1
So far I'm just starting on this, and I'm suspecting that there is no N>1.
Observations:
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 20041211 17:03:14 