My little home encyclopedia (Columbia Encyclopedia -- rather outdated edition, I think) has only 3 sentences on Sophie Germain: "French mathematician, 1776-1831. Although self-taught, she mastered mathematics and corresponded with J.L. Lagrance and C.F. Gauss. She is best known for her work on the vibrations of elastic plates."
This very inadequate entry wouldn't pass muster today. Feminism has changed the world too much for that! There are better biographies of Sophie Germain on the web, and of course encyclopedias of mathematics (science reference room, our library) never treated her so summarily.
She must have proved what is now called Germain's theorem when she was about 30. It has been described as the only significant progress on "Fermat's Last Theorem" in the hundred year period 1740-1840. This "theorem," which Fermat claimed in a marginal note to have proved, was really a conjecture, and proving it was a problem of widely acknowledged difficulty. Carl Friedrich Gauss, for example, the "prince of mathematicians," explicitly said he considered it too difficult to work on. Germain's other work, on the vibration of elastic plates, interestingly enough, had also been pronounced "too difficult," this time by another eminent mathematician, whom Germain knew personally, J.L. Lagrange. One gets a certain insight into Germain's character, and her ability, when one realizes that she attacked two problems, very different from each other, which had been described as exceptionally difficult by the best minds of the time, and made significant progress on both of them!
The Fermat conjecture (now Wiles' Theorem, but this is very recent!) says
has no nontrivial solution in integers if n is an integer greater than or equal to 3. (A trivial solution would be, for example, x= -y, z=0, for n odd. The conjecture is that no other, what one might call interesting solutions exist.) Euler managed to prove the conjecture in case n=3. Fermat himself proved it with an ingenious method in case n=4, and perhaps believed this method would generalize (it doesn't, though). There the conjecture lay at the time of Germain.
It is enough to prove
has no nontrivial solutions in integers for p a prime greater than or equal to 5, since a solution for a composite n would imply a solution for any of n's prime factors. One can also require that x, y, and z have no common factors, since you could always divide through by such factors. And you may have noticed that I moved the z term to the left hand side, which amounts to subtracting z^p from both sides, and then renaming z by -z, which is of course just another integer.
Germain's theorem and proof depend on a clever definition:
Definition: an odd prime p is a Germain prime if q=2p+1 is also prime. [Note that this implies p=(q-1)/2.]
Germain's Theorem: If p is a Germain prime and
,for (nontrivial) integers x, y, and z, then xyz=0 (mod p). Another way to state the conclusion is that one of x, y, or z must be a multiple of p.
Her proof depends on a simple observation, which she uses repeatedly, namely that if p is a Germain prime and q=2p+1, then for any integer x
This depends on q being prime, which it is, since p is Germain. Here is a reminder why this is so: x may be a multiple of q, in which case it is 0 (mod q) and any power of it is also 0. Otherwise it is relatively prime to q, and its (q-1)th power is 1 (mod q) by Fermat's Little Theorem. The (q-1)/2 power is therefore a square root of 1 (mod q) which can only be 1 or -1 (mod q).
The first step of the proof actually proves an easier version of the same theorem, which we might call Germain's Easy Theorem: If p is a Germain prime and
,for (nontrivial) integers x, y, and z, then xyz=0 (mod q). Another way to state the conclusion is that one of x, y, or z must be a multiple of q. (The only difference between the hard theorem and the easy one is the replacement of p by q in the conclusion.)
Proof of Germain's Easy Theorem: Reduce the equation (mod q). Then each of those pth powers is either 1, 0, or -1 (mod q), and they add to zero. The only two ways they can add to zero are (1) they are all zero, or (2) each of 1, 0, and -1 occurs exactly once. The first possibility is not of interest, since if all of them are 0 (mod q), i.e., multiples of q, we could divide out the common factor q until at least one term is different from 0 (mod q). Then we must be in the other case: exactly one of the three terms is 0 (mod q), say x^p=0 (mod q). But this means x=0 (mod q), since there are no divisors of zero in Zq.
Illustrate Germain's Easy Theorem by two examples:
(1) Find a solution to x^3+y^3+z^3=0 (mod 7), and point out which of x, y, or z is a multiple of 7 (there must be one).
(2) Find a solution to x^7+y^7+z^7=0 (mod 15) in which none of x, y, or z is a multiple of 15.
Note: In (1), since 3 is a Germain prime, Germain's Easy Theorem applies. In (2), since 7 is NOT a Germain prime, the conclusion of Germain's Easy Theorem may be false -- the assignment seems to imply that it is!