View Single Post
Old 2007-06-25, 21:12   #2
MatWur-S530113
 
MatWur-S530113's Avatar
 
Apr 2007
Spessart/Germany

2×34 Posts
Default

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
MatWur-S530113 is offline   Reply With Quote