Squares is greater or factorial?

For $n \geq 4$

$x^2=n!+10 \equiv 2 \pmod{4}$

But $x^2 \equiv 0,1 \pmod{4}$

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This implies $n<4$

Checking for $n=1,2,3$ we get the only solution for $n=3$ .

So the answer is $\boxed{1}$

For n>3 , n!+10 is a multiple of 2 , hence for it to be a perfect square it should have an even power of 2 .

But for n>6 , powers of 2 in n! add up to 16 or so. Thus for n>6 , n!+10 has only power of 2 i.e. 1

[We demand $x \times 2^{y}+10$ to be even power of 2 , which is not satisfied if $2^{y}>10$ ]

In other words, for if n>6 , n!+10 is divisible by 2 and not by 4 , so not a square.

Hence, we end up searching n between 1 and 6 .

"3" satisfies the condition.

Hence, the answer is $\boxed{1}$ .