Croatia_2018

Let $a$ be a positive integer such that $np+1=a^2$. Note that $a>1$. $$\text{We have }{a}^2-1=np,\text{ i.e. }(a-1)(a+1)=np.$$ Since $p$ is prime, it follows that $p\mid a-1$ or $p\mid a+1$. We treat these two cases separately:

i) Let $p\mid a-1$, i.e. let $a=kp+1$ for some integer $k$. $$\text{Then }1+np=(kp+1{)}^2=k^2{p}^2+2kp+1,\text{ i.e. }n=k^{2}p+2k.$$ Now we have: $$n+1=k^{2}p+2k+1=k^2+2k+1+(p-1)k^2=(k+1{)}^2+(p-1)k^2.$$ Notice that $a>1$ implies $k>0$, so $k+1$ and $k$ are both positive integers.

ii) Let $p\mid a+1$, so that $a=kp-1$ for some integer $k$. Analogously to the first case, we get $n+1=(k-1{)}^2+(p-1)k^2$. It remains to show that $k-1$ is indeed a positive integer, i.e. that we have $k\ge 2$. If we assume the contrary ($k=1$), it follows that $n+1=p-1$, i.e. $n=p-2$. This contradicts the given condition that $n\ge p-1$. We conclude that the assertion holds for any choice of $n$ and $p$ satisfying the conditions of the problem.