when $x^2+2$ splits $\operatorname{mod}p$

Question

Why $x^2+2$ splits iff $p \equiv 1$ or $3 \bmod 8$.

It is easy to check $p \equiv 1 \bmod 8 \quad x^2+2$ splits. In fact, if $\mathbb{F}_p$ contains a 8 th root of 1. Then $\left(\xi-\xi^{-1}\right)^2+2=0$, so $x^2+2$ splits in $\mathbb{F}_p$. so $8 \mid p-1, p \equiv 1 \pmod 8$.

I am confused about another situation when $p \equiv 3 \bmod 8$.

$2$ and $-1$ are not quadratic residues when $p\equiv3\pmod8$ — J. W. Tanner, Sep 05 '24 at 08:35
@J.W.Tanner I see.But why are 2 and -1 not quadratic residues when $p \equiv 3 \mathrm{mod} 8$? — Zaran Zin, Sep 05 '24 at 08:43
@J.W.Tanner I have seen such conclusion in wiki.But there is no proof. — Zaran Zin, Sep 05 '24 at 08:47
Quadratic reciprocity is proved in any book on elementary number theory, and also on this site. Have a look. $-1$ is a QR iff $p\equiv 1\bmod 4$ for an odd prime $p$. And $2$ is a quadratic nonresidue for $p\equiv \pm 3 \bmod 8$. — Dietrich Burde, Sep 05 '24 at 08:47
Also see proofs using quadratic Gauss sums in this Wikipedia article and also see here — J. W. Tanner, Sep 05 '24 at 08:56

J. W. Tanner · Accepted Answer · 2024-09-05T09:56:45.533

1

You are essentially asking if $-2$ is a quadratic residue modulo $p$, where $p\equiv 3\pmod8$. It is, because $-1$ and $2$ are not quadratic residues under that condition. Proofs of the latter abound in number theory textbooks, on Wikipedia here and here, and on this site here and here.

edited Sep 05 '24 at 09:56

answered Sep 05 '24 at 09:37

J. W. Tanner

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when $x^2+2$ splits $\operatorname{mod}p$

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