2022/047) Solve in integers $y^2 - 2001 = n!$

 We have 2001 is divisible by 3 and not 9. And if n Is above 5 then n! is divisible by 9. So $2001 + n!$ is divisible by 3 but not 9. so it cannot be a perfect square. so checking from 0 to 5 we get y = 45 and n =4