Question 15

The number of all integers n for which $$n^{2} + 96$$ is a perfect square, is

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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