Exercise 2.1.60 (Existence of solutions for $x^2 + 2 y^2 = p$.)

Question

Combine the results of the foregoing problems to show that a prime number $p$ can be expressed in the form $a^2 + 2 b^2$ if and only if the congruence $x^2 \equiv -2 \pmod p$ is solvable. (In Chapter 3 we show that this congruence is solvable if and only if $p=2$ or $p \equiv 1 	ext{ or } 3 \pmod 8$.)

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} We want to prove, combining the results of the foregoing problems, that for any prime $p$,
$$\exists (a,b) \in \Z^2,\ p = a^2 + 2b^2 \iff \exists x \in \Z,\ x^2\equiv -2 \pmod p.$$

If $p = 2$, since $2 = 0^2 + 2\cdot 1^2$, and $0^2 \equiv -2 \pmod 2$, the two members are true. Now, we may suppose that $p$ is an odd prime.

\begin{enumerate}
\item[$(\Rightarrow)$] Assume that $p = a^2 + 2b^2$ for some integers $a,b$. If $p\mid a$ and $p \mid b$, then $p^2 \mid a^2 + 2b^2 = p$, thus $p\mid 1$. Since $p$ is a prime number, this is impossible, so $p 
mid a$ or $p 
mid b$, i.e. $p$ does not divide both $a$ and $b$. By Exercice 59, $x^2 \equiv -2 \pmod p$ has a solution in $\Z$.
\item[$(\Leftarrow)$] Assume that there is some $x \in \Z$ such that $x^2 \equiv -2 \pmod p$. By Exercise 56, at least one of the equations $a^2 + 2b^2 = p,\ a^2 + 2b^2 = 2p$ has a solution. By Exercise 58, if $a^2 + 2b^2 = 2p$ has a solution, then $a^2 + 2b^2 = p$ has also a solution. In all cases $a^2 + 2 b^2$ has a solution.
\end{enumerate}
The equivalence is proven.
\end{proof}

Note: If we know the complements of the law of quadratic reciprocity and the Legendre's symbol , we obtain for an odd prime $p$
\begin{align*}
\exists x \in \Z,\ x^2\equiv -2 \pmod p &\iff \legendre{-2}{p} = 1\
&\iff \legendre{-1}{p} \legendre{2}{p} =1\
&\iff (-1)^{\frac{p-1}{2}}(-1)^{\frac{p^2-1}{8}} = 1.\
\end{align*}
If we list the values of $(-1)^{\frac{p-1}{2}}$ and $(-1)^{\frac{p^2-1}{8}}$, we obtain
$$
\begin{array}{|c|c|c|c|}
p 	ext{ mod } 8 & (-1)^{\frac{p-1}{2}} & (-1)^{\frac{p^2-1}{8}} &  \legendre{-2}{p}\
\hline
1 & 1 & 1 & 1\
3 & -1 & -1 & 1\
5 & 1 & -1 & -1\
7 & -1 & 1 & -1\
\hline
\end{array}
$$
This shows that, for every odd prime $p$,
$$
  \legendre{-2}{p} = 1 \iff p \equiv 1,3 \pmod 8.
$$
To conclude, for every prime $p$,
$$\exists (a,b) \in \Z^2,\ p = a^2 + 2b^2 \iff p = 2 	ext{ or } p \equiv 1,3 \pmod 8.$$

Exercise 2.1.60 (Existence of solutions for $x^2 + 2 y^2 = p$.)

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Exercise 2.1.60 (Existence of solutions for $x^2 + 2 y^2 = p$.)

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