Determine analytically all possible pairs of positive integers (m,n) for which 2^m+3ⁿ is a perfect square.

(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.

Posted by Joel on 2006-12-18 23:15:23