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.
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.