Interesting question regarding proof for a number to be composite.

Question

Prove that $n^4+4^n$ is not a prime for any natural number $n>1$.

My Approach:

Suppose $n^4+4^n$ is prime.

It is evident that $n\equiv 1 \pmod 2$ for $n^4+4^n$ to be prime. $$n\equiv 1\pmod 2\Rightarrow 4^n\equiv (-1)\pmod 5$$ Suppose $\gcd (n,5)=1$. Then by euler's identity:

$$\phi (5)=4\Rightarrow n^4\equiv 1\pmod 5$$ $$\therefore n^4+4^n\equiv 0\pmod 5$$ Thus $n^4+4^n$ is never a prime if $n$ is not a multiple of $5$.

Now I seem to be struck in proving the same for when $n$ is mutiple of $5$.

Please help and also please provide suggestions.

THANKS

$x^4 + 4 y^4 = (x^2 + 2xy + 2 y^2)(x^2 - 2xy + 2y^2)$ – Will Jagy Aug 18 '20 at 16:42 — Will Jagy, Aug 18 '20 at 16:42

score 1 · Answer 1 · answered Aug 18 '20 at 16:41

1

Actually, Sophie Germain's Identity kills it :

$n^4+4^n=((n+2^k)^2+4^k))((n-2^k)^2+4^k))$, when $n=2k+1$ and you can figure it out that $n$ must be odd in your case.

answered Aug 18 '20 at 16:41

Bastien Tourand

922

That definitely is a good way to solve this question. But is there a way to prove it without using sophie germain identity? – Devansh Kamra Aug 18 '20 at 16:42
This problem is usually proposed to see how to use this identity. So I never seen a solution without it... – Bastien Tourand Aug 18 '20 at 16:44

Interesting question regarding proof for a number to be composite.

1 Answers1