rewrite the equation as 2^T - 1 = 3^U. Now, if T is even (say, 2*N),
the left side is the difference of squares and can be factored as (2^N
- 1)(2^N + 1) = 3^U. But the right side has only threes in its
factorization while the left side's two factors can't both even be
multiples of three let alone powers of three! (Their difference is 2 so
they have different residues mod 3.) The only hope for even T is if one
of the factors is 1 (making the other 3) which happens when T=2, and
corresponds to the T=2,U=1 solution above.
So, to summarize, for even T, T=2 is the only solution to get around
the "both factors are powers of three" issue, and from my earlier post,
for odd T, T=1 is the only solution to get around the "2^T when T is
odd = -1 mod 3" problem. So, there is exactly one even T and one odd T
solution, both given in the statement of the problem, and these are the
only possibilities.
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Posted by Paul
on 2005-07-30 08:39:10 |
the rest of the solution
