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 2007-06-25, 21:12 #2 MatWur-S530113     Apr 2007 Spessart/Germany 2·34 Posts hmhmhmhm, something I made wrong. If $n=2^k+2$ then n is even, then n-1 is odd, then n-1 is not divisible by 2 as needed for $((n-1)/2)^2$. I made the division with a ShR-order, so the last bit was ignored. A corrected form of the first conjecture is: if and only if $n^{n-1}$ is divisible by $(floor((n-1)/2))^2$ then there is a non-negative Integer k with $n=2^k+2$ mfg Matthias