# How does one show that $(n-2)! = 2^{m-1} m! (n - 2m)!$ has only one solution for $n\ne 6$?

The obvious solution is 1, for $n=6$ there is another one - $m=3$. How does one show that for other $n$ there are no solutions but $m=1$?

This is to show that for $n\ne 6$ all automorphisms of $S_n$ are inner.

Thank you.

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Should $2^m$ be $2^{m-1}$? –  Will Orrick Apr 21 '12 at 20:28
If $n=2$, then $m=0$ is a solution. On the other hand, $m=0$ is not a solution for any larger $n$, and I suppose we want $1\le m\le n/2$.
Rewrite the equation to be solved as $$\frac{(n-2)!}{(n-2m)!(2m-2)!}=\frac{2^{m-1}m!}{(2m-2)!}$$ which is equivalent, when $m>1$, to $$\binom{n-2}{2m-2}=\frac{m}{(2m-3)!!}.$$ The left side is an integer for all valid $m$, but the right side is an integer only for $m=2$, $3$. We can rule out $m=2$ since then we must have $\binom{n-2}{2}=2$ which has no solution. Hence the only possible solution is $m=3$ which necessitates that $\binom{n-2}{4}=1$ or $n=6$.