Functional Equation simple problem

How do I show that if there are functions $f,g$ such that$$ f(g(x)+g(y))=bx+cy $$holds for all $b,c\in\mathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?

Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$ y^2+z^2=f(x,g(y-x)+g(z-x)). $$ Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?

For every pair $(x,y)$, $bx+cy=f(g(x)+g(y))=f(g(y)+g(x))=by+cx$.

Thus $b(y-x)=c(y-x)$. Take $x=0,y=1$.

Edit: for the second question, the answer is, in general, no. Take $g$ to be a bijection from $\mathbb{R}$ into a $\mathbb{Q}$-basis of $\mathbb{R}$, which we denote as $\mathcal{B}=(g(x))_x$.

Note that $g(a)+g(b)$ determines the unordered pair $(a,b)$ (because the $g(x)$ are all linearly independent over the rationals).

Take now $f(x,y)=0$ if $y \notin \mathcal{B} + \mathcal{B}$, and $f(x,y)=(a+x)^2+(b+x)^2$ otherwise where $y=g(a)+g(b)$.