Check if $X^3 -2$ is irreducible or not in $\mathbb{Z}_{31} [X]$

Question

Check if $f(X):= X^3 -2$ is irreducible or not in $\mathbb{Z}_{31} [X]$

I can note that if $f(X)$ is reducible I have one factor of degree $1$ and so there is at least one root of $2$ in $\mathbb{Z}_{31}$. I can somehow use Fermat to prove or not that there is a root?

Answer 1

Note that $2^{5}\equiv 1\pmod{31}$. So $2^6=(2^2)^3\equiv 2\pmod{31}$.

Answer 2

Note that for $a \neq 0 \pmod{31}$ if $X^3 \equiv a \pmod{31}$ then $$a^{10} \equiv x^{30} \equiv 1 \pmod{31}$$

Conversely if $a^{10} \equiv 1 \pmod{31}$, pick $g$ a primitive root $\pmod{31}$. Writing $a=g^n$ we get that $$g^{n \cdot 10} \equiv 1 \pmod{31} \Rightarrow 30 | n cdot 10 \Rightarrow 3|n \Rightarrow n=3k$$

Then $$a= (g^k)^3$$ and hence $g^k$ is a root of $$X^3-a$$

This shows that $$X^3-a \mbox{ is reducible } \pmod{31} \Leftrightarrow a^{10} \equiv 1 \pmod{31}$$