# 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)

Note by Ron Lauterbach
8 months, 3 weeks ago

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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 $$(x,y)=(2,4),(4,2)$$.

- 8 months, 2 weeks ago

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

Do you mean it has 3 solutions?

- 8 months, 3 weeks ago

I meant writing 3.

- 8 months, 3 weeks 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) .............

- 8 months, 2 weeks ago

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

- 8 months, 3 weeks ago

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

- 8 months, 3 weeks ago

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