Find all $k$ such that $x^p+k$ is reducible over $\mathbb{Z}$.

Question

Let $p$ is a prime number.

Find all $k\in\mathbb{Z}$ such that: $x^p+k$ is reducible over $\mathbb{Z}$.

When $p=2$, it is easy to see that $k=-a^2, a\in\mathbb{Z}$.

When $p>2$ i think the answer is $k=a^p, a\in\mathbb{Z}$ but i don't have any ideas to approach this problem.

Any suggestion?

$x+a$ is always a factor of $x^{2n+1}+a^{2n+1}$. This follows from induction on $x^2(x^{2n-1}+a^{2n-1})- a^{2n-1}(x^2-a^2)$ — sharding4, Jun 27 '17 at 03:16
I edited your title to be more in accord with your question, which I like! — Robert Lewis, Jun 27 '17 at 04:49

score 0 · Accepted Answer · answered Jun 27 '17 at 05:07

Here is my friend's solution: Consider $p>2$

Suppose that $x^p+k=g(x).h(x)$, $g,h\in\mathbb{Z}[x]$; $\textrm{deg }g,\textrm{ deg }h\geq 1$, $m<p$

Write $x^p+k$ in factored form $\prod\limits_{i=1}^{n}{(x-x_i)}$ and suppose $g(x)=\prod\limits_{i=1}^{m}{(x-x_i)}$

We have $|g(0)|^p=\prod\limits_{i=1}^{n}{|x_i|^p}$ and $x_i^p=-k$

So $|g(0)|^p=|k|^m$, but $(m,p)=1$, then we have $k=a^p$

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