# Why is Sophie-Germain's Identity popular?

Sophie Germain's identity is only about factorising $$a^4+4b^4$$ as product of two squares.It's not quite difficult. So,why is it so popular?

• This question is answered here hsm.stackexchange.com/questions/2786/… It is often useful in number theoretic arguments, especially related to divisibility, and so features in solutions to mathematical olympiad problems. Oct 28, 2015 at 18:16
• It is also the tip of an iceberg: a whole family of factorizations of certain perturbed versions of cyclotomic polynomials, often called Aurifeuillean factorizations (there's a Wiki entry, and I have some lecture notes on this: www-users.cse.umn.edu/~garrett/m/number_theory/… Apr 23, 2022 at 18:30

For any field $K$, $c \in K^\times$ and integer $n > 2$, when is $X^n - c$ irreducible over $K$? A necessary condition is that for each prime number $p$ dividing $n$, $c$ is not a $p$th power in $K$; otherwise $X^p - c$ would be reducible over $K$ (a linear factor), so $X^n - c$ would be reducible over $K$ too. It turns out that this necessary condition is sufficient except if $4\mid n$, in which case we also need to check $c$ is not of the form $-4d^4$ for some $d$ in $K$ precisely because of the "unexpected" factorization of $X^4+4d^4$ over $K$.
Thus the only subtlety in determining irreducibility of $X^n-c$ over $K$ is due to the type of factorization going back to Sophie Germain.
Lang's Algebra devotes a section to the study of $X^n - a$, and its irreducibility is the first main result he describes.
Because of it's simplicity I guess. The identity is $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$. The identity is easy to verify but difficult to arrive at in the first place. We are familiar with $a^2-b^2=(a+b)(a-b)$, but not with the kind $a^2+b^2$ splitting as two factors. Also knowing the sum can be factored by such a identity, it is helpful in number theory, specially factoring numbers, determining a large number is prime or not. You can check here for an example.