Given that each of x, y and z is a positive integer, it is observed that the equation 5^x + 12^y = 13^z is satisfied for x=y=z=2.

Does there exist any further solution?

It seems not.

There are two famiilies of solutions in which any power of 5 can (ultimately) appear:

1. z is even, say 2p, then 13^(2p)-12^(2p-5*4q)=r*5^x, once all the powers of 5 are collected in the first term on RHS.

2. z is odd, say (2p+1), then 13^(2p+1)-12^((2p+1)-5*(4q-2))=r*5^x, once all the powers of 5 are collected in the first term on RHS.

In either case, the multiple r is not itself a power of 5, unless x=y=z=2.

Edited on May 27, 2012, 1:04 pm

Posted by broll on 2012-05-27 13:01:20