What are the solutions for $ n(n+1)=p^2$ for n belongs to $N$ The following is my approach:

$ n^2+n =p^2$
$   n^2+n+\frac{1}{4} = p^2 + \frac{1}{4}$
$   (n+\frac{1}{2})^2 = p^2+\frac{1}{4}$
$    (n+\frac{1}{2}-p) (n+\frac{1}{2}+p) = \frac{1}{4}$

I am not able to proceed further from here. Any suggestion on what to do next?

Note that $\gcd(n, n+1) = \gcd(n, 1) = 1$ do eithet $n=p^2, n+1 = 1$ or $n+1 = p^2, n = 1$;  both of which are impossible. 
