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2007-06-26, 10:58   #3
maxal

Feb 2005

FC16 Posts

Quote:
 Originally Posted by MatWur-S530113 if and only if $n^{n-1}$ is divisible by $(floor((n-1)/2))^2$ then there is a non-negative Integer k with $n=2^k+2$
Consider two cases:

1) n is even then $m=floor((n-1)/2)=\frac{n-2}{2}$.
We have n=2m+2 and $0\equiv n^{n-1}\equiv (2m+2)^{2m+1}\equiv 2^(2m+1)\pmod{m}$, implying that $m=2^k$ for some k. Therefore, n=2*2^k+2=2^(k+1)+2.

2) n is odd then $m=floor((n-1)/2)=\frac{n-1}{2}$.
We have n=2m+1 and $0\equiv n^{n-1}\equiv (2m+1)^{2m}\equiv 1\pmod{m}$, implying that $m=1$ and $n=3=2^0+2$.