Re: Mersenne: Factoring

[Anonymous]

> Now, this implies that for no 0<N<2kp, 2^N=1 mod (2kp + 1)

That depends on whether 2 is a primitive root of 2kp+1. If 2kp+1=11, p=5, 2 is
a primitive root and 2^p is -1. If 2kp+1=7, p=3, it isn't, and 2^p is 1.

phma
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