Factorise $a^m + 1$

Question

I thought of maybe splitting it up: $$(a^m + 1)(a^{m-1} - \ldots - a^1 + 1).$$

But this method came to no avail as I had some extra terms left behind. We know nothing about $a$ or $m$ beforehand (i.e. prime or not, etc.).

Answer 1

If $m$ is not a power of $2$, then write $m=st$ with $s\ge 3$ odd. Then $$ a^m+1 = (a^t+1)(a^{(s-1)t}-a^{(s-2)t}+a^{(s-3)t}-\cdots + 1) $$

This doesn't work when $m$ is a power of $2$. Then it is possible that $a^m+1$ is prime, as it is for $2^1+1, 2^2+1, 2^4+1, 2^8+1$, and $2^{16}+1$. These are the only known exceptions for $a=2$; see https://en.wikipedia.org/wiki/Fermat_number#Generalized_Fermat_primes for other $a$s.