If $m^4 + 4^n$ is a prime, for $n,m\geq2$, then $m$ is odd and $n$ is even?

Question

One thing which i got is $m^4 + 4^n$ is congruent to $1 \pmod 8$ when both $n,m$ are odd...
Is it an iff condition?

Answer 1

$1^4+4^1=5$ is prime, but n is not even.

$m=1$ and $n=6$, $1+4^6=17\times 241,$ so is not prime.

Answer 2

It follows easily from $$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2 xy+2 y^2) \ . $$

Answer 3

There are primes when $m=1$ or when $n=0$ (indeed all but one of them are the same). You are probably excluding them.

For $n=1$ you have $m^4+4^1 = ((m-1)^2 +1) \times ((m+1)^2 +1)$ which is not prime unless $m=1$.

For $n \gt 1$ you have $m^4+4^n \equiv 0 \text{ or } 1 \mod 8$, which leads to your question

Answer 4

If you make some arithmetic modulo $\,4\,$:

$$m^4+4^n=m^4=0,1\pmod 4\Longrightarrow \,\,m\,\,\text{must be odd (why?)}$$

Also, if

$$n=2k+1\Longrightarrow 4^n=4\cdot (4^k)^2\,$$

and Pantelis's answer kicks in.