Thailand Mathematical Olympiad

Suppose that there are $a,b\in \mathbb{R}$ such that $f(a+b)\ne f(a)+f(b)$. By the assumption we have $f(a+b)=-f(a)-f(b)$. If $f(a+b)=0$ then $f(a)=-f(b)$. $$\begin{array}{rl}|f(2a+2b)|& =|f(a+(a+b+b))|\\ & =|f(a)+f((a+b)+b)|\\ & =|f(a)+f(a+b)+f(b)|\text{ or }|f(a)-f(a+b)-f(b)|\\ & =0\text{ or }|f(a)-(-f(a)-f(b))-f(b)|\\ & =0\text{ or }2|f(a)|.\end{array}$$ But $|f(2a+2b)|=|f((a+b)+(a+b))|=|2f(a+b)|=2|f(a+b)|\ne 0$. Therefore, $2|f(a)|=|f(2a+2b)|=2|f(a+b)|$ and thus $|f(a+b)|=|f(a)|$. Similarly, we can show that $|f(a+b)|=|f(b)|$. So we have $|f(a)|=|f(b)|$. That is $f(a)=f(b)$ or $f(a)=-f(b)$. If $f(a)=f(b)$, then $2|f(a)|=|f(a)+f(a)|=|f(a)+f(b)|=|f(a+b)|=|f(a)|$ and thus $f(a)=0$ which implies that $f(a+b)=-f(a)-f(b)=-2f(a)=0$, a contradiction. On the other hand, if $f(a)=-f(b)$, then $0=|f(a)+f(b)|=|f(a+b)|$, again a contradiction. So in any case we get a contradiction. $\square$