History of Mathematics: Sophie Germain and Fermat's Last Theorem Sophie Germain's greatest contribution to mathematics was in number theory. She discovered a special case of Fermat's Last Theorem which we now call the Germain Theorem.
Stated precisely:
The equation $x^p + y^p = z^p$ has no non-zero integer solutions where $p$ is a Germain prime ($p$ is a prime number if $2p+1$ is also prime) and $p$ does not divide $xyz$. Could someone prove this? I can't find an actual proof of this statement or Germain's surviving works in English translation.
 A: Sophie Germain's approach to the first case of Fermat's Last Theorem can be found in several textbooks that treat Fermat's Last Theorem. For example a very nice reference for her theorem is Kenneth Ireland and Michael Rosen's beautiful book A Classical Introduction to Modern Number Theory. There the theorem is proved in just about a page in Chapter 17, section 4, which is actually entitled "Sophie Germain's Theorem".
To add something to my comment let me mention a generalization of Sophie Germain's theorem which was found by Ernst Wendt.
Theorem (Wendt)
Let $p$ be an odd prime. If there exists an integer $k \geq 1$ such that the number $q := kp + 1$ is also prime and satisfies
$$
q \not \mid (k^k - 1)\operatorname{Res}{(X^k - 1, (X + 1)^k - 1)}
$$
then the first case of Fermat's Last Theorem is true for the exponent $p$, that is, there are no solutions to $x^p + y^p = z^p$ with $p \not \mid xyz$.
Here the notation $\operatorname{Res}{(f, g)}$ stands for the resultant of the polynomials $f$ and $g$.
From this theorem we can obtain Sophie Germain's theorem as an inmediate corollary because in that case we are basically dealing with the case in which $k = 2$. 
So we assume that $p$ is an odd prime such that $q = 2p + 1$ is also a prime number. Then we want to show that 
$$
q \not \mid (2^2 - 1)\operatorname{Res}{(X^2 - 1, (X + 1)^2 - 1)}
$$
Now since $p > 2$ then $q > 5$ so certainly $q \not \mid (2^2 - 1)$. Also
$$
\operatorname{Res}{(X^2 - 1, (X + 1)^2 - 1)} =  \operatorname{Res}{(X^2 - 1, X^2 + 2X)} =
$$
\begin{vmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1 \\
1 & 2 & 0 & 0 \\
0 & 1 & 2 & 0 \\
\end{vmatrix}
$$
= -3
$$
so $q \not \mid \operatorname{Res}{(X^2 - 1, (X + 1)^2 - 1)}$ either. Then by Wendt's theorem we can conclude that the first case of FLT is true for the prime $p$, and thus we have Sophie Germain's theorem as a corollary of Wendt's result.
Note A reference for this result is Henri Cohen's book Number Theory Volume I: Tools and Diophantine Equations, it appears in section 6.9.4, page 430.
Also observe that I used the determinant of the Sylvester matrix of the polynomials $X^2 - 1$ and $X^2 + 2X$ to compute their resultant.
