Squares in Fibonacci-like sequences

Question

After the answers to my last question, I wonder if there is any sequence of the form $A_0=0$, $A_1=1$, $$A_{n+2}=pA_{n+1}-qA_{n}$$ where $p$ and $q$ are fixed integers which contains infinitely many different square numbers.

Answer 1

Put $q=0$ and you get infinitely many squares.

Or put $p=q$.

Answer 2

You can take $p=2,q=1$, then $A_n=n$ for all $n$, and it follows that it is a square infinitely often.

Answer 3

Hint

The characteristic equation being $$r^2=p~r-q$$ its roots are $$r_{\pm}=\frac{1}{2} \left(p \pm \sqrt{p^2-4 q}\right)$$ so the easiest way to get rid of radicals is to set $q=\frac{p^2}{4}$ and, in such a case, the solution is $$a_n=n\Big(\frac{p}{2}\Big)^{n-1}$$ and you can see that this will lead to an infinite number of squares if $\cdots$.