0
$\begingroup$

Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$

How to prove that $(2,4)$ and $(4,2)$ are the only solutions of Diophantine equations ${x^y} = {y^x}$ for $x \ne y$?

  • 0
    Hah! Sorry, I can't read.2012-10-01

1 Answers 1

3

Taking logarithms:

$x\ln y=y\ln x$ $\frac{\ln y}y=\frac{\ln x}x$

For this to be true, the function has to take the same value at two different locations. Take the function

$f(x)=\frac{\ln x}x$ $f'(x)=\frac{1-\ln x}{x^2}$

It has a maximum at $x=e$, is decreasing for $x>e$ and increasing for $x. So if two values are equal, one has to be greater than $e$ and the other must be less. So for the lower value we only have $x=1,2$ as options. Our problem amounts to proving that there is no other number that gives the same value as $x=1$. Since $f(1)=0$, and $\ln(x)$ only has a single root at $1$, we know this can't happen, so we're done.

  • 0
    I used a very similar argument, but I was not convinced. Thanks.2012-10-01