, 2015

The answer is $a = 1, 2, 4$. For all $n \in \mathbb{N}$ we have
$$n^2 < n(n+1) < n(n+2) < (n+1)^2, \quad n^2 < n(n+4) < (n+2)^2, \quad n(n+4) \neq (n+1)^2.$$
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 $n = k^2l$ to obtain $n(a+n) = k^2l((2k+1)l+k^2l) = k^2l^2(k^2+2k+1) = (kl(k+1))^2$.