X^Y = Y^X

Date: 11/13/2002 at 19:38:47
From: Jeanella Clark
Subject: x^y=y^x

I'm in an anaysis class and as part of our assignments for the 
semester, we're supposed to find (with proof) all pairs of integers 
{x,y} that satisfy x^y=y^x. Obviously if x=y then the answer would be 
all integers, but when x does not equal y then the only pairs are 
{2,4} or {4,2} (not counting negative integers). I know this from 
inspection. How can I prove this? I've gotten as far as ln x/x=ln y/y.

Date: 11/14/2002 at 10:22:21
From: Doctor Floor
Subject: Re: x^y=y^x

Hi, Jeanella,

Thanks for your question.

If we consider the graph of f(x) = ln(x)/x, then we can observe the 
following:

* f(x)>0 if and only if x>1.

* the derivative of f(x) is f'(x) = (1 - ln(x)) / x^2, so that 
  f'(x) = 0 gives x = e. From that we conclude that f(x) is 
  increasing for 0<x<e and decreasing for x>e.

* Each horizontal line intersects the graph of f(x) twice.

* We know that f(2)=f(4).

* For any integer n>4 the horizontal line y=f(n) is "lower" than 
  the line y=f(2) and thus intersects the graph of f(x) as second 
  time for some 1<x<2, so that there is no integer m with f(m)=f(n).

* In the same way the line y=f(3) intersects the graph of f(x) a
  second time for some 2<x<e, so that there is no integer m with
  f(m)= f(3).

This yields that there are no unequal positive integer solutions other 
than 2^4=4^2.

When we started with x/ln(x) = y/ln(y), we restricted ourselves to 
positive x and y (by definition of ln). But the original question does 
not exclude negative solutions. It is easy to show that there are no 
solutions with one positive and one negative integer. Also, when 
x^y = y^x, then (-x)^(-y) = +/- (-y)^(-x).... 

Can you finish?

See also:

   Solutions to X^Y = Y^X - Dr. Math archives
   http://mathforum.org/library/drmath/view/53229.html

If you have more questions, just write back.

Best regards,
- Doctor Floor and Sarah, The Math Forum
  http://mathforum.org/dr.math/