Given an positive integer $a$. Assume there are two positive integers $k$ and $n$ s.t. $(1) \;\; n(n + a) = k^2$. According to equation (1) there exists three positive integers $p,q,r$, where $p$ is squarefree, s.t. $(2) \;\; (n,n+a,k) = (pq^2,pr^2,pqr)$, yielding $a = (n + a) - n = pr^2 - pq^2$, i.e. $(3) \;\; a = p(r^2 - q^2)$. Applying formula (3) we obtain the following result: $(4) \;\; (p,r) = (1,q+1) \;\;\; \Rightarrow \;\;\; a = 2q + 1$ $(5) \;\; (p,r) = (2,q+1) \;\;\; \Rightarrow \;\;\; a = 4q + 2$ $(6) \; \; (p,r) = (1,q+2) \;\;\; \Rightarrow \;\;\; a = 4(q+1)$. According to the formulas (4)-(6) equation (1) has a solution for all natural numbers $a$ except for $a \in \{1,2,4\}$, which follows from the fact that $r^2 - q^2 \geq 2^2 - 1^2 = 4 - 1 = 3$, yielding $a=4$ implies $p=1$ and $(r - q)(r + q) = 4$, an equation which has no solution in positive integers. Conclusion: The only positive integer $a$ for which $n(n + a)$ is not a perfect square for every positive integer $n$, is $a \in \{1,2,4\}$.

If $a > 1$ is odd, just take $n = \left(\frac{a - 1}{2}\right)^2$ so $n(a + n) = \left(\frac{a^2 - 1}{4}\right)^2$. If $a = 2^\alpha x$ where $\alpha > 0$ and $gcd(2, x) = 1$ take $n = 2^\alpha y$ so $n(a + n) = 2^{2\alpha}y(x + y)$ however if $x > 1$ then taking $y = \left(\frac{x - 1}{2}\right)^2$ $\Longrightarrow n(a + n)$ $= 2^{2\alpha}y(x + y) =$ $\left(\frac{2^{\alpha} (x^2 - 1)}{4}\right)^2$. If $x = 1$ i.e. $a = 2^\alpha$, if $\alpha > 2$ let $n = (2^{\alpha - 2} - 1)^2$ $\Longrightarrow n(a + n) =$ $(2^{2(\alpha -2)} - 1)^2$ Now, checking $a \in \{ 1, 2, 4\}$ it's easy to see these numbers works.