algebraic number theory - Does the equation $x^2+23y^2=2z^2$ have integer solutions?

I would like to show that the image of the norm map $\text N : \mathbb Z \left[\frac{1 + \sqrt{-23}}{2} \right] \to \mathbb Z$ does not include $2.$ I first thought that the norm map from $\mathbb Q(\sqrt{-23}) \to \mathbb Q$ does not have $2$ as its image either, so I tried to solve the Diophantine equation

$$x^2 + 23y^2 = 2z^2$$ in integers.

After taking congruences with respect to several integers, such as $2, 23, 4, 8$ and even $16,$ I still cannot say that this equation has no integer solutions. Then I found out that the map $\text{N}$ has a simpler expression and can be easily shown not to map to $2.$

But I still want to know about the image of $\text N,$ and any help will be greatly appreciated, thanks in advance.

Answer

$x^2+23y^2=2z^2\iff x^2+(5y)^2=2(z^2+y^2)$.

Since the solutions of the equation $X^2+Y^2=2Z^2$ are given by the identity

$$(a^2+2ab-b^2)^2+(a^2-2ab-b^2)^2=2(a^2+b^2)$$ we can try $(y,z)=(a,b)$ taking care of one of $a^2+2ab-b^2$ or $a^2-2ab-b^2$ be equal to $5b$.

Taking for example $(y,z)=(1,4)$ we get the solution $(x,y,z)=(3,1,4)$.

Thus the proposed equation have solutions (which can be parametrized but I stop here).