Numbers that makes $96$ are $\cdots$

Since we have,

$\large n^2 +96$ is perfect square.

Let the perfect square number be $\large p^2$ . Then $\large p^2 = n^2 +96 = p^2 -n^2 = 96 = (p+n)(p-n) = 96$ $\large 96$ can be written in the following ways

$\large (p+n)(p-n) = 96\times1$

$\large \Rightarrow p = \frac{97}{2}$ and $\large n = \frac{95}{2}$

$\large (p+n)(p-n) = 48\times2$ $\large \Rightarrow p = 25$ and $\large {\color{#3D99F6}n =23 \in N }$

$\large (p+n)(p-n) = 32\times3$

$\large \Rightarrow p =\frac{35}{2}$ and $\large n = \frac{29}{2}$

$\large (p+n)(p-n) = 24\times 4$

$\large \Rightarrow p = 11$ and $\large {\color{#3D99F6}n = 10 \in N}$

$\large (p+n)(p-n) = 16\times 6$

$\large \Rightarrow p = 11$ and $\large {\color{#3D99F6}n = 5 \in N}$

$\large (p+n)(p-n) = 12\times 8$ $\large p = 10$ and $\large {\color{#3D99F6}n = 2 \in N}$

$\large \Rightarrow S = 2 +10+23+5$

Therefore,

$\large S +6 = 40+6 = \boxed{46}$

$n^2<n^2+96$ $\normalsize$ Since $n^2+96<(n+10)^2=n^2+20n+100$ , $n^2+96=\begin{cases} (n+1)^2 \\ (n+2)^2 \\ (n+3)^2 \\ \vdots \\ (n+9)^2 \end{cases}$

$\normalsize$ Let's take a look at the options.

$\large n^2+96=(n+1)^2$ $\begin{aligned} n^2+96 & = (n+1)^2 \\ & = n^2+2n+1 \\ 96 & = 2n+1 \\ 95 & = 2n \\ n & =\dfrac{95}{2} \end{aligned}$

$\normalsize$ Since $\dfrac{95}{2} \not \in N$ , this case doesn't give any solution.

$\large n^2+96=(n+2)^2$ $\begin{aligned} n^2+96 & = (n+2)^2 \\ & = n^2+4n+4 \\ 96 & = 4n+4 \\ 92 & = 4n \\ n & =23 \end{aligned}$

$\normalsize$ Thus the unique solution in this case is $23$ .

$\large n^2+96=(n+3)^2$ $\begin{aligned} n^2+96 & = (n+3)^2 \\ & = n^2+6n+9 \\ 96 & = 6n+9 \\ 87 & = 6n \\ n & =\dfrac{29}{2} \end{aligned}$

$\normalsize$ Since $\dfrac{29}{2} \not \in N$ , this case doesn't give any solution.

$\large n^2+96=(n+4)^2$ $\begin{aligned} n^2+96 & = (n+4)^2 \\ & = n^2+8n+16 \\ 96 & = 8n+16 \\ 80 & = 8n \\ n & =10 \end{aligned}$

$\normalsize$ Thus the unique solution in this case is $10$ .

$\large n^2+96=(n+5)^2$ $\begin{aligned} n^2+96 & = (n+5)^2 \\ & = n^2+10n+25 \\ 96 & = 10n+25 \\ 71 & = 10n \\ n & =\dfrac{71}{10} \end{aligned}$

$\normalsize$ Since $\dfrac{71}{10} \not \in N$ , this case doesn't give any solution.

$\large n^2+96=(n+6)^2$ $\begin{aligned} n^2+96 & = (n+6)^2 \\ & = n^2+12n+36 \\ 96 & = 12n+36 \\ 60 & = 12n \\ n & =5 \end{aligned}$

$\normalsize$ Thus the unique solution in this case is $5$ .

$\large n^2+96=(n+7)^2$ $\begin{aligned} n^2+96 & = (n+7)^2 \\ & = n^2+14n+49 \\ 96 & = 14n+49 \\ 47 & = 14n \\ n & =\dfrac{47}{14} \end{aligned}$

$\normalsize$ Since $\dfrac{47}{14} \not \in N$ , this case doesn't give any solution.

$\large n^2+96=(n+8)^2$ $\begin{aligned} n^2+96 & = (n+8)^2 \\ & = n^2+16n+64 \\ 96 & = 16n+64 \\ 32 & = 16n \\ n & =2 \end{aligned}$

$\normalsize$ Thus the unique solution in this case is $2$ .

$\large n^2+96=(n+9)^2$ $\begin{aligned} n^2+96 & = (n+9)^2 \\ & = n^2+18n+81 \\ 96 & = 18n+81 \\ 15 & = 18n \\ n & =\dfrac{15}{18} \end{aligned}$

$\normalsize$ Since $\dfrac{15}{18} \not \in N$ , this case doesn't give any solution.

---

Therefor the answer is $S+6=23+10+5+2+6=\boxed{46}$ .