Can $ x^2-4a^n$ be a square?

Question

Let $ x,y,a,n \in {Z}>1$, Does the diophantine equation $ x^2-4a^n=y^2$ have any solutions? or How can I solve this?

Answer 1

Sure, let $x=a^n+1$ and $y=a^n-1$. There are many other solutions if $n$ is large and/or $a$ has many factors.

One can get all solutions in integers $x,y$ positive, negative, or $0$ by letting $x+y=2u$, $x-y=2v$ where $uv=a^n$, and solving for $x$ and $y$.

Answer 2

Note that if $n=2$ you can rewrite as $x^2=y^2+(2a)^2$ - which essentially means you are looking for a pythagorean triple since if positive integers satisfy $x^2=y^2+z^2$ at least one of $y, z$ must be even.