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

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 ?

## marked as duplicate by Travis, Shaun, kingW3, Martin Sleziak, CharlesJan 27 '15 at 15:06

• – Henry Jan 27 '15 at 12:39
• – Martin Sleziak Jan 27 '15 at 13:41

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 $>1$

• You may want to add that $n$ is odd $\implies\frac{n+1}{2}\in\mathbb{N}$ (and of course, that the expression above is in the form of $a^2-b^2=(a+b)(a-b)$). – barak manos Jan 27 '15 at 12:50
• @barakmanos, Please find the edited version – lab bhattacharjee Jan 27 '15 at 13:03

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