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

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