This is the second Challenge Problem I’ve made up.
Prove that $\displaystyle n^{4k}+4m^4$ is not prime for all positive integers $\displaystyle n,\,k,\,m$ with $\displaystyle m>1.$
NCA's identity has a name, its called Sophie Germain's Identity.
For real $\displaystyle m$ and $\displaystyle n$ and integer $\displaystyle k$:
$\displaystyle n^{4k}+4m^4=|n|^{4k}+4|m|^4$
So the restriction to positive integers for $\displaystyle n$ and $\displaystyle m$ is unnecessary.
Also if $\displaystyle m=0$ we have $\displaystyle n^{4k}+4m^4=n^{4k}$ which can never be prime for any integer $\displaystyle n$ (including $\displaystyle 0$).
So the problem can be rewordrd to for all positive integers $\displaystyle k$ and integers $\displaystyle n,m$ show that
$\displaystyle n^{4k}+4m^4$
is never prime.
(we may have to be explicit about what we want $\displaystyle 0^0$ to mean here or exclude the case where one or both of $\displaystyle n$ and $\displaystyle m$ are zero and $\displaystyle k$ is zero)
CB
That’s a very good point. However, the condition $\displaystyle m>1$ (or $\displaystyle |m|>1$ if you like) is absolutely vital, otherwise you could find a counterexample as NonCommAlg pointed out.
The way I originally did it was to factorize $\displaystyle n^{4k}+4m^4$ as $\displaystyle \left(n^{2k}+2mn^k+2m^2\right)\left(n^{2k}-2mn^k+2m^2\right)$ instead of writing the factors the way NonCommAlg did – and then it isn’t so clear that both factors are actually greater than 1. (One of them certainly is, but the other is iffy.) I then proceeded by considering the expressions $\displaystyle n^{2k}\pm2mn^k+2m^2-1$ as quadratics in $\displaystyle n^k.$ Their determinant is $\displaystyle 4m^2-8m^2+4=4(1-m^2)$ and the vital condition $\displaystyle m>1$ ensures that the determinant is negative and therefore that the expressions are positive.