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?
 A: 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. 
A: You are given
$$n(n+1) = p^2 \tag{1}\label{eq1}$$
with $n \in \mathbb{N}$ (I'm assuming positive integers is meant here). Note that $n^2 \lt n^2 + n = n(n+1) \lt n^2 + 2n + 1 = (n+1)^2$. Thus, $n \lt p \lt n + 1$, so there are no $p \in \mathbb{N}$ that solve \eqref{eq1}.
A: 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 $n=-1$.
A: 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.
