I have to show that $f(x) = x^4+x^3+x^2+x+1$ is a irreducible polynomial in $F_p$ with $p \equiv 2 \pmod{5}$ or $p \equiv 3 \pmod{5}$.
$f(x) \mid (x^5-1)$. This should be used for order of possible roots of $f(x)$ in $\mathbb{F}_p$ and $\mathbb{F}_{p^2}$.
I guess it is concerned with the order of subgroups of $p$. But I cant't find a solution. Any advice ?
$\newcommand{\F}[0]{\mathbb{F}}$Let $p$ be a prime, $p \equiv \pm 2 \pmod{5}$.
First note, as you are doing, that a root of $f$ has order $5$. This is because if $\alpha$ is a root of $f$, then $\alpha \ne 1$ (as $f(1) = 5 \not\equiv 0 \pmod{p}$). As $x^{5} -1 = (x - 1) f(x)$, we have $$ \alpha^{5} - 1 = (\alpha - 1) f(\alpha) = 0 $$ so that $\alpha^{5} = 1$, and $\alpha$ has order $5$.
Could it be that a root $\alpha$ is in $\F_{p}^{\star}$? The latter group has order $p-1$, so if it has an element $\alpha$ of order $5$, then $5 \mid p -1$. But $5$ does not divide $p-1 \equiv 1, 2 \pmod{5}$, so $f$ has no roots (factors of degree $1$) in $\F_{p}$.
Could it be that a root $\alpha$ is in $\F_{p^{2}}^{\star}$? The latter group has order $p^{2}-1$, so if it has an element $\alpha$ of order $5$, then $5 \mid p^{2} -1$. Now $p^{2} - 1 \equiv (\pm 2)^{2} - 1 \equiv 3 \pmod{5}$, so $5$ does not divide $p^{2} - 1$. Thus $f$ has no roots in $\F_{p^{2}}$ either, that is, $f$ is not the product of two irreducible polynomials of degree $2$ over $\F_{p}$.