Factorisation of numbers sum with exponent $12$ How can I prove that $n^{12}+64$ has at least four distinct factors other than $1$ and itself?
I applied $a^3+b^3$ identity.

Using the familiar factorization of $a^3+b^3$, we get the factors $n^4+4$ and $n^8-4n^4+16$. Now note the perhaps a little surprising factorization $$n^4+4=(n^2-2n+2)(n^2+2n+2).$$
 If $n\gt 1$, then $n^2-2n+2$, $n^2+2n+2$, $(n^2-2n+2)(n^8-4n^4+16)$ and $(n^2+2n+2)(n^8-4n^4+16)$ are obviously distinct non-trivial factors. There are others.
