Find all real-valued functions $f$ satisfying $f(2x+f(y))+f(f(y))=4x+8y$

Question

$ x,y \in\mathbb{R}$ and $f:\mathbb{R} \rightarrow \mathbb{R}$, find all functions satisfying,

$$f(2x+f(y))+f(f(y))=4x+8y$$ I couldn't seem to understand it.

Answer 1

Put $f(0)=k$. Then taking $x=y=0$ the given equation becomes $f(k)+f(k)=0$, so $f(k)=0$. So $f(f(0))=0$.

Now put $y=0$ and the original equation become $f(2x+k)=4x$. Putting $x=-k/2$ gives $k=f(0)=-2k$, so $k=0$. Hence the only solution is $f(x)=2x$.

Answer 2

Clearly, $f(f(0)) = 0$.

Next, set $y =0$ and $x=-\frac{1}{2}f(0)$ then it follows $f(0) = -2f(0)$ which means $f(0) = 0$. Thus, if we set $y=0$ we have \begin{align} f(2x) = 4x \ \ \implies \ \ f(z) = 2z. \end{align}