Does $x^2 + y^2 - \sqrt{2}z^2 = 0$ have solutions in $\mathbb{Z}[\sqrt{2}]$?

Question

We know there can be no solutions in the integers $x,y,z \in \mathbb{Z}$ for $$x^2 + y^2 - \sqrt{2}z^2 = 0 \tag{$\ast$}$$ if we adjoin enough elements we can solve it right away. So there really is hope. $$z = \frac{\sqrt{x^2 + y^2}}{\sqrt[4]{2}} $$ My question today is if there are any solutions if we adjoin the $\sqrt{2}$ itself so finding solutions to $(\ast)$ for $x,y,z \in \mathbb{Z}[\sqrt{2}]$

Answer 1

Trivial solution: (0, 0, 0)

if $z=0$, $x$ and $y$ must be 0.

if $z \ne 0$, go to $\mathbb{Q}[\sqrt2]$ and prove $x^2 +y^2 = \sqrt2$ has no solution in $\mathbb{Q}[\sqrt2]$

It is trivial: $a, b, c,d \in \mathbb{Q}, (a+b\sqrt2)^2 + (c+d\sqrt2)^2 = \sqrt2$ => $a^2 + 2b^2 + c^2 + 2d^2 = 0$ => $a=0, b=0, c=0, d=0$

So, (*) has only trivial solution (0,0,0)