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maxal · 2007-06-27, 05:35

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.