mersenneforum.org - View Single Post

MatWur-S530113 · 2007-06-25, 21:12

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

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