So far I'm just starting on this, and I'm suspecting that there is no N>1.
If there is such an N, then it must be true that:
2^N - 1 = k*N where k is an integer
Also, 2^N - 1 in binary is a series of N 1's, which suggests that
2^N - 1 is divisible by 2^m - 1 if N is divisible by m
Not sure if this helps.
Now I'll pause and wait for someone else to present an elegant proof.
Posted by Larry
on 2004-12-11 17:03:14