How to prove $y^{x}=x^{y}$ only has 2 solutions?

How do you prove, that $y^{x}=x^{y}$ only has 2 integral solutions, where $x$ and $y$ are not equal? These are the solutions: (2, 4) and (4, 2)

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Solving for integral solutions in $y^x=x^y$ for distinct $x$ and $y$ is equivalent to solving for integral solutions in $y^{\frac{1}{y}}=x^{\frac{1}{x}}$ for distinct $x$ and $y$ . Since $\dfrac{d}{dx}\big(x^{\frac{1}{x}}\big)=\dfrac{1-\text{ln}(x)}{x^2}\times x^{\frac{1}{x}}$ , we have a single maximum at $(e,e^{\frac{1}{e}})$ for $x^{\frac{1}{x}}$ over positive $x$ . Then since $x^{\frac{1}{x}}\rightarrow 0$ as $x\rightarrow 0^+$ , and $x^{\frac{1}{x}}$ decreases asymptotically to $1$ as $x\rightarrow \infty$ , all integral solutions must occur in the domain $[0,e]$ ; this leaves three possibilities for the value of $x$ : $0$ , $1$ , and $2$ . We have $0^{\frac{1}{0}}$ to be undefined, and $1^{\frac{1}{1}}$ to have no corresponding solution (since $x^{\frac{1}{x}}$ only asymptotically approaches $1$ ), leaving only $2^{\frac{1}{2}}$ to have the corresponding solution $4^{\frac{1}{4}}$ . Hence, the only solutions for distinct $x$ and $y$ over the positive integers are $(x,y)=(2,4),(4,2)$ .

Miles Koumouris - 3 years, 8 months ago

How do you prove, that $y^x = x^y$ only has 2 solutions

Do you mean it has 3 solutions?

Munem Shahriar - 3 years, 8 months ago

I meant writing 3.

Ron Lauterbach - 3 years, 8 months ago

I am satisfied with (2, 4) and (4, 2). Where (2, 4) is an obvious solution. (0,0) isn't a solution.

Munem Shahriar - 3 years, 8 months ago

I meant writing (1, 1). I was way to hurried.

Ron Lauterbach - 3 years, 8 months ago

This question would have 2 solutions only when x is not equal to y. In this question , it is not mention that x is not equal to y . So this question has infinite solutions . When x=y (not mention in this question , so considering ) , then there are infinitely many solutions of (x, y) i.e. (1,1) ; (2,2) ; (3,3) ; (4,4) .............

Raj Mantri - 3 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

How to prove yx=xyy^{x}=x^{y}yx=xy only has 2 solutions?

Comments

How to prove $y^{x}=x^{y}$ only has 2 solutions?