# solutions of $x^{y}=y^{x}$

The equation

 $\displaystyle x^{y}\;=\;y^{x}$ (1)

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

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

 $\displaystyle x\;=\;(1\!+\!u)^{\frac{1}{u}},\qquad y\;=\;(1\!+\!u)^{\frac{1}{u% }+1}$ (2)

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

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

where  $n=1,\,2,\,3,\,\ldots$
$3^{\circ}$. Consequently, the only integer solution of (1) is

 $2^{4}\;=\;16\;=\;4^{2}.$

Proof. $1^{\circ}$. Let  $(x,\,y)$  be a solution of (1) with  $1.  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

 $nz\;\equiv\;1\pmod{m}$

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

 $\displaystyle\left[\left(1\!+\!\frac{m}{n}\right)^{\frac{n}{m}}\right]^{z}\;=% \;\left(1\!+\!\frac{m}{n}\right)^{\frac{nz}{m}}\;=\;\left(1\!+\!\frac{m}{n}% \right)^{\frac{km+1}{m}}\;=\;\left(1\!+\!\frac{m}{n}\right)^{k}\left(1\!+\!% \frac{m}{n}\right)^{\frac{1}{m}}.$ (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 $m^{\mbox{th}}$ 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 $m^{\mbox{th}}$ 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

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

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

 $\left(\frac{9}{4}\right)^{\frac{27}{8}}\;=\;\left[\left(\frac{9}{4}\right)^{% \frac{1}{2}}\right]^{\frac{27}{4}}\;=\;\left(\frac{3}{2}\right)^{\frac{27}{4}}% \;=\;\left[\left(\frac{3}{2}\right)^{3}\right]^{\frac{9}{4}}\;=\;\left(\frac{2% 7}{8}\right)^{\frac{9}{4}}\!.$

## References

• 1 P. Hohler & P. Gebauer:  Kann man ohne Rechner entscheiden, ob $e^{\pi}$ oder $\pi^{e}$ grösser ist? $-$ Elemente der Mathematik 36 (1981).
• 2 J. Sondow & D. Marques:  Algebraic and transcendental solutions of some exponential equations.  $-$ Annales Mathematicae et Informaticae 37 (2010); available directly at http://arxiv.org/pdf/1108.6096.pdfarXiv.
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