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Solution
Let $$n^2+96=k^2$$
$$k^2-n^2=96$$
$$\left(k+n\right)\left(k-n\right)=96$$
$$\left(k+n\right)\left(k-n\right)=2\times48,\ 4\times24,\ 6\times16,\ 8\times12,\ 12\times8,\ 16\times6,\ 24\times4,\ 48\times2$$
n can take 8 values, i.e. -23, -10, -5, -2, 2, 5, 10 and 23.
The answer is option C.
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