Exercise 5.13

Proof.

$\text{∙}$ As $a^6+1=(a^2+1)(a^4-a^2+1)$, $p\mid a^4-a^2+1$ implies $p\mid a^6+1$, thus $\left(\frac{-1}{p}\right)=1$ and $p\equiv 1(\mathit{mod}4)$.

$\text{∙}$ $p\mid 4a^4-4a^2+4={(2a-1)}^2+3$, so $\left(\frac{-3}{p}\right)=1$.

As $-3\equiv 1(\mathit{mod}4)$, $\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)$, therefore $\left(\frac{p}{3}\right)=1$, thus $p\equiv 1(\mathit{mod}3)$.

$4\mid p-1$ and $3\mid p-1$, with $3\wedge 4=1$, thus $12\mid p-1$ :

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