--- Jon Perry <

perry@...> wrote:

> The proof I have is much simpler. sorry about the Hardy credit - it's in his

> book, and I thought I had read something somewhere which said it was

> Hardy's.

Your original wording offers you a get out of jail card - it probably was

proved by Hardy, whether any other number of mathematicians proved it before

that is irrelevant.

> This is from Number Theory by Naoki Sato, which is freely available as a

> PDF.

>

> There exists 'a' such that a^2=-1modp

There may exist 2.

> Consider the set of integer ax-y, x,y integers, o<=x<sqrt(p).

I assume you mean 0 not o.

> The number of possible pairs (x,y) is then

infinite, as there's no restriction on y. I assume you meant 0<=x,y<sqrt(p)

> [floor(sqrt(p))+1]^2>[sqrt(p)]^2=p

>

> So, by the pigeonhole principle, there exist 0<=x1,x2,y1,y2<sqrt(p) such

> that

>

> ax1-y1 = ax2 - y2 modp.

>

> Let x=x1-x2 and y=y1-y2. At least one of x and y is non-zero.

The pidgeon hole principle cannot immediately on its own guarantee that

both x and y are non-zero, but ax-y == 0 (mod p) implies that if one is

zero, the other is too, a contradiction. Therefore you conclude that both

are non-zero.

> Thus x^2 + y^2 is a multiple of p, and 0<x^2+y^2<sqrt(p)^2+sqrt(p)^2=2p,

>

> Hence x^2+y^2=p.

I like it. It's less constructive than the Fermat version, but nonetheless,

you can't doubt its verity. I quite like pigeonhole proofs, often they're

very elegant, this is no exception.

Phil

=====

The answer to life's mystery is simple and direct:

Sex and death. -- Ian 'Lemmy' Kilminster

__________________________________________________

Do you Yahoo!?

Yahoo! Mail Plus - Powerful. Affordable. Sign up now.

http://mailplus.yahoo.com