The set $A = \{a^2 + 2b^2\mid a,b \in \Bbb Z\setminus\{0\}\}$

Question

Let $A = \{a^2 + 2b^2\mid a,b \in \Bbb Z\setminus\{0\}\}$ and $p$ be a prime number. Prove that if $p^2 \in\ A$, then $p \in A$.

Answer 1

The quadratic ring $Z(\sqrt{-2})$ is a unique factorization domain. A rational prime is one of our usual list 2,3,5,... (an integer). By a ring prime I'll mean a prime element of $Z(\sqrt{-2})$. Some rational primes are also ring primes, and others are not but factor in $Z(\sqrt{-2})$. (There is a characterization mod 8 of which primes do which. We don't need that here.)

Recall from fmat's post that

$A = \{a^2 + 2b^2\mid a,b \in \Bbb Z\setminus\{0\}\}$.

If $p^2$ is in $A$ then $p^2=a^2+2b^2=(a+b\sqrt{-2})(a-b\sqrt{-2})$.

If here $p$ were a ring prime this would contradict unique factorization, since the units of $Z(\sqrt{-2})$ are $-1,1$ and we're assuming $a,b$ nonzero.

And if $p$ is not a ring prime it factors as $p=(x+y\sqrt{-2})(x-y\sqrt{-2})$ [A standard result for $Z(\sqrt{-2})$ and similar rings]. But then $p=x^2+2y^2$ on multiplying out, so that we have $p$ in $A$ as desired.