<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] You are actually very close to the answer. All you really need to do is multiply both sides of $(n+{1\over2}-p)(n+{1\over2}-p)={1\over4}$ by $4$ and then note that the only integer solutions of $ab=1$ is $a=b=\pm1$. Here is a complete presentation:
Assuming only $n,p\in\mathbb{Z}$, we have
$$\begin{align}
n(n+1)=p^2&\iff4n^2+4n=4p^2\\
&\iff(2n+1)^2=(2p)^2+1\\
&\iff(2n+1)^2-(2p)^2=1\\
&\iff(2n+1-2p)(2n+1+2p)=1\\
&\iff2n+1-2p=2n+1+2p=\pm1\\
&\iff p=0\land n\in\{0,-1\}
\end{align}$$
The requirement $n\in\mathbb{N}$ eliminates $n=-1$, leaving $(n,p)=(0,0)$ as the only solution.
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