If $x, y, z, u \in \Bbb{N}$ are distinct numbers, do any set of numbers exist to fulfil the following property $$x^y =y^z =z^u.$$

It is certainly true for only three numbers $x, y, z$. Is there any formula to find a set of these numbers?

  • $\begingroup$ Do you want it to be equal to $u^x$ also? $\endgroup$ Apr 8 '17 at 13:46
  • $\begingroup$ $$ 2^{16} = 16^4 = 4^8, $$ $$ 64^8 = 8^{16} = 16^{12}, $$ $$ 256^2 = 2^{16} = 16^4, $$ $$ 256^4 = 4^{16} = 16^8 $$ $\endgroup$
    – Dando18
    Apr 8 '17 at 14:45

Not a complete answer, but here is some python code that can generate some of these (note: not even close to optimized):

from itertools import product
import math

max = 260
xyz = list(product(range(2, max), repeat=3))
for (x, y, z) in xyz:
    if x != y and y != z and x != z:
        if x ** y == y ** z:
            u = (z * math.log(y, z))
            if u.is_integer():
                u = int(u)
                print("x={}, y={}, z={}, u={} --> {}^{} = {}^{} = {}^{}".format(x, y, z, u, x, y, y, z, z, u))

$$ 2^{16}=16^4=4^8, \\ 64^{8}=8^{16}=16^{12}, \\ 256^2=2^{16}=16^{4}, \\ 256^4=4^{16}=16^8 $$


The equation $$(2^{128})^8=8^{256}=256^{96}$$ holds. I found this by assuming $x,y,z$ would all be powers of the same small prime (picking $2$ for connivence) and doing a little trial and error. $y$ and $z$ need to be reasonably space powers of the base number, so then it was just a matter of doing some trial and error and exponent manipulation. A major issue with generating things like this is making sure all the solutions are integers, but by working with powers of a prime we can make that happen easily by just moving around parentheses. All of these numbers are just different ways to group exponents for $2^{1024}$.

I expect arbitrarily long chains are possible. The key to them is to try to keep the exponent of the next term to the right a power of $2$ as long as possible, with likely requires the entire sum to equal a number of the form $2^{2^{2^\ldots}}$ or close to it. Writing these numbers as nested powers of $2$ might prove illuminating in that regard as well.


$$(x, y, z, u) = \left(r^{br^i}, r^b, r^{b+i}, \frac{br^{b+i}}{b+i}\right),$$ where

$r, b \in \mathbb{N}$ and $i \in \mathbb{Z}$ satisfy:

  • $r$ is not a perfect power, i.e., there is no natural number $k > 1$ such that $r^{1/k} \in \mathbb{N}$.
  • either $i > 0$ or $i < 0$ and $r^{-i}$ divides $b$.
  • $(b+i)$ divides $br^{b+i}$.

These conditions don't QUITE guarantee uniqueness. For example, if $r=2$ and $b=i=1$ then $x=z$. There may be other corner cases, as well.

To see that this works, the original equation suggests that $x$, $y$, and $z$ can all be written as powers of some common integer $r$, which we may assume to be minimal. So, we can write $$(x, y, z) =(r^a, r^b, r^c)$$ for some natural numbers $a, b, c$.

We substitute into the original equations and take $\log_r$ to get $$ar^b = br^c = cu.$$

If we take $r$ and $b$ as fixed, and set $i=c-b$ then we can solve for the other values, and the necessary conditions fall out.

  • $\begingroup$ This is excellent! One critique: I think you mean $r^{1/k}\in\mathbb N$ $\endgroup$ Apr 29 '17 at 11:55
  • $\begingroup$ Right you are. I have made the edit. $\endgroup$ Sep 7 '18 at 21:56

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