I have been stuck on this one for months, really simple to state, really giving me trouble.
Show that if $m^4 + 4^n$ is prime, $m>0$, $n>0$, then $m$ is odd and $n$ is even, except when $m=n=1$.
The case $m=n=1$ gives $5$, so that is why it is excluded. Clearly $m$ is odd, for if it was even, the number would be divisible by at least $4$, but I can't seem to get rid of the case where $n$ must be odd.
The only thing I've managed to try is to write $m$ as an odd number, say $m = 2k+1$, and $n = 2\ell + 1$, then get
$$
m^4 + 4^n = (2k+1)^4 + 4^{2\ell+1} = 16k^4 + 32k^3 + 24k^2 + 8k + 1 + 4^{2 \ell + 1}
$$
and this is congruent to $0 \mod 5$ unless $k \equiv 2 \mod 5$, and now it gives you an even more ugly polynomial if I write $k = 5j+2$, and it's not working. I've tried other inconclusive approaches, like trying to see if two numbers of some form generate numbers of the same form.. (for instance primes of the form 8k+1 and 8k+7 always generate primes of the same form when multiplied together, stuff like that).
Any ideas? Even just ideas you haven't gave a thought… they're welcome!
Best Answer
If $m$ even then obviously it is divisible by $2$ and since it is greater than $2$, so not prime. Now if $n$ odd, say $n=2k+1$. Then it can be written as $m^4+4.(2^k)^4$ which, by Sophie Germain Identity is not a prime for $m,n>1$.