How to show $a^2+2B^2=p$ has integer solutions for all primes p with $(\frac{−2}{p})=1$

Question

How to show $a^2+2b^2$=p has integer solutions for all primes p with $(\frac{−2}{p})=1$ (legendre symbol)

Partial solution:

$(\frac{−2}{p})=1$ $\Rightarrow$ p $\ |$ $a^2+2$=$(a-\sqrt{-2})$$(a+\sqrt{-2})$ (since $\mathbb{Z}[\sqrt{-2}]$ is UFD).

Hence, p $\ |$ $(a-\sqrt{-2})$ and $(a+\sqrt{-2})$ Thus, p $\ |$ $a-\sqrt{-2}-a+\sqrt{-2}$=$2\sqrt{-2}$ $\Rightarrow$ $Norm(p)\ | Norm(2\sqrt{-2})=8$ I have to find a contradiciton, from the norms. then, conclude that p is not irreducible. so reducible. but why?

Could you please clarify the points indicated.thank you.

Answer 1

You proved that $p$ is not prime in $\mathbb Z[\sqrt{-2}]$ (as $p|(a-\sqrt{-2})(a+\sqrt{-2})$ and $p$ does not divide both multipliers). Since $\mathbb Z[\sqrt{-2}]$ is UFD it implies that $p$ is also not irreducible. Thus there are integers $x,y,u,v$ such that $p=(x+y\sqrt{-2})(u+v\sqrt{-2})$ with $x^2+2y^2$ and $u^2+2v^2$ not equal $1$. Taking the norm we get $p^2=(x^2+2y^2)(u^2+2v^2)$. Clearly, $p^2$ has only two different factorizations in $\mathbb Z$ into two positive integers, $p^2=1\cdot p^2$ and $p^2=p\cdot p$. The former is excluded, therefore $x^2+2y^2=u^2+2v^2=p$.