Show that if $m^4 + 4^n$ is prime, $m>0$, $n>0$, then $m$ is odd and $n$ is even, except when $m=n=1$.
This question has an answer here but I am asking for alternative proof(s). All of these duplicates have the same factorisation (the same solution )
My attempt: it is clear that $m$ should be odd , any number can be written in one of these forms: $5k+1 , \ 5k+2 , \ 5k-2, \ 5k-1 , \ 5k $, for odd numbers in the form of $5k \pm 2$ note that $4^n \mod 10 =4 $ for all odd numbers. Let $4^n= 4+ 10 q$ $$(m)^4 + 4^n = 5(125 k^4 \pm 8\cdot 25 k^3 + 24\cdot 5 k^2 \pm 8k )+ 16 + 4 + 5 \cdot 2 \cdot q $$
It is clear that this is always a multiple of $5$.
The same approach will work for $5k \pm 1 $.
I couldn't find a way to prove this for multiples of $5$ i.e $ 5k$.
