Prove or disprove that $x_1 = x_2= \cdots = x_n$, if $x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1}$, $x_i > 1, \forall 1 \le i \le n$.

Question

The case $n=3$ is from here. It's straightforward to prove it's true:

First we notice that if any two of $x_1, x_2, x_3$ are equal then all must be equal.

Denote $a=x_1^{x_2}=x_2^{x_3}=x_3^{x_1}$ then $$x_1 = a^{1/x_2}, x_2=a^{1/x_3}, x_3=a^{1/x_1}$$ WLOG assume $x_1 \ge x_2, x_3$. There are two cases:

• $x_1 \ge x_2 \ge x_3 \implies \frac{1}{x_2} \ge \frac{1}{x_3} \ge \frac{1}{x_1} \implies x_1 \ge x_3 \ge x_2 \implies x_2=x_3 \implies x_1=x_2=x_3$.

• $x_1 \ge x_3 \ge x_2 \implies \frac{1}{x_2} \ge \frac{1}{x_1} \ge \frac{1}{x_3} \implies x_3 \ge x_1 \ge x_2 \implies x_1=x_3 \implies x_1=x_2=x_3$.

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If $n$ is even, then $x_{2i-1} = 2, x_{2i}=4$ is a counterexample.

When $n=5$ the above method will need to examine $4!=24$ cases. Basically we map $x_i$ to $x_{(i \mod 5) +1}$ and reverse the order. In many of cases we can deduce $4$ or all $5$ of the $x_i$'s are equal. For example, if $$x_1 \ge x_3 \ge x_2 \ge x_5 \ge x_4 \tag 1$$ then $$ x_5 \ge x_1 \ge x_3 \ge x_4 \ge x_2 \tag 2 $$ Since the order of $x_2$ and $x_5$ reversed from $(1)$ to $(2)$, they must be equal and so are everything in between them from both $(1)$ and $(2)$ hence all $x_i$'s are equal.

There are other cases that are different.

Example 2: $$x_1 \ge x_3 \ge x_4 \ge x_2 \ge x_5 \tag 3$$ then $$x_1 \ge x_3 \ge x_5 \ge x_4 \ge x_2 \implies x_1 \ge x_3 \ge x_2=x_4=x_5 \tag 4$$

Example 3: $$x_1 \ge x_3 \ge x_4 \ge x_5 \ge x_2 \tag 5$$ then $$x_3 \ge x_1 \ge x_5 \ge x_4 \ge x_2 \implies x_1=x_3 \ge x_2=x_4=x_5 \tag 6$$

But they all lead to the conclusion that $x_1=x_2=x_3=x_4=x_5$.

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Now my questions:

Question #1: Is it always true if $n>1$ and $n$ is odd?

Question #2: What if we allow $x_i>0$ instead of $x_i>1$?

Answer 1

Consider the question as a sequence. It starts with two values and has the recursive definition $x(n) = x(n-1)\frac{\log x(n-2)}{\log x(n-1)}$. For $x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1} $ to hold, this sequence has to be periodic with some period length. This gives rise to the hypothesis (to be proved) that the sequence will indeed be monotonously rising / falling for odd/even $n$, hence they cannot be periodic. If this can be established, indeed $x_1 = x_2= \cdots = x_n$ is the only solution.

Let $x(1) = z > 1$ and $x(2) = z^q$. Then $x(3) = \frac{z^q}{q}$. For $x(3) = x(1)$ to hold we require that $z = q^{\frac{1}{q-1}}$. One can for example choose $q=2$ which gives $z=2$, hence we obtain $x(1)= 2$ and $x(2) =4$ which gives the series $2,4,2,4, \cdots$ which OP already quoted. Infinite other choices of $q$ are possible as well. For $q=3$ we would have $z = \sqrt 3$ and hence $x(1)= \sqrt 3$ and $x(2) =\sqrt{27}$ which gives the series $\sqrt 3, \sqrt{27}, \sqrt 3, \sqrt{27}, \cdots$. All of the series constructed this way are periodic with (even) period length 2 or integer multiples thereof.

Now let's prove that there are no series possible with odd period length. Start with the situation that the initially (for $x(1)$ and $x(2)$) chosen $z$ and $q$ with $z \ne q^{\frac{1}{q-1}}$. Then the first case is that $x(3) > x(1)$ and hence we have $x(4) = x(2)\frac{\log x(1)}{\log x(3)} < x(2)$. This gives $x(5) = x(3)\frac{\log x(2)}{\log x(4)} > x(3)$. Continuing like this, the odd sequence is rising whereas the even sequence is falling. If initially $x(1) > x(2)$, as the odd sequence values will always be larger than $x(1)$, and the even sequence values will always be smaller than $x(2)$, hence no periodicity will be possible. If initially $x(1) < x(2)$, hence the odd sequence (rising) and the even sequence (falling) will converge to some $x^*$ in between $x(1)$ and $x(2)$, which means no periodicity will be possible.

In the second case, $x(3) < x(1)$. With the same reasoning as above, $x(4) = x(2)\frac{\log x(1)}{\log x(3)} > x(2)$ and $x(5) = x(3)\frac{\log x(2)}{\log x(4)} < x(3)$. Continuing like this, the odd sequence is falling whereas the even sequence is rising. This again means that the sequences either diverge from each other or converge to some $x^*$ in between $x(1)$ and $x(2)$. In both cases, no periodicity will be possible.

Note that the condition $x_i >1$ is not guaranteed with all initial conditions. The lower values can fall rapidly, consider $x(1) = 1.1, x(2) = 2, x(3) = 0.275007, \cdots$

In summary, in the cases where $x(1) = z > 1$ and $x(2) = z^q$ with $z = q^{\frac{1}{q-1}}$, the period is $2$ or any integer multiple of that, hence even. In all other cases, there is no periodicity, hence a constant $x(i) =z$ is the only solution.

Answer 2

After reading this answer to another question, as well as Andreas's answer I decide to write up the following solution. I believe John Omielan had already figured out how to do it, though he chose to leave that to the one who asked the case $n=3$ question last year. Then Andreas's answer made everything clear to me.

Assume $n=2m+1$. First notice $$x_2 \ln x_1 = x_3 \ln x_2 = \cdots = x_n \ln x_{n-1} = x_1 \ln x_n$$ Then there are only three possibilities: 1) $x_i = 1, \forall i$; 2) $x_i > 1, \forall i$; 3) $x_i < 1, \forall i$.

For case 2), if not all $x_i$'s are equal, WLOG we assume $x_1 > x_2$.

Then $x_2 \ln x_1 = x_3 \ln x_2 \implies x_2 < x_3$.

Then $x_3 \ln x_2 = x_4 \ln x_3 \implies x_3 > x_4$.

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$x_{2m+1} \ln x_{2m} = x_1 \ln x_{2m+1} \implies x_{2m+1}>x_1$

$x_1 \ln x_{2m+1} = x_2 \ln x_1 \implies x_1 < x_2$, a contradiction. Therefore all $x_i$'s must be equal.

For case 3), since $\ln x_i < 0$ we'd have $x_i > x_{i+1}$, and $x_{2m+1}>x_1$, also a contradiction (in fact this also works when $n$ is even).

Answer 3

It can be proved to be false for n = 2.

Let x1^x2 = x2^x1. Then we raise both sides with power 1/(x1*x2).

This means x1^((x2/x2)*x1) = x2^((x1/x1)*x2). This means x1^(1/x1) = x2^(1/x2).

This means we have to check if the function f(x) = x^(1/x) is one-one or not. It turns out it is not.

A function whose derivative changes sign is not one-one.

f(x) = x^(1/x) has a maxima at x=e.

So this proves the existence of distinct values of x1,x2 for n=2.