I have a question about number theory:

> 
> How many pairs of positive integers $(x,y)$ satisfy the following equation? $$x^2 - 10! = y^2$$
> 
> 
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My attempt:

Move the $y^2$ from right to the left and 10! From left to the right such that:

$$x^2-y^2= 10!$$

$$(x-y)(x+y)=10!$$

$$(x-y)(x+y)= 10.9.8.7.6.5.4.3.2$$

Until this step, I think I will get many possibility to get the answer, such that:

$$(x-y)= 10.9$$ and $$(x+y)= 8!$$

Also

$$(x-y)= 10.9.8$$ and $$(x+y)= 7!$$

And another possibilities, like $$(x-y)=10.9.8.7$$ and $$(x+y)=6!$$

And so on until $$(x-y)=10.9.8.7.6.5.4.3.2$$ and $$(x+y)=1$$

And I think solving all the possibility is a bit tedious work. Can somebody help me to find a better solution to solve this kind of problem? 

Thanks