China Mathematical Competition

Let $f(x)=2x^3+5x-2$. Then we have $f^{\prime }(x)=6x^2+5>0$, 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 $a_1<a_2<\cdots <a_n<\dots$ and $b_1<b_2<\cdots <b_n<\dots$ satisfying $$r^{a_1}+r^{a_2}+r^{a_3}+\cdots =r^{b_1}+r^{b_2}+r^{b_3}+\cdots =\frac{2}{5}$$ Deleting the terms that appear at both sides of the expression, we have $$r^{s_1}+r^{s_2}+r^{s_3}+\cdots =r^{t_1}+r^{t_2}+r^{t_3}+\dots$$ 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 $$\begin{array}{rl}{r}^{s_1}<r^{s_1}+r^{s_2}+\dots & =r^{t_1}+r^{t_2}+\dots ,\\ 1<r^{t_1-s_1}+r^{t_2-s_1}+\dots & \le r+r^2+\dots \\ & =\frac{1}{1-r}-1<\frac{1}{1-\frac{1}{2}}-1=1.\end{array}$$ It is a contradiction. This proves that $\{a_n\}$ is unique.