From: Bill Daly <bill.daly@tradition-ny.com>
Subject: Re: Sophie Germaine's proof
Date: Thu, 21 Dec 2000 22:49:25 GMT
Newsgroups: sci.math
Summary: [missing]

In article <91r8l4$nur$1@nnrp1.deja.com>,
  Will Self <wself@msubillings.edu> wrote:
> I am sure that I have neither the brains nor the time to understand
> Wiles proof of Fermat's Last Theorem.  I have read that Sophie Germaine
> proved it in the case that the exponent is what is now called a Sophie
> Germaine prime, a prime number p such that 2p+1 is also prime.  I think
> she did this around 1825.  I would be very interested in looking at her
> proof if someone could give me a reference.
>
> Thanks,
>
> Will Self
> direct email replies appreciated
>
> Sent via Deja.com
> http://www.deja.com/
>

The proof is pretty simple, so let me try to reconstruct it from memory.
The theorem can be stated as follows. Suppose that there is a solution
x^p + y^p + z^p = 0 with xyz <> 0 mod p and x,y,z pairwise coprime.
Suppose further that there is a prime q such that

 1. x^p + y^p + z^p = 0 mod q implies xyz = 0 mod q
 2. w^p = p mod q has no rational solution

Then we reason as follows. We can factor x^p + y^p + z^p = 0 in three
different ways to reach the conclusion that

 x+y = a^p
 x+z = b^p
 y+z = c^p
 (x^p+y^p)/(x+y) = A^p
 (x^p+z^p)/(x+z) = B^p
 (y^p+z^p)/(y+z) = C^p

This is justified because we can write -z^p = (x+y) * (x^p+y^p)/(x+y) =
(x+y) * (x^(p-1)+x^(p-2)*(-y)+x^(p-3)*(-y)^2+...+(-y)^(p-1)), and since
z is not divisible by p, neither is either factor on the right, thus the
factors on the right are coprime and each must be the p-th power of an
integer. (Note that the gcd of the two factors is either 1 or p.)

Now, since x^p + y^p + z^p = 0 mod q implies xyz = 0 mod q, at least one
of x,y,z is divisible by q. Suppose without loss of generality that x is
divisible by q. Then 2x = (x+y) + (x+z) - (y+z) = a^p + b^p - c^p = 0
mod q, thus also abc = 0 mod q and at least one of a,b,c must be
divisible by q. If x is divisible by q, then neither y nor z can be
divisible by q, by the coprimality of x,y,z, thus c = y+z = 0 mod q,
i.e. -z = y mod q. But we also have (y^p+z^p)/(y+z) = y^(p-1) +
y^(p-2)*(-z) + y^(p-3)*(-z)^2 + ... + (-z)^(p-1) = C^p, thus
substituting -z = y mod q, y^(p-1) + y^(p-2)*y + y^(p-3)*y^2 + ... +
y^(p-1) = p*y^(p-1) = C^p mod q. Furthermore, since x = 0 mod q, A^p =
(x^p+y^p)/(x+y) = y^(p-1) mod q, thus p*A^p = C^p mod q, i.e. p =
(C/A)^p mod q, where the division is justified because A is not
divisible by q. But this contradicts the premise that w^p = p mod q has
no rational solution.

We can thus conclude that if, for a given p, there exists a prime q with
the requisite properties, then there can be no solution of x^p + y^p +
z^p = 0 with xyz <> 0 mod p.

If q = 2p+1 is prime, then q satisfies the conditions of the theorem.
Note that in this case x^p = 0, 1 or -1 mod q, thus the only solutions
of x^p + y^p + z^p = 0 mod q are those with xyz = 0 mod q. Furthermore
w^p = p mod q implies p = 0, 1 or -1 mod q, which is never the case.

More generally, if q = 2kp+1 is prime, and furthermore

 1. q does not divide the resultant of x^2k - 1 with (x-1)^2k - 1.
 2. q does not divide (2k)^2k - 1.

then q satisfies the conditions. (Note that the resultant is 0 if k is
divisible by 3, thus we can only consider values of k not divisible by
3.) In particular, the theorem works whenever 4p+1 or 8p+1 or 16p+1 is
prime, and also whenever 10p+1 or 14p+1 is prime except for p = 3.


Regards,

Bill


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