Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$
I have found $2^4 = 4^2$ by trial and error,
What is the general solution? I have no idea of where to start.
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type it into wolfram alpha and then learn about the http://en.wikipedia.org/wiki/Lambert_W_function – binn Sep 26 '12 at 04:05
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This looks like a duplicate of both $x^y = y^x$ for integers $x$ and $y$ and Determine the number of solutions of the equation $n^m = m^n$ – robjohn Sep 26 '12 at 06:41
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@robjohn how did you find that, sorry that it is a duplication. – yiyi Sep 26 '12 at 07:22
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I had answered Determine the number of solutions of the equation $n^m = m^n$ and that is commented to be a duplicate of $x^y = y^x$ for integers $x$ and $y$ – robjohn Sep 26 '12 at 07:53
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You have found the only solution in integers except $p=q.$ For rationals, let $p=\frac ab, q=\frac cd$, both in lowest terms. Now use the laws of exponents and you will find a one dimensional set of solutions. Continuity extends this to the reals.
Added: You want $\left(\frac ab\right)^{\frac cd}=\left(\frac cd\right)^{\frac ab}$. Think about the primes that might divide $a,b,c,d$
Ross Millikan
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For the case when $q/p$ is rational, let $q/p = r$, so $q = p r$. Substituting this in $p^q = q^r$, $p ^ {pr} = (pr)^p$ so, taking the $p$-th root, $p^r = pr$ sp $p^{r-1} = r$ or $p = r^{1/(r-1)}$ and $q = p r = r^{r/(r-1)}$.
If $r = 2$, $p = 2$ and $q = 4$.
If $r = 3$, $p = 3^{1/2}$ and $q = 3^{3/2}$.
If $1/(r-1) = n$, where $n$ is an integer, $r = 1+1/n = (n+1)/n$ so $r/(r-1) = n+1$ and $p = ((n+1)/n)^n$ and $q = ((n+1)/n)^{n+1}$.
Note: This is definitely not original with me.
That's all for now.
marty cohen
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