x!!=y!, any integer solutions but trivial ones? Question says it all, are there any non trivial solutions? Wolfram gives those:
x=
But I wonder, if in fact there aren't any, is there any way we can prove that? I ask only becuase I am curious. Thank you.

No there are not.  If $x$ is odd, then $x!!$ is odd so you only have $y=0$ or $y=1$ and $x=1$.  If $x$ is even, say $x=2k$, then $x!!=2^k\cdot k!$.  So that means that $k<x$.  But then $y!=2^k\cdot k!$ implies that $k+1, k+2, \ldots y$ have only factors of $2$, which is impossible because $2^k$ is too large.  So you only have the trivial solution $2!!=2!$.
