What Are the Positive Rational Solutions to x^y = y^x?

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Positive Rational
ehrenfest
Messages
2,001
Reaction score
1

Homework Statement


Determine all positive rational solutions of x^y=y^x.

Homework Equations


The Attempt at a Solution


Obviously, x=y will always work. I think that is the only solution. If I can show that x^y must be rational, I think it will be easy because then both x and y must have the same primes in the numerator and the denominator. I tried writing out x=r/s, y = t/u, and manipulating, which leads to
r^{ts} u^{ru} = s^{ts} t^{ru}
which is really not helpful.
 
Physics news on Phys.org
ehrenfest said:

Homework Statement


Determine all positive rational solutions of x^y=y^x.

The Attempt at a Solution


Obviously, x=y will always work. I think that is the only solution...

How about x = 2 , y = 4 ?
 
> I think that is the only solution.

What about 2^4 = 4^2...
 
Hmmm. Maybe that is the only exception. But can we prove that we have found them all...
 
This has been discussed in these forums a few times before, but I can't seem to find the threads. I did find this however.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top