How to solve the equation $‎x^2-2y=z^2$? Consider the following equation
$$‎x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can help me. Thanks in advance(
I tried to rewrite the equation as follows
\begin{align}
2y=x^2-z^2 ‎\Rightarrow 2y=(x-z)(x+z)‎\Rightarrow y=\frac{(x-z)(x+z)}{2}
\end{align}
but it did not work).


Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bounds: $-10\le x\le10$, $-10\le y\le10$, $-10\le z\le10$, and $x$, $y$, $z$ are all integers with $xy\ne0$.

I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{x^2 - 2*y == z^2, 
   x*y != 0 && -10 <= x <= 10 && -10 <= y <= 10 && -10 <= z <= 
     10}, {x, y, z}, Integers]]

Running the code gives:
Out[1]={{x -> -6, y -> 10, z -> -4}, {x -> -6, y -> 10, z -> 4}, {x -> -5, 
  y -> 8, z -> -3}, {x -> -5, y -> 8, z -> 3}, {x -> -4, y -> -10, 
  z -> -6}, {x -> -4, y -> -10, z -> 6}, {x -> -4, y -> 6, 
  z -> -2}, {x -> -4, y -> 6, z -> 2}, {x -> -4, y -> 8, 
  z -> 0}, {x -> -3, y -> -8, z -> -5}, {x -> -3, y -> -8, 
  z -> 5}, {x -> -3, y -> 4, z -> -1}, {x -> -3, y -> 4, 
  z -> 1}, {x -> -2, y -> -6, z -> -4}, {x -> -2, y -> -6, 
  z -> 4}, {x -> -2, y -> 2, z -> 0}, {x -> -1, y -> -4, 
  z -> -3}, {x -> -1, y -> -4, z -> 3}, {x -> 1, y -> -4, 
  z -> -3}, {x -> 1, y -> -4, z -> 3}, {x -> 2, y -> -6, 
  z -> -4}, {x -> 2, y -> -6, z -> 4}, {x -> 2, y -> 2, 
  z -> 0}, {x -> 3, y -> -8, z -> -5}, {x -> 3, y -> -8, 
  z -> 5}, {x -> 3, y -> 4, z -> -1}, {x -> 3, y -> 4, 
  z -> 1}, {x -> 4, y -> -10, z -> -6}, {x -> 4, y -> -10, 
  z -> 6}, {x -> 4, y -> 6, z -> -2}, {x -> 4, y -> 6, 
  z -> 2}, {x -> 4, y -> 8, z -> 0}, {x -> 5, y -> 8, 
  z -> -3}, {x -> 5, y -> 8, z -> 3}, {x -> 6, y -> 10, 
  z -> -4}, {x -> 6, y -> 10, z -> 4}}

So, using these bounds I get $36$ number of solutions. Extending the bounds to $-10^3$ and $10^3$ gives $12716$ number of solutions. Extending the bounds, again, to $-10^4$ and $10^4$ gives $173364$ number of solutions.
