Prove that 2^T-3^U=1 has only two integer solutions: T=1 and U=0, and T=2 and U=1.

2^T-3^U=1 means 2^T-1+2^T-2+...+2+1=2^T-1=3^U = 2^U+...+2U+1 where the binomial theorem has been employed to get the last expression. But

2^T-1+2^T-2+...+2+1=2^U+...+2U+1

is impossible for U>1 since there will be a gap in the powers of 2 being summed on the rhs while no such gap appears in the powers of 2 being summed on the lhs, although I cannot yet give a rigorous proof.

Edited on July 29, 2005, 9:04 pm

Posted by Richard on 2005-07-29 20:56:26