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

(2^(m/2))^2 + (3^(n/2))^2 = c^2 --> 2^(m/2), 3^(n/2) and c are a pythagorean triple. So let 2^(m/2) = 2pq and 3^(n/2)= p^2 - q^2. --> p is a power of 2 (say p=2^k) and 3^(n/2)=(p-q)(p+q). So let p-q=3^v and p+q=3^w. It follows that p=(3^v + 3^w)/2. So 2^(k+1) = 3^v + 3^w. Both v and w cannot be one or greater since in those cases 3 would be a factor of 2^(k+1). Either (v=w=0) or (w=1 and v=0). The first case --> p=1 and q=0 --> n=0 so this case is not viable. Finally w=1 and v=0 --> p=2 and q=1 --> m=4 and n=2 only.

Posted by Dennis on 2006-12-18 21:37:51