# Can a double-factorial be a perfect square?

The title says it, basically. My question is $-$ for $n \ge 2$, can $n!!$ be a perfect square, where $!!$ represents the double-factorial? My conjecture is no, but I can't seem to be able to find a good proof for this.

• Can a factorial be a perfect square? – Inceptio Oct 7 '14 at 18:42

No, for odd $n,$ there is a prime between $(n-1)/2$ and $n.$ The exponent of this prime in factoring $n!!$ is one, that is, odd.
Edit. looking at even numbers and the definition, it appears $$(2n)!! = 2^n n!$$ in which case we ignore the exponent of $2$ and concentrate on $n!,$ which cannot be a square either for this $n \geq 3,$ also because of an odd prime. See here.
• I don't understand your dividing-out strategy. Are you claiming that if $n$ is even, then $n!!=2^mk!!$ for some integer $m$ and odd $k$? – TonyK Oct 7 '14 at 18:57
• @TonyK, never used $n!!$ for even $n$ before. The definition seems to say $(2n)!! = 2^n n!$ Then the same argument applies to $n!$ – Will Jagy Oct 7 '14 at 19:00