Exercise 1.3.34* (Primes of the form $m^4 + 4^n$)

beginproof Assume that $m^4+4^n$ is prime, with $m,n\in \mathbb{N}$ and $(m,n)\ne (1,1)$.

For the sake of contradiction, suppose that $m$ is even. Then $m^4$ is even. If $n\ge 1$, $4^n$ is even, so $m^4+4^n$ is even. This is possible only if $m^4+4^n=2$, which gives $m=n=1$. This proves that $m$ is odd, except when $m=n=1$.

Now suppose that $n$ is odd, so that $n=2k+1$ for some integer. Then

(For instance, if $m=35,n=17$, then $m^4+4^n=17181369809=114377\times 150217$.)

Moreover $m^2-2^{k+1}m+2\cdot 4^k={(m-2^k)}^2+4^k$. If $k>0$, then $m-2^k$ is odd, so ${(m-2^k)}^2\ge 1$, therefore ${(m-2^k)}^2+4^k\ge 1+4^k>1$. This proves that $m^4+4^n$ is composite.

If $k=0$, then $n=1$, and $m^4+4^n=m^4+4$ is composite, unless $m=1$ (see Exercise 32).

To conclude, if $m^4+4^n$ is prime, then $m$ is odd and $n$ is even, except when $m=n=1$.