Functional equation $f(f(x)+3y)=12x + f(f(y)-x)$ I found this problem on a French exchange forum : 

Find all the $f : \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)+3y)=12x + f(f(y)-x)$

In fact I solved the problem when $f$ is supposed to be continuous. Then it can be shown that $f(x)=3x+\lambda$ for some real constant $\lambda$.
What about the general case, when we do not suppose $f$ to be continuous?

It is easy to see that such a function will be surjective. For the sake of proving injectivity, assume $f(a)=f(b)$ and set $n$ so that $f(n)=a+b$.  Then $$f(f(a)+3n)=12a+f(f(n)-a)=12a+f(b)$$ and $$f(f(b)+3n)=12b+f(f(n)-b)=12b+f(a)$$ so $a=b$, proving that the function is bijective. Then if we substitute $x=0$, we see that $f(3y+f(0))=f(f(y))$ which by injectivity, means that $f(y)=3y+c$ for some constant $ c=f(0)$. 
