Re: [PrimeNumbers] Re: This has to be flaky

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• ... [SNIP - somewhat canny maths] ... Indeed. Cornaccia-Smith (p=x^2+d.y^2) owes a lot to its ideas, methinks. Phil ===== The answer to life s mystery is
Message 1 of 14 , Jan 4, 2003
> Here I try to set the history straight.
>
> Theorem [Fermat] : Every prime p which is congruent
> to 1 modulo 4 can be expressed as a sum of two squares.
>
> Jon credited this to Hardy.
>
> Fermat proved it by a "method of descent", roughly like this:

[SNIP - somewhat canny maths]

> 3) Keep on trucking until you get to
>
> x^2+y^2=p
>
> [End sketch of Fermat's proof]
>
> When I saw this, in a little book by
> T.H. Jackson, more than a quarter of a century
> ago, it took my breath away;
> Fermat descent is a truly wonderful thing.

Indeed. Cornaccia-Smith (p=x^2+d.y^2) owes a lot to its ideas, methinks.

Phil

=====
The answer to life's mystery is simple and direct:
Sex and death. -- Ian 'Lemmy' Kilminster

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• Neither Bookfinder nor Amazon could find a sellable copy of Jackson: http://www.amazon.co.uk/exec/obidos/ASIN/0710079982 It was fine little book, since Walter
Message 2 of 14 , Jan 4, 2003
Neither Bookfinder nor Amazon could find a sellable copy of Jackson:

http://www.amazon.co.uk/exec/obidos/ASIN/0710079982

It was fine little book, since Walter Ledermann, the series editor,
demanded the highest standards of pedagogy, aimed at good high-school

The 1975 price was 1.5 GPB ~ \$2.

David
• 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. This
Message 3 of 14 , Jan 4, 2003
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.

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

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

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

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

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.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
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• ... 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
Message 4 of 14 , Jan 4, 2003
--- 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

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• Jackson taught Fermat s original proof by descent for an excellent pedagogic reason: to prepare the student for Lagrange s tour de force, by descent, in
Message 5 of 14 , Jan 4, 2003
Jackson taught Fermat's original proof by descent for an excellent
pedagogic reason: to prepare the student for Lagrange's
tour de force, by descent, in proving

Theorem [Lagrange]: Every natural number can be expressed
as a sum of four integer squares

whose proof proceeds along the same lines as the one I sketched.

Short proofs are not always the most instructive.
• From: http://turnbull.dcs.st-and.ac.uk/~history/Mathematicians/Fermat.html Fermat described his method of infinite descent and gave an example on how it could
Message 6 of 14 , Jan 5, 2003
From:

http://turnbull.dcs.st-and.ac.uk/~history/Mathematicians/Fermat.html

'Fermat described his method of infinite descent and gave an example on how
it could be used to prove that every prime of the form 4k + 1 could be
written as the sum of two squares. For suppose some number of the form 4k +
1 could not be written as the sum of two squares. Then there is a smaller
number of the form 4k + 1 which cannot be written as the sum of two squares.
Continuing the argument will lead to a contradiction. What Fermat failed to
explain in this letter is how the smaller number is constructed from the
larger. One assumes that Fermat did know how to make this step but again his
failure to disclose the method made mathematicians lose interest. It was not
until Euler took up these problems that the missing steps were filled in.'

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/