Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$?

Question

I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with $n\le 12\ 000$.

• If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.

• If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pmod 5$.

  So, $n$ must have the form $10k+5$.

  For $n=35$ and $n=55$, the number $4^n+n^4$ splits into two primes with almost the same size.

  So, is there an obvious reason (like algebraic factors) that there is no prime I am looking for ?

Answer 1

If $n$ is even, so will be $4^n+n^4$ and the later will definitely be $>2$ and hence composite

Else

$$4^n+n^4=(2^n)^2+(n^2)^2=(n^2+2^n)^2-2\cdot2^n\cdot n^2$$

$$=(n^2+2^n)^2-(n2^{\frac{n+1}2})^2$$

As $n$ is odd $\iff n+1$ is even $\implies\dfrac{n+1}2$ is an integer

$$4^n+n^4=(n^2+2^n+n2^{\frac{n+1}2})(n^2+2^n-n2^{\frac{n+1}2}) $$

Establish that both factors are $>1$

Answer 2

An alternative to @labbhattacharjee’s exemplary response (for the case of odd $n$ only):

$(1) \, X^4 + 4=(X^2-2X+2)(X^2+2X+2)$

$(2) \, X^4+4a^4=(X^2-2aX+2a^2)(X^2+2aX+2a^2)$

$(3) \, X^4+4^{2k+1}=X^4+4\cdot2^{4k}$.