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.

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    $\begingroup$ Can a factorial be a perfect square? $\endgroup$ – 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.

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    $\begingroup$ Just want to note that he used Bertrand's postulate. Very nice argument. $\endgroup$ – BeaumontTaz Oct 7 '14 at 18:25
  • $\begingroup$ Beautiful; thank you very much! $\endgroup$ – Ahaan S. Rungta Oct 7 '14 at 18:43
  • $\begingroup$ 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 $k$? $\endgroup$ – TonyK Oct 7 '14 at 18:57
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    $\begingroup$ @TonyK, never used $n!!$ for even $n$ before. The definition seems to say $(2n)!! = 2^n n!$ Then the same argument applies to $n!$ $\endgroup$ – Will Jagy Oct 7 '14 at 19:00

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