Let 1<x=y.
Then a^z*a^z=z^z for some a,z, and if so, z=a^2:
a^(a^2)*a^(a^2)=(a^2)^(a^2);
so a^(a^2) = x^x; x is a multiple of a, say k*a
But (a^(a^2))^(1/(k*a)) = k*a yields a^(a/k) = ak, which does not produce nontrivial integer solutions. So x cannot equal y.
Along similar lines, let z^z=a^zb^z, with b exceeding a. Since z=ab,
a^(a^2+an)*(a+n)^(a^2+an) =(a^2+an)^(a^2+an); eligible candidates for z^z soon become large.
a^z and b^z cannot be decomposed so that a^z=x^x and b^z = y^y:
Say b^z=d^d; then for some x, b^x=d since every factor of d is in b. Now (a+n)^(a^2+an) = ((a+n)^x)^((a+n)^x)
Only the powers need be considered: (a(a+n)) = (x(a+n)^x); but then a = x(a+n)^(x1), and both a and x must = 1.
So if there is a solution at all, it involves switching factors between a^zb^z to produce x^xy^y; a restricted exercise considering that the latter both have to remain perfect powers after the transfer.
Edited on July 5, 2014, 3:25 am

Posted by broll
on 20140705 03:02:24 