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Is the function $f(x) = x^x$ unique over the positive rationals? That is, do there exists two different positive rational numbers $x$ and $y$ so that $x^x = y^y$ ?

user21820
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archedleon
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    I think you mean "injective" by "unique". – Mohsen Shahriari Oct 11 '20 at 06:33
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    Yes, $x=1/2$ and $y=1/4$, and infinitely many more. –  Oct 11 '20 at 06:33
  • @ProfessorVector As the OP asks for pairs of rationals, I wonder how one can come up with "infinitely many more" cases. – Mohsen Shahriari Oct 11 '20 at 07:02
  • @Mohsen Shahriari Just there are infinitely many pairs of rationals $(x,y)$ with $x^x=y^y$. –  Oct 11 '20 at 07:08
  • @ProfessorVector Could you provide more examples? It is not clear for me that there are more rational solutions. What I know is that the only solution of the form $x = 1/a, y =1/b$ with $a, b\in \mathbb{Z}_{>0}$ is $(1/2, 1/4)$. – Seewoo Lee Oct 11 '20 at 07:10
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    This question has been changed so much as to be unrecognizable, and to invalidate the existing answers. Please don't do that, you should revert your question so that the answer makes sense, and then you can ask your new question in a new post. – Lee Mosher Oct 11 '20 at 19:02

2 Answers2

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Here's a family of such rational solutions: $$ x = \left(\frac{m-1}{m}\right)^{m}, \quad y = \left(\frac{m-1}{m}\right)^{m-1} $$ for integers $m \geq 2$.

How to get these solutions: assume that $x = b/a, y = c/d$ with $\gcd(a, b) = \gcd(c, d)= 1$. Then the equation $x^x = y^y$ gives that the prime factors of $a$ (resp. $b$) should coincides with $c$ (resp. $d$). Also, setting $a = p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}, b = q_{1}^{f_{1}}\cdots q_{s}^{f_{s}}, c = p_{1}^{e_{1}'}\cdots p_{r}^{e_{r}'}, d = q_{1}^{f_{1}'}\cdots q_{s}^{f_{s}'}$ we can prove $e_i' / e_i = f_i' / f_i = bc/ad$ for all $i$, so that $y = x^t$ for some $t\in \mathbb{Q}_{>0}$. From this, we have $x = t^{-(1/(t-1))}$, and $t = (m-1)/m$ makes $x, x^t \in \mathbb{Q}$.

Seewoo Lee
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    And these are in fact all the solutions. See here. Note that letting $ u = \frac 1 x $ and $ v = \frac 1 y $, $ x ^ x = y ^ y $ is equivalent to $ u ^ v = v ^ u $. – Mohsen Shahriari Oct 11 '20 at 08:01
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I followed the idea of @ProfessorVector. The resulting pairs should be the same as given by @SeewooLee.

Consider the form $y=x^r$ and suppose $r\in\mathbb{Q}$ and $r>1$ without loss of generality.

Claim that if $n:=\frac1{r-1}\in\mathbb{N}$, then $$x=r^{-n},\,y=r^{-rn}$$ is a pair in need.

At first, by $rx^{r-1}=1$, we know that $$y^y=x^{rx^r}=x^x.$$

It is clear that $x\in\mathbb{Q}$. For $y$, we note that $$y=x^r=r^{\frac{r}{1-r}}=\frac1r\left(\frac1r\right)^n\in\mathbb{Q}.$$ This proves the claim. $\square$

From this result, we know that $r=2,\,\frac32,\,\frac43$... could be possible choices, in which cases $(x,y)=(\frac12,\frac14),\,(\frac49,\frac8{27}),\, (\frac{27}{64},\frac{81}{256})$...

Yufeng Lu
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