Austria2019

Answer. The solutions are all pairs $(a,b)$ with $a=0$ or $b=0$ or $a=b$ or both $a$ and $b$ integers.

Let $a_0=\lfloor a\rfloor$ and $a_i$ be the binary digits of the fractional part of $a$ such that $a=a_0+{\sum }_{i=1}^{\infty }\frac{a_i}{2^i}$ with $a_0\in \mathbb{Z}$ and $a_i\in \{0,1\}$ for $i\ge 1$. Similarly, let $b=b_0+{\sum }_{i=1}^{\infty }\frac{b_i}{2^i}$ with $b_0\in \mathbb{Z}$ and $b_i\in \{0,1\}$ for $i\ge 1$. In the case of a non-unique binary expansion, we choose the expansion ending on infinitely many zeros. Now choose $n=2^k$ and $m=2^{k-1}$ in the given equation. We get the equations $$\begin{array}{rl}a\left(2^k{b}_0+\sum _{i=1}^k{b}_i{2}^{k-i}\right)& =b\left(2^k{a}_0+\sum _{i=1}^k{a}_i{2}^{k-i}\right),\\ a\left(2^{k-1}{b}_0+\sum _{i=1}^{k-1}{b}_i{2}^{k-i-1}\right)& =b\left(2^{k-1}{a}_0+\sum _{i=1}^{k-1}{a}_i{2}^{k-i-1}\right).\end{array}$$ The first equation for $k=0$ and the difference of the first equation and the doubled second equation for $k\ge 1$ yields $$a{b}_k=b{a}_k\text{\hspace{1em}(1)}$$ for $k\ge 0$. Now, we consider three cases. If one or both of $a$ and $b$ are zero, then the original equation is clearly satisfied. If both fractional parts are zero, then both numbers are integers and again, the original equation is satisfied. So, finally, we consider the case that $a,b\ne 0$ and that there is a $k\ge 1$ with $a_k=1$. The equation (1) shows that $b_k$ cannot be zero, so we get $b_k=1$ and thus from the same equation $a=b$. This clearly satisfies the original equation. (Of course, $b_k=1$ leads to the same conclusion.) Therefore, the solutions are exactly the pairs listed in the answer.