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.
 A: $x^2+23y^2=2z^2\iff x^2+(5y)^2=2(z^2+y^2)$.
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).
A: To solve such problems, it is necessary to use the formula.  Solving a Diophantine equation of the form $x^2 = ay^2 + byz + cz^2$ with the constants $a, b, c$ given and $x,y,z$ positive integers
The formula is written in a General form so I will have to solve the equation. For the equation.
$$23x^2+y_1^2=2z^2$$
Will do the replacement.  $y_1=x+ay$ The equation will look like.
$$(23+a^2)x^2+2axy+y^2=2z^2$$
And we need to find this number  $a$.
Using these formulae. And we will need to solve the Pell equation. $3$ use the formula.
$$q=\sqrt{4a^2+4(23+a^2)(2-1)}=2\sqrt{23+2a^2}$$
This equation Pell.  $q=10$  : $a=1$
So you need to solve the equation.
$$24x^2+2xy+y^2=2z^2$$
Using the formula the solution will be.  Reducing a common divisor.
$$x=(40\mp54)p^2+4(1\mp5)ps\mp{s^2}$$
$$y=-(230\pm54)p^2+4(1\mp5)ps+(5\mp1)s^2$$
$$z=54(5\mp1)p^2+2(27\mp10)ps+(5\mp1)s^2$$
Remember that  $y_1=x+y$  For the equation.
$$23x^2+y_1^2=2z^2$$
Finally the solution can be written as.
$$x=(40\mp54)p^2+4(1\mp5)ps\mp{s^2}$$
$$y_1=-(190\pm108)p^2+8(1\mp5)ps+(5\mp2)s^2$$
$$z=54(5\mp1)p^2+2(27\mp10)ps+(5\mp1)s^2$$
