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Old 2007-06-27, 05:35   #5
maxal
 
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Feb 2005

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Quote:
Originally Posted by MatWur-S530113 View Post
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 View Post
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.
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