Find all triples of continuous functions $f, g, h$ from $\mathbb{R}$ to $\mathbb{R}$ such that $f(x+y)=$ $g(x)+h(y)$ for all real numbers $x$ and $y$.
The answer is $f(x)=c x+a+b, g(x)=c x+a, h(x)=c x+b$, where $a$, $b, c$ are real numbers. Obviously these solutions work, so we wish to show they are the only ones.

First, put $y=0$ to get $f(x+0)=g(x)+h(0)$, so $g(x)=f(x)-h(0)$. Similarly, $h(y)=f(y)-g(0)$. Therefore, the functional equation boils down to $f(x+y)=$ $f(x)+f(y)-(g(0)+h(0))$. By shifting and appealing to Cauchy's functional equation (with $f$ continuous) we get $f(x)=c x+g(0)+h(0), g(x)=c x+g(0)$ and $h(x)=$ $c x+h(0)$.