China Hong Kong Mathematical Olympiad

$r$ can be any number of the form $1 + \frac{1}{k}$, where $k$ is an integer different from $0$, $-1$.

If $r > 1$, let $r = 1 + \frac{m}{n}$ where $m, n \in \mathbb{Z}^+$ and $(m, n) = 1$. Since
$$r^{\frac{1}{r-1}} = \left(\frac{n+m}{n}\right)^{\frac{n}{m}}$$
and $(n+m, n) = (m, n) = 1$, we must have $n+m = a^m$ and $n = b^m$ for some $a, b \in \mathbb{Z}^+$. This gives $a^m - b^m = m$. Clearly, $a \ge b + 1$. Therefore, we have
$$m = a^m - b^m = (a-b)(a^{m-1} + a^{m-2}b + \dots + ab^{m-2} + b^{m-1}) \ge (1)(1+1+\dots+1) = m.$$
Equality must hold. This means $m = 1$ (since otherwise $a = b = 1$, contradiction) and $a - b = 1$. Conversely, when $m = 1$, $r = 1 + \frac{1}{n}$, so that $r^{\frac{1}{r-1}} = \left(\frac{n+1}{n}\right)^n \in \mathbb{Q}$.

If $r < 1$, let $r = 1 - \frac{m}{n}$ where $m, n \in \mathbb{Z}^+$ and $(m, n) = 1$. Since
$$r^{\frac{1}{r-1}} = \left(\frac{n-m}{n}\right)^{-\frac{n}{m}} = \left(\frac{n}{n-m}\right)^{\frac{n}{m}}$$
and $(n, n-m) = (n, -m) = 1$, we must have $n = a^m$ and $n-m = b^m$ for some $a, b \in \mathbb{Z}^+$. This gives $a^m - b^m = m$. Again, the only solution is $m = 1$ and $a-b = 1$.

When $m = 1$, $r = 1 - \frac{1}{n}$. Since $r > 0$, we need $n > 1$. In that case, we have
$$r^{\frac{1}{r-1}} = \left(\frac{n}{n-1}\right)^n \in \mathbb{Q}\$.$$
Therefore, $r = 1 + \frac{1}{n}$ for $n \ge 1$ or $r = 1 - \frac{1}{n}$ for $n > 1$.