Baltic Way shortlist

Answer: $n={\left(\frac{p+1}{2}\right)}^2$. Assume that $\sqrt{n^2-pn}=m$ is a positive integer. Then $n^2-pn-m^2=0$, and hence $$n=\frac{p\pm \sqrt{p^2+4m^2}}{2}.$$ Now $p^2+4m^2=k^2$ for some positive integer $k$, and $n=\frac{p+k}{2}$ since $k>p$. Thus $p^2=(k+2m)(k-2m)$, and since $p$ is prime we get $p^2=k+2m$ and $k-2m=1$. Hence $k=\frac{p^2+1}{2}$ and $$n=\frac{p+\frac{p^2+1}{2}}{2}={\left(\frac{p+1}{2}\right)}^2$$ is the only possible value of $n$. In this case we have $$\sqrt{n^2-pn}=\sqrt{{\left(\frac{p+1}{2}\right)}^4-p{\left(\frac{p+1}{2}\right)}^2}=\frac{p+1}{2}\sqrt{{\left(\frac{p^2+1}{2}\right)}^2-p}=\frac{p+1}{2}\cdot \frac{p-1}{2}.$$