Factoring Factorial

It's easy to check that $n=3$ is the only solution $.$ Let $n!+10=j^{2}$ for some integer $j$

Note that, $n!+10$ is always even for $n > 1,$ which implies $j$ is also even for $n > 1$

Again note that, if $n\geq4,\:$ then $n! \equiv 0\: (mod\: 4)$ and $10 \equiv 2\: (mod\: 4).$ Which implies $j^{2} \equiv n!+10 \equiv 0+2 \equiv 2 \: (mod\: 4)$

But we know, $j^{2} \equiv 0\: (mod\: 4)\: [as\: j\: is\: even]$

Here we have a contradiction as $0\not\equiv2\:(mod\:4),$ which proves that for $n\geq4,$ the value of $n!+10$ can't be a perfect square.

Now we are left with $n=0,1,2\: and\: 3.$ By Trial and Error it's easy to check that $n=3$ is the only solution.

Thus the answer is $1+1=\boxed{2}$

For the first few integers, we can check

1! + 10 = 11, NO

2! + 10 = 12, NO

3! + 10 = 16, YES YES YES YES

4! + 10 = 34, NO

After this, we can change n! + 10 to

2(5 + n!/2)

But after 4! every factorial will yield at least a zero at the end. Therefore, it will look something like

2(.......5), so we will never get another 2 to make it a perfect square. Hence, there are no more such integers. And, thus the answer is 1 + 1 = $\boxed 2$

I would appreciate Tasmeem's solution because I didn't get it!!! ;)