Mongolian National Mathematical Olympiad

Answer: $f(x)=Ax+B$ for real numbers $A$ and $B$. For $x\in \mathbb{R}$, let $a=\frac{x-1}{2}$, $b=\frac{x+1}{2}$ and $c=\frac{1-x}{2}$. Then we have $$\{\begin{array}{l}a+b=x\\ b+c=1\\ c+a=0\end{array}$$ and $$\{\begin{array}{l}a-b=-1\\ b-c=x\\ c-a=1-x\end{array}$$ , therefore we have $-f(x)+xf(1)+(1-x)f(0)=0$. Denoting $A=f(1)-f(0)$ and $B=f(0)$, we get $f(x)=Ax+B$.

Conversely, for any $A$ and $B$, the function $f(x)=Ax+B$ satisfies $$(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(c+a)=\sum _{\text{cyc}}A(a^2-b^2)+B(a-b)=0.$$