Quarantine Integers!

$2n-3$ should be a prime. otherwise, there exists a prime $p|2n-3$ , such that $p \leq \frac{2n-3}{2} < n$ . So, that $p$ would satisfy $p|n!$ and, consequently $p \nmid 15(n!)^2+1$ .

Nest, we take $p=2n-3$ to be a prime.

$15 \cdot (\frac{p+3}{2}!)^2+1 \equiv 0 \ (mod \ p) \implies 15\cdot (\frac{p+3}{2}!)^2 \equiv -1 \ (mod \ p) \implies 15\cdot (\frac{p+3}{2})^2 \cdot (\frac{p+3}{2}-1)^2 \cdot (\frac{p-1}{2}!)^2 \equiv -1 \ (mod \ p)$

There is a theorem, its name I don't really know, that says for primes $p \geq 3$

$(\frac{p-1}{2}!)^2 \equiv (-1)^{\frac{p+1}{2}} \ (mod \ p)$

Hence, when $p\geq 3$

$15\cdot (\frac{p+3}{2})^2 \cdot (\frac{p+3}{2}-1)^2 \cdot (\frac{p-1}{2}!)^2 \equiv -1 \ (mod \ p) \implies 15\cdot (\frac{p+3}{2})^2 \cdot (\frac{p+3}{2}-1)^2 \cdot (-1)^{\frac{p+1}{2}} \equiv -1 \ (mod \ p)$

$\implies 15\cdot (\frac{p+3}{2})^2 \cdot (\frac{p+3}{2}-1)^2 \equiv (-1)^{\frac{p+3}{2}} \ (mod \ p)$

$\implies 15\cdot (\frac{p-1}{2}+2)^2 \cdot (\frac{p-1}{2}+1)^2 \equiv (-1)^{\frac{p+3}{2}} \ (mod \ p)$

note that

$\frac{p-1}{2}+2 \equiv -(\frac{p-1}{2}-1) \ (mod \ p)$

therefore,

$\implies 15\cdot (\frac{p-1}{2}-1)^2 \cdot (\frac{p-1}{2}+1)^2 \equiv (-1)^{\frac{p+3}{2}} \ (mod \ p)$

$\implies 15\cdot \big((\frac{p-1}{2})^2-1 \big)^2 \equiv (-1)^{\frac{p+3}{2}} \ (mod \ p)$

Multiplying both sides by $4^2$ is legal, since $2$ and $p$ are coprimes.

$\implies 15\cdot \big((p-1)^2-4 \big)^2 \equiv 4^2 \cdot (-1)^{\frac{p+3}{2}} \ (mod \ p)$

$\implies 15\cdot 9 \equiv 4^2 \cdot (-1)^{\frac{p+3}{2}} \ (mod \ p)$

if $p \equiv 1 \ (mod \ 4)$ , then

$p| 135 - 16= 119=7\cdot 17$

but only $17 \equiv 1 \ (mod \ 4)$ , which gives $n=\frac{p+3}{2} = 10$

if $p \equiv 3 \ (mod \ 4)$ , then

$p| 135 + 16=151$

$151 \equiv 3 \ (mod \ 4)$ is a prime and gives $n=\frac{p+3}{2} = 77$

Note that we have considered $p\geq 3 \implies n=\frac{p+3}{2} \geq 3$ . $n=1,2$ both make the fraction an integer.

So, the final sum is $1+2+10+77=90$