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2007-06-27, 05:35   #5
maxal

Feb 2005

22×32×7 Posts

Quote:
 Originally Posted by MatWur-S530113 Many thanks for your reply. But 2 things I don't understand: 1) if $0\equiv 2^{2m+1}\pmod{m}$ why does this imply that $m=2^k$ for some k? (Probably I miss an easy thought...)
Because $0\equiv 2^{2m+1}\pmod{m}$ means that m divides a power of 2, namely, $2^{2m+1}$. Therefore, $m=2^k$ for some k.

Quote:
 Originally Posted by MatWur-S530113 2) if n is even I still don't see why n must have the form $n=2^k+2$, if $n^{n-1}$ is divisible by $m^2=(\frac{n-2}{2})^2$ (not only by $m=\frac{n-2}{2}$).
If $n^{n-1}\equiv 0\pmod{m^2}$ then $n^{n-1}\equiv 0\pmod{m}$, and I showed that the latter congruence implies n=2^(k+1)+2 for some k.