Find all values of $n$ for which $n^2+96$ is a perfect square

Question

There can be infinite values of $n$. The statement is true for $n=2,5$. How to find out others? Please tell if there is any formula.

Answer 1

$$ n^2+96 = m^2 \tag{1}$$ is equivalent to: $$ 96 = (m-n)(m+n) \tag{2}$$ Now notice that $(m-n),(m+n)$ have the same parity and $96=3\cdot 2^5$, so $$ ((m-n),(m+n))\in\left\{(2,48),(4,24),(6,16),(8,12)\right\} \tag{3}$$ leads to: $$ (n,m)\in\left\{(23,25),(10,14),(5,11),(2,10)\right\}. \tag{4}$$

Answer 2

As $n^2+96=(-n)^2+96,$ if $n_1$ is a solution, so will be $-n_1$ and vice versa

So, we can safely start with $n>0$

Let $n^2+96=(n+a)^2\iff a(2n+a)=96$

If $a$ is odd, so will be $a(2n+a)$

So, $a$ must be a even $=2c$(say) $\implies n^2+96=(n+2c)^2\iff c(c+n)=24\ \ \ \ (1)$

If the roots of $c^2+cn-24=0$ are $c_1,c_2;$ using Vieta's formula, $c_1\cdot c_2=-24$

So, the signs of $c_1,c_2$ are opposite

WLOG $c_1>0$ and $c_1^2<c_1(c_1+n)=24\implies c_1^2<24\implies0<c_1<5$

By $(1),c_1+n=\dfrac{24}{c_1}\iff n=\dfrac{24}{c_1}-c_1$

If $c_1=1, n=23$

If $c_1=2, n=10$

If $c_1=3, n=5$

If $c_1=4, n=2$