China Mathematical Competition

Let $f(x) = 2x^3 + 5x - 2$. Then we have $f'(x) = 6x^2 + 5 > 0$, which means $f(x)$ is strictly increasing. Furthermore, $f(0) = -2 < 0$, $f(\frac{1}{2}) = \frac{3}{4} > 0$. Therefore, $f(x)$ has a unique real root $r \in (0, \frac{1}{2})$. From $2r^3 + 5r - 2 = 0$, we have
$$\frac{2}{5} = \frac{r}{1 - r^3} = r + r^4 + r^7 + r^{10} + \dots$$
Therefore, sequence $a_n = 3n - 2$ ($n = 1, 2, \dots$) satisfies the required condition.

Assume there are two different positive integer sequences
$a_1 < a_2 < \dots < a_n < \dots$ and $b_1 < b_2 < \dots < b_n < \dots$
satisfying
$$r^{a_1} + r^{a_2} + r^{a_3} + \dots = r^{b_1} + r^{b_2} + r^{b_3} + \dots = \frac{2}{5}$$
Deleting the terms that appear at both sides of the expression, we have
$$r^{s_1} + r^{s_2} + r^{s_3} + \dots = r^{t_1} + r^{t_2} + r^{t_3} + \dots$$
where $s_1 < s_2 < s_3 < \dots$, $t_1 < t_2 < t_3 < \dots$ with all the $s_i$ and $t_j$ different from each other.
We may as well assume that $s_1 < t_1$. Then
$$\begin{align*} r^{s_1} < r^{s_1} + r^{s_2} + \dots &= r^{t_1} + r^{t_2} + \dots, \\ 1 < r^{t_1 - s_1} + r^{t_2 - s_1} + \dots &\le r + r^2 + \dots \\ &= \frac{1}{1 - r} - 1 < \frac{1}{1 - \frac{1}{2}} - 1 = 1. \end{align*}$$
It is a contradiction. This proves that $\{a_n\}$ is unique.