solutions of x^y=y^x

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

∘

. The only

positive

solutions of the equation (1) with

1

<

x

<

y

are in a

parametric form

x

=

(

1

+

u

)

1

u

,

y

=

(

1

+

u

)

1

u

+

1

(2)

where

u

>

0

.

2

∘

. The only

rational

solutions of (1) are

x

=

(

1

+

1

n

)

n

,

y

=

(

1

+

1

n

)

n

+

1

(3)

where

n

=

1

,

2

,

3

,

…

3

∘

. Consequently, the only

integer

solution of (1) is

2

4

=

16

=

4

2

.

Proof.

1

∘

. Let

(

x

,

y

)

be a solution of (1) with

1

<

x

<

y

.  Set

y

=

x

+

δ

(

δ

>

0

).  Now

x

x

+

δ

=

(

x

+

δ

)

x

,

from which we obtain easily

x

=

(

1

+

δ

x

)

x

δ

:=

(

1

+

u

)

1

u

,

where

u

=

δ

x

.  Then

y

=

x

+

δ

=

x

⁢

(

1

+

δ

x

)

=

(

1

+

u

)

1

u

⁢

(

1

+

u

)

=

(

1

+

u

)

1

u

+

1

.

2

∘

. The

unit fractions

u

=

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

:=

m

n

with

coprime

positive integers

m

,

n

where

m

>

1

.  Make the antithesis that

x

=

(

1

+

m

n

)

n

m

∈

ℚ

.

Because the integers coprime with

m

form a

group with

respect to the

multiplication

modulo

m

(cf.

prime residue classes

), the

congruence

n

⁢

z

≡

1

(

mod

m

)

has a solution

z

.  Thus we may write

n

⁢

z

=

k

⁢

m

+

1

and rewrite the

rational number

[

(

1

+

m

n

)

n

m

]

z

=

(

1

+

m

n

)

n

⁢

z

m

=

(

1

+

m

n

)

k

⁢

m

+

1

m

=

(

1

+

m

n

)

k

⁢

(

1

+

m

n

)

1

m

.

(4)

This product form tells that

(

1

+

m

n

)

1

m

is rational.  But the number

(

1

+

m

n

)

1

m

=

m

+

n

n

m

cannot be rational without the coprime integers

m

+

n

and

n

both being

m

th

powers (

http://

planetmath

.org/GeneralAssociativity

).  If we had

n

=

v

m

,  then by

Bernoulli inequality

,

(

v

+

1

)

m

>

v

m

+

m

⁢

v

≧

n

+

m

,

i.e.

m

+

n

could not be a

m

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

=

9

4

,

y

=

27

8

,  whence (1) reads

(

9

4

)

27

8

=

(

27

8

)

9

4

.

(5)

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

(

9

4

)

27

8

=

[

(

9

4

)

1

2

]

27

4

=

(

3

2

)

27

4

=

[

(

3

2

)

3

]

9

4

=

(

27

8

)

9

4

.

References

1

P. Hohler & P. Gebauer

:  Kann man ohne Rechner entscheiden, ob

e

π

oder

π

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.pdf

arXiv.

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

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