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

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?

• Nice descriptive title. – Git Gud Sep 14 '13 at 20:11
• I laughed in real life. – Git Gud Sep 14 '13 at 20:16
• @GitGud, good. I do try to add a little humor...also it is quite true – Will Jagy Sep 14 '13 at 20:17
• I think, mucinex is an answer. – Moishe Kohan Sep 14 '13 at 20:23
• Looks like a kind of last fermat jagy theorem showing up^^ – GarouDan Sep 17 '13 at 23:05

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 $x^{(y^z)}= y^{(z^x)}$.

• Thank you. I'll take a careful look in a few hours. – Will Jagy Oct 2 '13 at 23:49
• I would say instead that $h(x,y,z) = x^{\left( y^z \right)} - y^{\left( z^x \right)}$ is $0$ along the line $x=y=z,$ meaning that its gradient, if defined, is orthogonal to the vector $(1,1,1)$ along that line. My suspicion is that the $0$ level surface of $h$ is a wiggly variant of a helicoid en.wikipedia.org/wiki/Helicoid wrapping around that line, so are the other two pairwise equality surfaces, and the three meet only along that line. perhaps offset by $120^\circ$ rotations. – Will Jagy Oct 3 '13 at 2:49
• Sounds plausable, Will. I tried to think of the problem geometrically as well but soon relaized that the level surfaces were a bit too exotic for me to wrap my head around. Recreational math should not make the brain hurt :-) – Doc Oct 3 '13 at 3:35
• What's the domain of $f$? In particular, why does it follow that it's identically $0$ from the assumption that you have a single $(x,y,z)$ off of the line $x=y=z$ for which $f(y,z)= 0$? – Jason DeVito Oct 3 '13 at 18:14
• @Jason, You're right, I cannot make that assumption. My "proof" is not a proof. Back to the drawing board. – Doc Oct 4 '13 at 13:04

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.

• Does dividing through by $p^{p^p}$ work? In particular, it isn't the case that $x^{y^z}/p^{p^p}=x/p^{y/p^{z/p}}$. – Oscar Cunningham Sep 26 '13 at 16:21
• It might be easier to look at if you take logs twice, and you get $z\ln y+\ln\ln x = x\ln z+\ln\ln y = y\ln x+\ln\ln z$ – Empy2 Sep 26 '13 at 16:45
• To first order, if $x=p+a,y=p+b,z=p+c$, and $a,b,c$ are small compared to $p$, I think you need $c\ln p+b+a/(p\ln p)$ to equal its cyclic permutations. Form a 3x3 matrix, multiply its inverse by the vector $(1,1,1)$ – Empy2 Sep 26 '13 at 17:01
• hm, you get a multiple of $(1,1,1)$. Unless the determinant is zero. Can you solve $(\ln p)^3+1+1/(p\ln p)^3 = 3/p$? – Empy2 Sep 26 '13 at 17:08
• Doc, I hope this works out. I got nowhere. – Will Jagy Sep 26 '13 at 17:20