Puzzling Equation: x^y=y^x - Are There Solutions?

[Ynaught?]

Is there a proof that shows that there are no positive integers, other than 2 and 4, that satisfy the equation x^y=y^x?

Thanks

 

[Pere Callahan]

Just take x=y and you have infinitely many solutions

So the best you can hope for is to find a proof for your assertion under the additional constraint that x and y be distinct.

 

[ramsey2879]

Is there a proof that shows that there are no positive integers, other than 2 and 4, that satisfy the equation x^y=y^x?

Thanks

Try using the prime factorization of x and y for a start

 

[Ynaught?]

Thanks. By converting the equation to ln(x)/x = ln(y)/y I think I have it figured out.

 

[al-mahed]

Is there a proof that shows that there are no positive integers, other than 2 and 4, that satisfy the equation x^y=y^x?

Thanks

I think on this proof:

consider y > x

$$x^y = y^x ==> y = kx ==> x^y = (kx)^x ==> x^{y-x} = k^x ==> x^{x(k-1)} = k^x ==> x^{k-1} = k$$

as $$x^{k-1} = k$$ is true for x > k if and only if k = 1, and k = 1 ==> x=y, this is a contradiction with our initial consideration that y > x, then k >= x

by $$x^{k-1} = k$$ it is easy to see that k and x have exactly the same prime numbers as factors.

proof: k >= x, supose that there is one factor in k not in x, so let's write $$k = w*x^n ==> x^{k-1 -n} = w$$ that is true if and only if w = 1 and n = k - 1

as x and k have exactly the same factors $$x^{k-1} = k$$ is true if and only if x=k ==> $$x^{x-1} = x ==> x = 2$$

x = k = 2, as y = kx, then y = 4 and the proof is finish