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Let $\sqrt{n^2+pn}=m$, $m\in {\mathbb{N}}_0$. Then $n^2+pn-m^2=0$. The solutions of this quadratic equation are $$n=\frac{1}{2}\left(-p\pm \sqrt{p^2+4m^2}\right).$$ The solutions are integers, so $p^2+4m^2=w^2$, for some $w\in {\mathbb{N}}_0$. Since $w+2m\ge 0$ and $w+2m\ge w-2m$, from $p^2=(w-2m)(w+2m)$, we obtain the following possibilities: $$\{\begin{array}{l}w-2m=1\\ w+2m=p^2\end{array}\text{or}\{\begin{array}{l}w-2m=p\\ w+2m=p\end{array}$$ In the first case we obtain $w=\frac{p^2+1}{2}$, $m=\frac{p^2-1}{4}$, so it follows: $$\begin{array}{rl}n& =\frac{1}{2}\left(-p\pm \sqrt{p^2+4{\left(\frac{p^2-1}{4}\right)}^2}\right)\\ n_1& ={\left(\frac{p-1}{2}\right)}^2,n_2=-{\left(\frac{p+1}{2}\right)}^2.\end{array}$$ These numbers are integers only when $p$ is odd, i.e. only if $p>2$.

In the second case we obtain $w=p$, $m=0$, so it follows that $n=\frac{1}{2}(-p\pm p)$ or $n_3=0$ and $n_4=-p$.