Exercise 4.23

Proof. If $x^2\equiv -1(\mathit{mod}p)$, then $\bar{x}$ has order 4 in ${𝔽}_p^{\text{∗}}$, hence from Lagrange’s theorem, $4\mid p-1$.

Conversely, suppose $4\mid p-1$, so $p=4k+1,k\in {\mathbb{N}}^{\text{∗}}$. From proposition 4.2.1, as $2\mid p-1$, $-1$ is a square modulo $p$ iff ${(-1)}^{(p-1)∕2}\equiv 1(\mathit{mod}p)$, which is true because ${(-1)}^{(p-1)∕2}={(-1)}^{2k}=1$.

If $x^4\equiv -1(\mathit{mod}p)$, then ${\bar{x}}^8=1\in {𝔽}_p^{\text{∗}}$, and ${\bar{x}}^4\ne 1$, so $\bar{x}$ has order $8$ in ${𝔽}_p^{\text{∗}}$, so $8\mid p-1$.

Conversely, if $p\equiv 1(\mathit{mod}8)$, $p=8K+1,K\in {\mathbb{N}}^{\text{∗}}$. From Prop.4.2.1, as $4\mid p-1$, there exists $x\in \mathbb{Z}$ such that $-1=x^4$ iff ${(-1)}^{(p-1)∕4}\equiv 1(\mathit{mod}8)$, which is true because ${(-1)}^{(p-1)∕4}={(-1)}^{2K}=1$.

Conclusion :

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