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Let $p$ be a prime number such that $p \equiv 1 \mod 3$. Prove that there is an $x \in \mathbb{Z}$ such that $ x^2 \equiv -3 \mod p$. Any hints or solutions? Thanks in advance.

(I already proved that $\mathbb{Z}\backslash p \mathbb{Z}$ has an element of order $3$ and that $a^2 + a + 1 = 0$ in $(\mathbb{Z}\backslash p \mathbb{Z})^{\times}$. So maybe that is useful.)

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    Can't we just use Quadratic Reciprocity for $$\left(\frac{-3}{p} \right)=1$$ – 2017-02-20
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    So by Quadratic Reciprocity: $\left(\dfrac{3}{p}\right)\left(\dfrac{p}{3}\right) = (-1)^{(p-1)/2}$. We know $\left(\dfrac{p}{3}\right)=\left(\dfrac{1}{3}\right)=1 $ and that $(-1)^{(p-1)/2} \left(\dfrac{3}{p}\right) = \left(\dfrac{-3}{p}\right)$. Then the result follows. Is this OK? – 2017-02-20
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    Yes. It is OK to do so. – 2017-02-20

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