Find all functions $f:\mathbb{Q}\to \mathbb{Q}$ such that for any $x,y\in{}\mathbb{Q}$ we have $$\{f(x)\}+\{f(y)\}=\{f(x+y)\}\text.$$
Note that $\{t\}$ denotes the fractional part of $t$ for instance $\{1.5\}=0.5$
Progress: Since $t-\lfloor t\rfloor=\{t\}$, I found out that
\begin{align*} f(x)+f(y)-\lfloor f(x)\rfloor-\lfloor f(y)\rfloor &= f(x+y)-\lfloor f(x+y) \rfloor. \\ f(x)+f(y)-f(x+y)&=\lfloor f(x)\rfloor +\lfloor f(y) \rfloor -\lfloor f(x+y) \rfloor. \end{align*}
So one trivial solution is $f(x)=c$ where $c$ is an integer.
Hint: Put $g(x)=\{f(x)\} \in [0,1[$. Then you have $g(x+y)=g(x)+g(y)$ for all $x,y$. Show that for $n\in \mathbb{N}$, and all $x\in \mathbb{Q}$, we have $g(nx)=ng(x)$.