Diophantine equation $x^2+101^2y^2=2z^2$

Question

I tried find solutions in integers to the equation $$x^2+101^2y^2=2z^2$$ and I believe there doesn't exist one, but I keep missing how to prove it. I tried Looking square residue and similar, but haven't managed to find the proof. Any tip will help.

Answer 1

Notice that $101$ is a prime.

There are two cases:

$1)$ If $101\nmid z$, then $x^2\equiv 2z^2\pmod{101}$, $(x\cdot z^{-1})^2\equiv 2\pmod{101}$ (modulo multiplicative inverse).

But $101$ is a prime of the form $8k+5$, so this congruence/equivalence has no solutions by Quadratic Reciprocity.

$2)$ If $101\mid z$, then $101\mid x^2$, by Euclid's lemma $101\mid x$.

Let $x=101a$, $z=101b$, $a,b\in\mathbb Z$.

$a^2+y^2=2b^2$. This equation has infinitely many different integer solutions. This equation is in this Math.SE question with answers.