Calculator Killer

Note that $7!\times 5! \times 3! = 10!$ so the equation became $\dfrac{x!}{10!} = x^2 - 12$ Now the first member is integer so $x \geq 10$ . Note that $x = 10$ is not a solution so we can be sure that $x \geq 11$ . Grouping this way $x \left( \dfrac{(x-1)!}{10!} - x\right) = -12$ we have the member in parenthesys is integer (couse $x-1 \geq 10$ ) and so $x$ is a (positive) divisor of twelve. Since it must be greater that $10$ the only divisor of twelve greater than $10$ is $12$ itself. Necessarily $x = 12$ . Since (by mere prove) $12$ is also a solution, we found it is indeed the only solution. :)