Actually this fact is much easier to prove (once you correct it by adding 4 - 3 = 1). First you show that 3^(2^(n-1)) = d2^n + 1 where d is odd, from which it follows that 3 is a primitive root (mod 2^n). If 3^a = 2^b +/- 1 then 3^a == +/-1 (mod 2^b) => a >= 2^(b-2) => 3^(2^(b-2)) <= 3^a <= 2^b + 1, and this last inequality can only be satisfied for b <= 3. J.P. On Sun, 22 Dec 2002 asimovd@aol.com wrote:
This question remindes me of the news item that's been posted on the homepage of Eric Weisstein's MathWorld for a number of months:
<< A paper submitted by P. Mihailescu on April 18 [2002] allegedly proves Catalan's 158-year-old conjecture.
Am I right that this is the conjecture that the only positive integral powers of 2 and 3 that differ by 1 are 3 - 2 = 1 and 9 - 8 = 1 ???
Does anyone know whether this proof has been accepted by experts? If so, can someone please give a sketch of the proof?
Thanks,
Dan