(In reply to
re: unique solution? by Dennis)
In fact, I can prove that you only need to consider even m and n.
if 2^m+3^n = x^2
x^2 cannot be divisible by 2 or 3 so x cannot be divisible by 2 or 3, so either x=6y-1 or 6y+1. Thus x^2 = 36y^2-12y+1 or 36y^2+12y+1.
in either case x^2 mod 12 = 1
lets look at 2^m mod 12 and 3^n mod 12.
m/n : 1, 2, 3, 4, 5, ...
2^m mod 12 : 2, 4, 8, 4, 8, ...
3^n mod 12 : 3, 9, 3, 9, 3, ...
The only combination that = 1 mod 12 is 4 + 9 i.e. even m and even n. So, I think pythagorean triple analysis is justified as the only possibilities.
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Posted by Joel
on 2006-12-18 23:15:23 |
re(2): unique solution?
