<s>[INST] 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?
 [/INST] There are only $2$ integer solutions. Neither are greater than $0$.

Assume $n\ge0$.
Without loss of generality, $p\ge0$. Then since $n\ge0$, $p^2=n^2+n\implies p\ge n$. Note that
$$
n(n+1)=p^2\implies n=p^2-n^2\tag1
$$
If $p=n$, then $n=p^2-n^2=0$.
If $p\gt n$, then
$$
n(n+1)=p^2\implies n=\overbrace{\ (p-n)\ }^{\ge1}\overbrace{\ (p+n)\ }^{\gt2n}\gt2n\implies n\lt0\tag2
$$
Thus, $n\lt0$, which contradicts $n\ge0$.
Therefore, the only non-negative solution is $n=0$. Symmetry then says the only other integer solution is $n=-1$.
</s>