If p is prime then 2^{p}  2 is divisible by 2p. So if q divides 2^{p}  1, gcd(q1, 2^{p}  2) is divisible by 2p.
However, 2^{n}  2 is divisible by 2 but not by 4 if n > 1. If p is prime and the smallest prime factor q of 2^{p}  1 is congruent to 1 (mod 4), q1 does not divide 2^{p}  2.
Consulting a table of factorizations of 2^{n}  1, I find that p = 29 (already noted) is the smallest p with this property. The next is p = 53 (q = 6361).
