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

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The equation x^y = y^x has positive rational solutions, with x = y being a trivial solution. A potential non-trivial solution is x = 2 and y = 4, as both satisfy the equation (2^4 = 4^2). The discussion raises the challenge of proving whether these are the only solutions. Participants express uncertainty about finding all solutions and reference previous discussions on the topic. The quest for a comprehensive proof remains a focal point in the conversation.
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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.
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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