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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 ?

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$\newcommand{\F}[0]{\mathbb{F}}$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 $\F_{p}$.

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  • $\begingroup$ Thank you for your Explanation. Unfortunately I am not so familiar with roots and irreducible polynomials. So I have some questions left: why hasa root of f order 5 ? Why do you divide p-1 by 5 ? Why do you examine p^2-1 ? $\endgroup$ – ready4math Feb 27 '15 at 11:14
  • $\begingroup$ I have added some extra explanation, but I feel you have not exposed to enough theory, despite the way the problem is formulated. $\endgroup$ – Andreas Caranti Feb 27 '15 at 11:34

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