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What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

For the case when $x=y$ it is clear, what about the case when $x\ne y$?

Servaes
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Steve Phoenix
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    Then we may assume $x>y$, for positive integers this implies $x^x>y^y$. – Dietrich Burde Feb 17 '21 at 19:56
  • When graphing on Desmos, it appears solutions where $x\neq{y}$ is only on $x\in[0,1]$ – Cotton Headed Ninnymuggins Feb 17 '21 at 20:01
  • @DietrichBurde only when $x\gt1$ – Cotton Headed Ninnymuggins Feb 17 '21 at 20:08
  • @CottonHeadedNinnymuggins Positive integers are $\ge 1$ and even if $y=1$, then certainly $x>1$, since $x>y$. So $x>1$ anyway in the comment. – Dietrich Burde Feb 17 '21 at 20:09
  • @DietrichBurde makes sense. I think you meant to comment "since $x\gt{y}$" – Cotton Headed Ninnymuggins Feb 17 '21 at 20:11
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    There are no integer solutions. All real solutions with $x>y$ can be obtain in a parametric way: $$x=\left(\frac{t}{t+1}\right)^t,\quad y=\left(\frac{t}{t+1}\right)^{t+1}$$ for $t>0$. If $t$ is integer, that gives all rational solutions. – NoNames Feb 17 '21 at 20:21
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    @NoNames: You can consider $0^0 = 1^1$ to be a solution, but I have no intention of starting a flame war in the comments about whether or not we should define $0^0$ :) – Joe Feb 17 '21 at 20:23
  • @Joe in this precise case it is not undefined as $\lim_0 x^x=1$. – zwim Feb 17 '21 at 20:26
  • @NoNames: I used your parameterisation to finish off the argument that there are infinitely many rational solutions. As such, my answer is really a team effort, and so I have turned into a community wiki post. Feel free to edit it as you see fit. – Joe Feb 17 '21 at 20:41
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    @Joe What if I'm wrong? Then, you will get the downvotes for that! ;-) – NoNames Feb 17 '21 at 20:43
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    @NoNames: Thankfully, downvotes to community wiki posts don't harm one's reputation points, which are surely the most valuable commodity in this entire universe. – Joe Feb 17 '21 at 20:44
  • @Joe You completed the NoNames solution, except for the part " If t is integer, that gives all rational solutions." which is not trivial. – jjagmath Aug 17 '22 at 15:57
  • @NoNames "If t is integer, that gives all rational solutions." which not trivial. – jjagmath Aug 17 '22 at 15:57

3 Answers3

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The only integer solutions are $x=y$ and also $0^0=1^1$, provided that $0^0$ is defined. This is because $x^x$ is strictly increasing on $(1/e,\infty)$:

Graph of x^x

On the interval $[0,1]$, there are infinitely many real solutions of $x^x=y^y$ where $x \neq y$. This is because when $x=1/e$, $y=e^{-1/e}$. Then, when $x=1$, $y=1$. Since $x^x$ is continuous on $[1/e,1]$, by the intermediate value theorem, $x^x$ must take on every real value between $e^{-1/e}$ and $1$. Similarly, on the interval $[0,1/e]$, $x^x$ is also continuous, and so must take on every real value between $e^{-1/e}$ and $1$. This means that for every $a \in (e^{-1/e},1]$, there are two possible solutions for $x^x = a$. So there are infinitely many real solutions to $x^x=y^y$, $x \neq y$.

As NoNames has pointed out, there are also infinitely many rational solutions, since the solutions can be parameterised as $$ x=\left(\frac{t}{t+1}\right)^t,\quad y=\left(\frac{t}{t+1}\right)^{t+1} \, , $$ for $t>0$. If $t$ is an integer, then this gives us a valid rational solution.

Joe
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  • Could the downvoter please explain their reason for downvoting? If there is something unsound about my argument, I am happy to correct it. – Joe Feb 17 '21 at 20:25
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    It is not at all obvious that there are infinitely many rational solutions. If $r$ is a rational between $0$ and $1/e$, there is some $y$ between $1/e$ and $1$ with $r^r = y^y$, but there is no reason to think $y$ is rational. – Robert Israel Feb 17 '21 at 20:29
  • @RobertIsrael: You're right. I'll edit out that part of my answer. Thanks for correcting my error. – Joe Feb 17 '21 at 20:32
  • Thank you very much Joe and NoNames – Steve Phoenix Feb 17 '21 at 21:45
  • You're welcome @StevePhoenix. It's my pleasure. – Joe Feb 17 '21 at 21:58
  • When $x = 1/e$, $x^x = e^{-1/e}$, which is not $1/e^2$. – Robert Israel Feb 18 '21 at 00:29
  • @Robert Israel: Thanks again. I’ve fixed it. Feel free to edit this post if you spot something that is not right. – Joe Feb 18 '21 at 00:45
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$x^x=y^y$ is obviously equivalent to $(1/y)^{1/x}=(1/x)^{1/y}$, and that's a very popular problem at MSE (e.g. $x^y = y^x$ for integers $x$ and $y$ containing also the general parametric solution).

NoNames
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"OP" asked the query:

$x^y=y^x$

Above has numerical integer solution:

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

Sam
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    This was not the question that Steve was asking. He is looking for solutions to $x^x = y^y$. – Joe Feb 17 '21 at 20:39
  • Although this answer misquotes the actual question, it is easily seen to be equivalent. See for instance, the answer below by NoNames. – Erick Wong Feb 17 '21 at 21:51