Borbas wrote: Find all fuctions $f,g:\mathbb{R}\rightarrow \mathbb{R}$ such that: $f(x-3f(y))=xf(y)-yf(x)+g(x) \forall x,y\in\mathbb{R}$ and $g(1)=-8$ Let $P(x,y)$ be the assertion $f(x-3f(y))=xf(y)-yf(x)+g(x)$ Let $a=f(0)$ Note that $f\equiv 0$ would imply $g\equiv 0$, impossible since $g(1)=-8$ So $\exists u,v$ such that $f(u)=v\ne 0$ Then comparaison of $P(u,x)$ with $P(u,y)$ implies $f(x)$ injective If $a=0$ : $P(x,0)$ $\implies$ $f(x)=g(x)$ $P(x,x)$ $\implies$ $f(x-3f(x))=g(x)$ Si $f(x)=f(x-3f(x))$ and so, since injective, $f(x)=0$; impossible. So $a\ne 0$ $P(0,x)$ $\implies$ $f(-3f(x))=-ax+g(0)$ and so, since $a\ne 0$, $f(x)$ is bijective. Let then, since surjective, $u$ such that $f(u)=0$. $P(u,u)$ $\implies$ $g(u)=0$ $P(u,x)$ $\implies$ $f(u-3f(x))=uf(x)$ and so, since surjective : $f(u-3x)=ux$ And so $f(x)=cx+a$ $\forall x$ Plugging this back in original equation, we get $c=1$ (using $g(1)=-8$) and $a=3$ And so $\boxed{f(x)=x+3\text{ and }g(x)=-2x-6\quad\forall x}$ Which indeed is a solution

pco wrote: Then comparaison of $P(u,x)$ with $P(u,y)$ implies $f(x)$ injective Can someone explain this part?

Submathematics wrote: Can someone explain this part? If $f(x)=f(y)$, subtracting $P(u,x)$ from $P(u,y)$ gives $v(x-y)=0$ and since we choosed $u$ such that $v=f(u)\ne 0$, this implies $x=y$ and so $f(x)$ injective.

Ans: $f(x)=x+3, g(x)=-2x-6 \ \ \forall x\in \mathbb{R}.$ It is easy to see that the above functions do satisfy the given equation. Now, let $P(x,y)$ be the given assertion, since $f(x)\ne 0 \ \ \forall x\in \mathbb{R}$ (otherwise, $P(1,x)\implies g(1)=0$ a contradiction), we can choose $x\in \mathbb{R}$ such that $f(x)\ne 0$ and let $f(a)=f(b)$, \[P(x,a) \implies f(x-3f(a))=xf(a)-af(x)+g(x)\]\[P(x,b) \implies f(x-3f(b))=xf(b)-bf(x)+g(x)\]and equating these two equations gives $a=b$, so $f$ is injective. We claim that $f$ is a bijection, otherwise we use \[P(0,x)\implies f(-3f(x))=-xf(0)+g(0)\]we have $f(0)=0$ and then \[P(x,0)\implies f(x-3f(0))=f(x)=xf(0)+g(x)=g(x)\]and \[P(x,x) \implies f(x-3f(x))=g(x)=f(x) \implies f(x)=0 \ \ \forall x\in \mathbb{R}\]by injectivity which leads to a contradiction. Thus, there exist $k\in \mathbb{R}$ such that $f(k)=0$ and \[P(k,k) \implies g(k)=0\]\[P(x,k)\implies f(x)=-kf(x)+g(x) \implies g(x)=(1+k)f(x)\]and \[P(x+3f(1+k),1+k) \implies f(x)=xf(1+k)+3f(1+k)^2\]so $f$ is linear. Letting $f(x)=mx+c$ and $g(1)=-8$, we find that $\boxed {f(x)=x+3, g(x)=-2x-6}.$