nx+flip(ny) |z-oz|>0.0001 } .... I think. My brain is running very sluggishly right now. I'm down to 20wpm and it took quarter of an hour to figure out the equation for a line in terms of rotation and offset. I wouldn't be completely surprised if all that were totally bogus. Morgan L. Owens "Quarter of an hour. That's bad...."
At 14:20 22/08/2002, Ricardo M. Forno wrote:
Morgan L. Owens "All a mathematician needs a computer for is emacs, TeX, and mail."
Well, the computer is also useful to find counterexamples, uh?
But all that's _needed_ is to show that counterexamples must exist. Morgan L. Owens "it's just more fun to construct one."
participants (5)
-
Andrew Coppin -
Guy Marson -
John Lewis -
Morgan L. Owens -
Ricardo M. Forno