Possible distinct positive real $x,y,z \neq 1$ with $x^{(y^z)} = y^{(z^x)} = z^{(x^y)}$ in cyclic permutation?

Question

Can we have distinct positive real $x,y,z \neq 1$ with $$ x^{\left( y^z \right)} = y^{\left( z^x \right)} = z^{\left( x^y \right)} $$ in cyclic permutation?

It does not work well if any variable is 1. Also, it obviously works if all three are equal. I think if two are equal, probably the third must match as well. If there is anything else, one would expect a curve of some sort..I guess from what I am asking, one might as well demand $x < y < z.$ NO, not the same as $x < z < y$ as not cyclically equivalent, so maybe drop that.

Suggested by what is the largest number here?

Answer 1

It is an interesting problem. I think that it gave you your cold. I hope by now you are fully recovered.

I believe it never happens.

First, we can establish the result over the integers. Supposing this double equality ever occurs, consider a minimal example. Then any prime divisor $p$ of $x$ must divide both $y$ and $z$. It means that $p^{(p^p)}$ divides each of $x^{(y^z)}$, $y^{(z^x)}$, and $z^{(x^y)}$. Division of all terms by $p^{(p^p)}$ now gives a smaller example, so a contradiction.

Moving to the rational numbers is now easy ... just a matter of clearing out the denominators and repeating the same basic argument.

To get the result over the reals is a bit more subtle, and I haven't worked it out entirely yet. I'm thinking that a necessary first step is to realize each of $x$, $y$ and $z$ as the limit of a sequence of rational numbers. You can try this yourself, as I'll be preoccupied for the next three days.

Best of luck with this interesting problem. I will return to it when I get back to my home base.

Answer 2

I think I've solved it, but I just don't trust myelf. So please hop on and correct me if I've done something stupid..

By way of contradiction, let's assume the equality $x^{(y^z)}=y^{(z^x)}=z^{(x^y)}$ holds. Now define the function $f(y,z)= x^{(y^z)}- y^{(z^x)}$. On one hand, $f(y,z)$ is identically zero, so all its first partials vanish. On the other hand, we may compute them formally as $$f_y=\frac{x^{(y^z)}zy^z\ln{x} -y^{(z^x)}z^x}{y}\quad \mbox{and} \quad f_z=\ln{y}\left(x^{(y^z)}y^z\ln{x}-y^{(z^x)}xz^{x-1}\right)$$ Setting each of these to zero, and using the fact that $x^{(y^z)}=y^{(z^x)}$, we obtain $$zy^z\ln{x}=z^x\quad \mbox{and} \quad y^z\ln{x}=xz^{x-1}$$ From here it is fairly trivial to show that either $z=0$ or $x=1$, but each of these possibilities is ruled out by hypothesis.

Note that in the proof, I only used the asumption that $x^{(y^z)}= y^{(z^x)}$.