Ukrainian Mathematical Olympiad

a) Suppose that for some real numbers $x$ and $y$ the equality $\{x\}+\{2y\}=\{y\}+\{2x\}$ holds. Let $\{x\}=\alpha$, $\{y\}=\beta$, $\alpha ,\beta \in [0;1)$. It suffices for $\alpha ,\beta \in [0;1)$ to consider the equality $\alpha +\{2\beta \}=\beta +\{2\alpha \}$.

If $0\le \gamma <\frac{1}{2}$, then $\{2\gamma \}=2\gamma$, and for $\frac{1}{2}\le \gamma <1$, $\{2\gamma \}=2\gamma -1$. Note that for the cases $0\le \alpha <\frac{1}{2}\le \beta <1$ and $0\le \beta <\frac{1}{2}\le \alpha <1$, the equality $\alpha +\{2\beta \}=\beta +\{2\alpha \}$ cannot hold. If $0\le \alpha ,\beta <\frac{1}{2}$ or $\frac{1}{2}\le \alpha ,\beta <1$, then the equality $\alpha +\{2\beta \}=\beta +\{2\alpha \}$ takes the form $\alpha =\beta$.

b) Answer: no, such integers do not exist. Take $x=0$, $y=\frac{1}{n-1}$. Then $\{x\}\ne \{y\}$, but, as is easy to see, $\{x\}+\{ny\}=\{y\}+\{nx\}$.