Why is the Sophie Germain Identity called thus?

Several authors (z.B.: Arthur Engel in his Problem-Solving Strategies, Alexander Bogomolny in this entry of the Cut the Knot website) refer to the following (straightforward) consequence of the difference of squares factorization formula as the Sophie Germain Identity:

$A^{4}+4B^{4} = (A^{2}+2B^{2}+2AB)(A^{2}+2B^{2}-2AB)$

Can anybody here explain why it is that this identity was christened thus? Do we actually have any evidence that it can be found somewhere in the Nachlaß of M. Le Blanc or the said identity is yet another instance of the famed Stigler law of eponymy?

• See this source for a comment. Sep 12 '15 at 21:18

This is one of those cases where the significance of a result is not in its content but in its use. The identity itself is trivial to verify, but behind it is a trick that is unreasonably useful in number theoretic reasoning. It is that that Sophie Germain pointed out in her famous analysis of the last Fermat theorem, and it is that that apparently prompted Dickson to assign the authorship to her. The identity has since found many other applications, including to olympiad problems like showing that $2015^{4}+4^{2015}$ isn't prime or solving $3^r+4^s=5^t$ in non-negative integers, more here.
Sophie Germain does not quite state it in Dickson's form however. She writes "No number of the form $p^4+4$ except $5$ is a prime number, because $p^4+4=(p^2-2)^2+4p^2$ and consequently these numbers are representable in more than one way as a sum of two squares. In general $p^4+q^4=(p^2-q^2)^2+2p^2q^2=(p^2+q^2)^2-2p^2q^2$. Thus when $p^4+q^4$ is a prime one is sure that the number is only thus expressible as the sum of a square $+$ or $-$ twice a square. The numbers $2^{2^i}+1$ are just a special case of the form $p^4+q^4$". The identity used here follows from $p^4+4q^4=(p^2+2q^2)^2-4(pq)^2$ by factorising the difference of two squares.
Dickson also found a prior occurence of the same identity in a letter from Euler to Goldbach in 1742, but only in a special case $1+4x^4=(2x^2+2x+1)(2x^2-2x+1)$. See the original manuscript and references at Theorem of the Day.