Is there any pair of natural numbers $\{ k, m \}$ satisfying: $4+(2k)! = m^2$?
I tried simplifying this into
$$ 2^k k!(2k-1)!! = (m-2)(m+2) \text , $$
where !! denotes the double factorial, i.e., $1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k-1) $.
I suppose that this equation has no solutions but could anyone please help me solve it or suggest how to approach it?