# How to solve $x^2-4y=m^2$ where $m$ is given?

Respected all

Kindly help me to solve the following diophantine equation.

The equation is given by $x^2-4y=m^2$ where $m\in \mathbb Z$ is given. How to solve this equation in integers?

I have read Pell's equation but not sure how to solve this equation . Please show me the path

Thank you

Note that $x$ should have the same parity as $m$. For those $x$, $(x,y) = (x, \frac{x^2-m^2}{4})$ is a solution. Of course, these are all solutions.