Let $R = \{a+b\sqrt{-2}\ |\ a,b \in \mathbb{Z}\}$
Rational primes $p \geq 3$ of the form $p = a^2 + 2b^2$ factorize in $R$ as a >product of two irreducibles which are not associate. Such primes $p$ are $\equiv 1, 3 \pmod 8$. Rational primes $q \equiv 5, 7 \pmod 8$ remain prime in $R$.
My answer so far: $p = a^2+2b^2 = (a+b\sqrt{-2})(a-b\sqrt{-2})$ where $p\geq3$ is a rational prime. $a±b\sqrt{-2}$ are irreducible since their norm $(a^2+2b^2=p)$ is a prime and I know every prime is irreducible. They are not associates since one is not a unit times the other (the units are $±1$).
Next bit I am not sure about. I wrote a square number is $0,1,4 \pmod 8$ (not sure if I have to prove this or how to). Since the residues $\{0,1,2,3,4,5,6,7\}$ square to give residues $\{0,1,4,9,16,25,36,49\}\pmod8 = \{0,1,4\}$. But $p\geq3$ so we have only odd primes. A square of an integer is always $1$ or $4 \pmod8$. So $a^2+2b^2$ can only be $\equiv 1,3,4,6 \pmod8$. But $a^2+2b^2$ is prime so it cannot be $4,6 \pmod8$. So we have p are $\equiv 1,3 \pmod8$. And primes of the form $q \equiv 5,7 \pmod8$ remain prime in R since the are not of the form $a^2+2b^2$. Is this correct and how can it be improved?
