Argentine National Olympiad 2015

The answer is $a=1,2,4$. For all $n\in \mathbb{N}$ we have $$n^2<n(n+1)<n(n+2)<(n+1{)}^2,n^2<n(n+4)<(n+2{)}^2,n(n+4)\ne (n+1{)}^2.$$ Hence $n(a+n)$ is never a perfect square for $a\in \{1,2,4\}$.

On the contrary, for each $a\ne 1,2,4$ there is an $n\in \mathbb{N}$ such that $n(a+n)$ is a perfect square. Suppose first that such an $a$ is a power of $2$. Hence $a$ is divisible by $8$ since $a\ne 1,2,4$; let $a=8k$. To obtain $n(a+n)$ as a perfect square it suffices to take $n=k$, because then $n(a+n)=k(8k+k)=(3k{)}^2$.

If $a$ is not a power of $2$, it has an odd prime divisor greater than $1$; let $a=(2k+1)l$ with $k\ge 1,l\ge 1$. Take $n=k^{2}l$ to obtain $n(a+n)=k^{2}l((2k+1)l+k^{2}l)=k^2{l}^2(k^2+2k+1)=(kl(k+1){)}^2$.