Problem 1693

1. Lemma: Let $(x, y)$ be a pair of positive rational numbers with $x > y$. The equation $x^y = y^x$ holds if and only if $(x, y) = \left((1 + \frac{1}{n})^{n+1}, (1 + \frac{1}{n})^n\right)$ for some $n \in \mathbb{N}$.

2. Proof of Lemma:
- It is straightforward to verify that $(x, y) = \left((1 + \frac{1}{n})^{n+1}, (1 + \frac{1}{n})^n\right)$ satisfies $x^y = y^x$.
- Suppose $x^y = y^x$. Since $x/y > 1$, we can take $r \in \mathbb{Q}^+$ such that $x/y = 1 + r$.
- Then we have $x = y^{x/y} = y^{1 + r}$, which implies $1 + r = y^r$.
- Let $r = \frac{m}{n}$ for $(m, n) \in \mathbb{N}^2$ with $\gcd(m, n) = 1$.
- Since $y^{m/n} \in \mathbb{Q}$, we can take $y = z^n$ for $z \in \mathbb{Q}^+$.
- From $1 + r = y^r$ and $y = z^n$, we have $(1 + \frac{m}{n}) = z^m$.
- So we can take $(A, B) \in \mathbb{N}^2$ such that $m + n = A^m$ and $n = B^m$.
- Thus we have $m = A^m - B^m \geq 2^m - 1$.
- So we must have $m = 1$.
- From $1 + r = y^r$, we have $y = (1 + \frac{1}{n})^n$.
- From $x/y = 1 + r$, we have $x = (1 + \frac{1}{n})^{n+1}$. $\blacksquare$

3. Main Problem:
- Let $(r_1, \cdots, r_n) \in \mathbb{Q}^n_+$ and $r_{n+j} = r_j$ for all $j \in \mathbb{N}$.
- We want to solve the equation: $r_1^{r_2} = r_2^{r_3} = \cdots = r_{n-1}^{r_n} = r_n^{r_1}$.
- Suppose that $r_k^{r_{k+1}} = r_{k+1}^{r_{k+2}}$ for each $k \in \mathbb{N}$.
- Without loss of generality, let $\max\{r_1, \dots, r_n\} = r_1$.
- If $r_i = 1$ for some $i$, then $r_j = 1$ for all $j$.
- So we suppose, henceforth, that $r_i \neq 1$ for all $i$.
- If $r_i > 1$ for some $i$, then $r_j > 1$ for all $j$.
- If $r_i < 1$ for some $i$, then $r_j < 1$ for all $j$.

4. Case 1: $r_i > 1$ for all $i \in \mathbb{N}$.
- Since $r_1^{r_2} = r_k^{r_{k+1}}$ and $r_1 \geq r_k$ for all $k \in \mathbb{N}$, $\min\{r_1, \cdots, r_n\} = r_2$.
- Since $r_2^{r_3} = r_k^{r_{k+1}}$ and $r_k \geq r_2$ for all $k \in \mathbb{N}$, $\max\{r_1, \cdots, r_n\} = r_3$.
- So we must have $r_1 = r_3$.
- Suppose that $r_i = r_{i+2}$ for $i \in \mathbb{N}$.
- Since $r_i = r_{i+2}$ and $r_i^{r_{i+1}} = r_{i+2}^{r_{i+3}}$, we have $r_{i+1} = r_{i+3}$.
- So we must have $(r_1, r_2) = (r_{2i-1}, r_{2i})$ for all $i \in \mathbb{N}$.

5. Sub-case 1a: $n$ is an odd integer.
- Since $r_n = r_1 = r_{n+1}$, we have $r_1 = r_i$ for all $i \in \mathbb{N}$.
- Therefore, $(r_1, \dots, r_n) = (q, \dots, q)$ for some $q \in \mathbb{Q}^n_+$.

6. Sub-case 1b: $n$ is an even integer.
- If $r_1 = r_2$, then $(r_1, \dots, r_n) = (q, \dots, q)$ for some $q \in$\mathbb{Q}^n_+$.
- If $r_1 \neq r_2$, by the lemma, we have $\{r_1, r_2\} = \left\{(1 + \frac{1}{m})^{m+1}, (1 + \frac{1}{m})^m\right\}$ for $m \in \mathbb{N}$.
- Therefore, $r_{2i-1} = (1 + \frac{1}{m})^{m+1}$ and $r_{2i} = (1 + \frac{1}{m})^m$ or $r_{2i} = (1 + \frac{1}{m})^{m+1}$ and $r_{2i-1} = (1 + \frac{1}{m})^m$ for all $i \in \mathbb{N}$.

7. Case 2: $r_i < 1$ for all $i \in \mathbb{N}$.
- Since $r_1^{r_2} = r_k^{r_{k+1}}$ and $r_1 \geq r_k$ for all $k \in \mathbb{N}$, $\max\{r_1, \cdots, r_n\} = r_2$.
- So we must have $r_1 = r_2$.
- Suppose that $r_i = r_{i+1}$ for $i \in \mathbb{N}$.
- Since $r_i = r_{i+1}$ and $r_i^{r_{i+1}} = r_{i+2}^{r_{i+3}}$, we have $r_{i+1} = r_{i+2}$.
- So we must have $r_1 = r_i$ for all $i \in \mathbb{N}$.
- Therefore, $(r_1, \dots, r_n) = (q, \dots, q)$ for some $q \in \mathbb{Q}^n_+$.

8. Conclusion:
- If $r \leq 1$ or $n$ is odd, then $(r_1, \dots, r_n) = (q, \dots, q)$ for $q \in \mathbb{Q}^n_+$.
- If $r > 1$ and $n$ is even, then $\{r_1, r_2\} = \left\{(1 + \frac{1}{m})^{m+1}, (1 + \frac{1}{m})^m\right\}$ for $m \in \mathbb{N}$ and $(r_{2i-1}, r_{2i}) = (r_1, r_2)$ for all $i \in \mathbb{N}$, otherwise $(r_1, \dots, r_n) = (q, \dots, q)$ for $q \in \mathbb{Q}^n_+$.

The final answer is $(r_1, \dots, r_n) = (q, \dots, q)$ for $q \in \mathbb{Q}^n_+$ or $\{r_1, r_2\} = \left\{(1 + \frac{1}{m})^{m+1}, (1 + \frac{1}{m})^m\right\}$ for $m \in \mathbb{N}$ and $(r_{2i-1}, r_{2i}) = (r_1, r_2)$ for all $i \in \mathbb{N}$.