Nice polynomial reducibility: $x^n+4$

Question

Problem: Find all $n\in \mathbb{N}$ such that $f(x)=x^n+4$ is reducible in $\mathbb{Z}[x]$.

It seems $n=4k$ is the only one (the factorization follows easily from Sophie Germain's identity in this case), but I can't prove it. I can prove, however, that if $f(x)=g(x)h(x)$ for non-constant integer polynomials $g(x),h(x)$, then their constant terms, say $a_0,b_0$, must satisfy $a_0=b_0=\pm 2$.

EDIT: I am still looking for an elementary solution which only uses olympiad tools. It will be viewed in higher regard than one which uses linear/abstract algebra. Thanks!

Answer 1

It is a consequence of the following

Theorem $\ $ Suppose $\rm\:c\in F\:$ a field, and $\rm\:0 < n\in\mathbb Z\:.$

$\rm\quad x^n - c\ $ is irreducible over $\rm\:F \iff c \not\in F^p\:$ for all primes $\rm\:p\: |\: n,\:$ and $\rm\ c\not\in -4\:F^4$ when $\rm\: 4\ |\ n. $

A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6, or Lang's Algebra (Galois Theory).

Answer 2

i think:

if $x^n+4=g(x)h(x)$ , we can see n>3, and $x^n+4$ don't have integer roots therefore $deg h(x)=k>1,deg g(x)=s>1$ we have : $g(0)h(0)=4$ if $|g(0)|=4$ then $|h(0)|=1$, assume $x_1,x_2,...,x_k$ are roots of $h(x)=0$ then $x_1,x_2,...,x_k$ are roots of $x^n+4=0$ and $h(x)=(x-x_1)(x-x_2)...(x-x_k)$. We have $|h(0)|=|(0-x_1)(0-x_2)...(0-x_k)|=|x_1x_2...x_k|=1$. And $ |x_i|^n=4 \Rightarrow |x_1x_2...x_k|^n=4^k\Rightarrow 1=4^k$ (contradict)

Hence $h(0)=g(0)=2$ . if $h(0)=2$ we have : $2=|h(0)|=|(0-x_1)(0-x_2)...(0-x_k)|=|x_1x_2...x_k|$ and $2^n=|x_1x_2...x_k|^n=4^k=2^{2k}$ show that $n=2k$. If $n=4s$ we have $x^{4s}+4=(x^{2s}+2x^s+2)(x^{2s}-2x^s+2)$

if $n=4s+2$ i think this polyminal is irreducible but i can't prove it. Can you help me prove it? or show me a simple solution for this problem? Thank you !