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^(2p5*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*(4q2))=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 20120527 13:01:20 