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Let $p >3 $ be a prime number and $\Bbb F_p$ denote the finite field of order $p$. Prove that the polynomial $X^2 +X +1$ is reducible in $\Bbb F_p[X]$ if and only if $p ≡ 1\pmod 3.$

Attempt: $X^2 +X +1=\dfrac{X^3-1}{X-1}.$ Now if $x^3=1$ for some $x\in \Bbb F_p$; $x\in \Bbb F_p^*\implies x^{p-1}=1\implies 3\mid p-1\implies p\equiv 1(\mod 3)$.

How to do the converse?

  • 2
    You need to assume $x \ne 1$ to conclude $3 \mid p-1$. – 2017-01-15
  • 0
    The "ultimate" duplicate may be [this](http://math.stackexchange.com/q/425683/11619)? – 2017-01-15

2 Answers 2

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If $q$ is prime, any finite group of order divisible by $q$ has an element of order $q$.

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Other approach: since our polynomial has degree two, being reducible $\mathbb{F}_p[X]$ is equivalent with having a root in $\mathbb{F}_p$. Now since $4$ is invertible in our field, this equation is equivalent with $$4X^2+4X+4 = 0 \Leftrightarrow (2X+1)^2 = -3 $$ So this equation has a solution if and only if $-3$ is a square in $\mathbb{F}_p$. Now writing this in terms of legendre symbols and using quadratic reciprocity, we get that $$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{p}{3}\right) (-1)^{\frac{p-1}{2}} = \left(\frac{p}{3}\right)$$ So $-3$ is a square in $\mathbb{F}_p$ iff $p$ is a square in $\mathbb{F}_3$, that is, iff $p\equiv 1 \mod{3}$.