Prime factors of numbers of form $a^2+2b^2$ We are just being introduced to proofs in our class, and my teacher has offered a bounty. If anyone can prove the following statement, at any point during the year, we will get no homework for that week. 

Call a prime number $2$-expressible if it can be expressed in the form
  $a^2+2b^2$ and $2$-inexpressible if it cannot. If two numbers $x$ and
  $y$ have no common factor, $x^2+2y^2$ either has no $2$-inexpressible
  prime factors or more than 1.

I've managed to prove this through a sort of convoluted and hard to follow argument using modular congruence, but I'm convinced there is a simple proof that I am missing. Is there any proof of this that uses only elementary number theory? 
 A: Hint:
$\Bbb Z[\sqrt{-2}]$ is an Unique Factorization Domain. In fact, it is an Euclidean Domain.
A: I think I fixed the proof, and it is rather elementary, but not so simple...
We're working with $N>1$  ($N=1=1^2+2\times0^2$ has no inexpressible prime factor)
Notice that according to Brahmagupta–Fibonacci identity, $(a^2+2b^2)(c^2+2d^2)=(ac+2bd)^2+2(ad-bc)^2$.
Therefore, the set of all numbers of the form $x^2+2y^2$ is closed under multiplication.
Let's suppose $N=x^2+2y^2$ has exactly one $2$-inexpressible prime factor.
Then $N$ can be expressed as  $N=x^2+2y^2=(X^2+2Y^2)I$ where $I$ is the only inexpressible prime factor. $I$ is necessarily odd since $2=0^2+2\times1^2$ is $2$-expressible. If $2|N$, then $2|x^2$, and so $2|x$ since $2$ is a prime, so $x$ would be even. Assuming this is the case, dividing by $2$ yields $\frac N2=N'=\frac 12 (4x'^2+2y^2)=2x'^2+y^2=(2X'^2+Y^2)I$. It is then possible to iterate the same reasoning until $N$ becomes odd (since you mentioned that $x$ and $y$ have no common factors, then once at most is sufficient). If $N$ is odd and $I$ is odd then $2X'^2+Y^2$ has to be odd too.
Since squares are congruent to $0,1$ or $4 \pmod 8$ and $N$ is odd,  $N=x^2+2y^2\equiv1$ or $3 \pmod 8$, and $2X'^2+Y^2\equiv1$ or $3\pmod 8$ too, so $I\equiv 1,3$ or $\frac 13 \pmod 8\equiv1$ or $3\pmod 8$.
So $I$ is an odd prime such that $I\equiv 1$ or $3\pmod 8$.  However, it can be shown (relatively elementarily, although the proof is long and tedious...) that such primes are always $2$-expressible. This is a contradiction.
Therefore, either $N$ has no $2$-inexpressible factors, or it has more than $1$.
