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PostPosted: Wed, 2 Dec 2009 00:25:58 UTC 
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Solve the equation x^y=y^x in the set of the positive rational numbers x,y.

(It's easy to find the solutions (x,y)=(2,4), (4,2) in the set of the positive integer numbers).


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PostPosted: Wed, 2 Dec 2009 01:09:57 UTC 
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?

This seems odd,

If I write my two rational numbers x,y\in\mathbb{Q}

in reduced form as

$x={p\over q}\quad y={r\over s}

then there's a problem (one I'm certain is a consequence of Gauß' lemma, but am too lazy to check).

$\left({p\over q}\right)^{r\over s}=\left({r\over s}\right)^{p\over q}

since we're dealing with the positive integers, we can assume the raising to the ${1\over n},\, n\in\mathbb{N} is 1-1.

Then I raise both sides to the ${s\over r} power. (ignore r=0 as that clearly gives a trivial result)

I get:

p=q\left({r\over s}\right)^{ps\over qr}.

Now p\in\mathbb{N}, so this means that our fraction is also a natural number. But then as we chose x reduced, this implies q=1, wherein x was secretly a positive integer named p in the first place.

The argument is symmetric in showing y=r, so there are no extra rational answers, only the integer ones.

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PostPosted: Wed, 2 Dec 2009 17:59:50 UTC 
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What about the values x=\frac{9}{4} and y=\frac{27}{8} ??

Look carefully! There are many rational solutions and you can find all of them!


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PostPosted: Wed, 2 Dec 2009 20:49:02 UTC 
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mathemagics wrote:
What about the values x=\frac{9}{4} and y=\frac{27}{8} ??

Look carefully! There are many rational solutions and you can find all of them!


Okay, I believe that. I made the mistake of concluding that the second term in my product was an integer when that's clearly crazy talk.

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PostPosted: Thu, 3 Dec 2009 04:20:46 UTC 
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x =\left(\frac{n+1}{n}\right)^{n+1},\ y=\left(\frac{n+1}{n}\right)^{n}

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