Number of Integer Solutions for Perfect Square

Let $n^2 + 96 = k^2 \Rightarrow k^2 - n^2 = 96 \Rightarrow (k-n)(k+n) = 96$. Since $96$ is even, both $(k-n)$ and $(k+n)$ must be even. Factor pairs of 96 into two even factors: $(2, 48), (4, 24), (6, 16), (8, 12)$. Each pair gives 2 solutions for $n$ (positive and negative). Thus $4 \times 2 = 8$ integer values of $n$.