# $x^5+a$ is reducible in $\mathbb{Z}_5[x]$ for each $a\in\mathbb{Z}_5$

Question: Show that $x^3+a$ is reducible in $\mathbb{Z}_3[x]$ for each $a\in\mathbb{Z}_3$, and that $x^5+a$ is reducible in $\mathbb{Z}_5[x]$ for each $a\in\mathbb{Z}_5$

So I got these two as my homework problems, and I guess I can "show" by listing all the outcomes. But, is it true that when $p$ is prime, $x^p+a$ is reducible in $\mathbb{Z}_p[x]$ for each $a\in\mathbb{Z}_p$?

What I've done so far: Suppose that $x^p+a$ is irreducible for some $a_0\in\mathbb{Z}_p$, so $x^p+a_0$ is irreducible. That means $x^p+a_0=0$ has no root in $\mathbb{Z}_p$.Then I don't know how to proceed.

Any idea would be appreciated. Thanks!

• See Fermat's little theorem... – Luiz Cordeiro Nov 13 '14 at 1:33
• @LuizCordeiro OK so Fermat again... – 3x89g2 Nov 13 '14 at 1:35
• Do you mean for each non-identity element? – Mr.Fry Nov 13 '14 at 6:58