solutions of $x^y=y^x$

The equation

$x^y=y^x$ (1)

has trivial solutions on the line  $y=x$.  For other solutions one has the

Theorem.  $1^{\circ }$. The only positive solutions of the equation (1) with    are in a parametric form

$x={(1+u)}^{\frac{1}{u}},y={(1+u)}^{\frac{1}{u}+1}$ (2)

where  $u>0$.
$2^{\circ }$. The only rational solutions of (1) are

$x={\left(1+{\displaystyle \frac{1}{n}}\right)}^n,y={\left(1+{\displaystyle \frac{1}{n}}\right)}^{n+1}$ (3)

where  $n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }$
$3^{\circ }$. Consequently, the only integer solution of (1) is

$$2^4=\mathrm{\hspace{0.33em}16}={\mathrm{\hspace{0.33em}4}}^2.$$

Proof. $1^{\circ }$. Let  $(x,y)$  be a solution of (1) with  .  Set  $y=x+\delta$  ($\delta >0$).  Now

$$x^{x+\delta }={(x+\delta )}^x,$$

from which we obtain easily

$$x={\left(1+\frac{\delta }{x}\right)}^{\frac{x}{\delta }}:={(1+u)}^{\frac{1}{u}},$$

where  $u=\frac{\delta }{x}$.  Then

$$y=x+\delta =x\left(1+\frac{\delta }{x}\right)={(1+u)}^{\frac{1}{u}}(1+u)={(1+u)}^{\frac{1}{u}+1}.$$

$2^{\circ }$. The unit fractions  $u=\frac{1}{n}$  yield from (2) rational solutions (3). Further, no irrational value of $u$ cannot make both $x$ and $y$ of (2) rational, since otherwise the ratio $1+u$ of the latter numbers would be irrational (cf. rational and irrational).  Accordingly, for other rational solutions than (3), we must consider the values

$$u:=\frac{m}{n}$$

with coprime positive integers $m,n$ where  $m>1$.  Make the antithesis that

$$x={\left(1+\frac{m}{n}\right)}^{\frac{n}{m}}\in \mathbb{Q}.$$

Because the integers coprime with $m$ form a group with respect to the multiplication modulo $m$ (cf. prime residue classes), the congruence

$$nz\equiv \mathrm{\hspace{0.33em}1}(modm)$$

has a solution $z$.  Thus we may write  $nz=km+1$  and rewrite the rational number

${\left[{\left(1+{\displaystyle \frac{m}{n}}\right)}^{\frac{n}{m}}\right]}^z={\left(1+{\displaystyle \frac{m}{n}}\right)}^{\frac{nz}{m}}={\left(1+{\displaystyle \frac{m}{n}}\right)}^{\frac{km+1}{m}}={\left(1+{\displaystyle \frac{m}{n}}\right)}^k{\left(1+{\displaystyle \frac{m}{n}}\right)}^{\frac{1}{m}}.$ (4)

This product form tells that ${\left(1+\frac{m}{n}\right)}^{\frac{1}{m}}$ is rational.  But the number

$${\left(1+\frac{m}{n}\right)}^{\frac{1}{m}}=\sqrt[m]{\frac{m+n}{n}}$$

cannot be rational without the coprime integers $m+n$ and $n$ both being $m^{\text{th}}$ powers (http://planetmath.org/GeneralAssociativity).  If we had  $n=v^m$,  then by Bernoulli inequality,

$${(v+1)}^m>v^m+mv\geqq n+m,$$

i.e. $m+n$ could not be a $m^{\text{th}}$ power.  The contradictory situation means, by (4), that the antithesis is wrong.  Therefore, the numbers (3) give the only rational solutions of (1).

Note.  The value  $n=2$  in (3) produces  $x=\frac{9}{4}$,  $y=\frac{27}{8}$,  whence (1) reads

${\left({\displaystyle \frac{9}{4}}\right)}^{\frac{27}{8}}={\left({\displaystyle \frac{27}{8}}\right)}^{\frac{9}{4}}.$ (5)

The truth of the equality (5) may also be checked by the calculation

$${\left(\frac{9}{4}\right)}^{\frac{27}{8}}={\left[{\left(\frac{9}{4}\right)}^{\frac{1}{2}}\right]}^{\frac{27}{4}}={\left(\frac{3}{2}\right)}^{\frac{27}{4}}={\left[{\left(\frac{3}{2}\right)}^3\right]}^{\frac{9}{4}}={\left(\frac{27}{8}\right)}^{\frac{9}{4}}.$$

Title	solutions of $x^y=y^x$
Canonical name	SolutionsOfXyYx
Date of creation	2014-12-16 16:29:29
Last modified on	2014-12-16 16:29:29
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Result
Classification	msc 26D20
Classification	msc 26B35
Classification	msc 26A09
Classification	msc 11D61
Synonym	equation $x^y=y^x$
Related topic	CatalansConjecture
Related topic	RationalAndIrrational
Related topic	PerfectPower