Question

The number of positive integers $n$ for which $n^2+96$ is a perfect square is

• A. $8$
• B. $12$
• C. $4$
• D. Infinite

Answer 1

The correct option is C $4$

Let $k^2=n^2+96$ , where $n,k$ is a positive integer

$\Rightarrow k^2-n^2=96$

$\Rightarrow (k-n)(k+n)=96$

Both $k,n$ must have the same parity. Hence we will try to find two factors of $96$ such that both are even.

$96$ can be factorized as $2\times 48,4\times 24,6\times 16,8\times 12$

The smaller factor $=k-n$ and the bigger factor $=k+n$

Hence, the values of $(n,k)$ are $(23,25),(10,14),(5,11),(2,10)$

Hence, there are $4$ possible values of $n$