Thread: January 2017
View Single Post
Old 2017-02-17, 01:20   #4
Jan 2017

79 Posts

It's pretty easy to directly show starting from the question rationals of that form are answers and are the only rational answer:

x^y = y^x
Let 'a' be such that x^a = y, and replace y by that in above
x ^ x ^ a = x ^ (a*x)
Take x-based logarithm
x ^ a = a*x
Since x^a = y, this means a = y/x, thus rational too
Let 'b' be such that a = x^b
x ^ x ^ b = x^b * x
Take x-based logarithm
x^b = b + 1
x^b equals a, thus b = a-1 and rational
Let x = p/q, b = n/m where p,q,n,m integers
(p/q)^(n/m) = (n+m)/m
If the value on the left is rational (as per above it equals a and must be rational for a valid solution), it must be n:th power of one. Thus the value on the right must be too. So (n+m) and m are n:th powers. But it's not possible for two integers n apart (m and n+m) to both be n:th powers unless n=1. Thus:
(p/q)^(1/m) = (1+m)/m
p/q = ((1+m)/m)^m
And this gives a solution for any integer m.

Last fiddled with by uau on 2017-02-17 at 01:20
uau is offline   Reply With Quote