The original problems includes part a too This version is just part b. Part a: Prove $f(x)$ is surjective and $g(x)$ is injective Edit: This problem is a bit harder than ISL 2000 A3

a_507_bc wrote: Find all functions $f, g: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f (0)=2022$ and $f (x+g(y)) =xf(y)+(2023-y)f(x)+g(x)$ for all $x, y \in \mathbb{R}$. Let $a=2022=f(0)$, $u=g(a+1)$ and $v=f(a+1)$ Let $P(x,y)$ be the assertion $f(x+g(y))=xf(y)+(a+1-y)f(x)+g(x)$ $P(0,x)$ $\implies$ $f(g(x))=(a+1-x)a+g(0)$ (and so, since $a\ne 0$, $f(x)$ is surjective and $g(x)$ is injective) $P(0,a+1)$ $\implies$ $f(u)=g(0)$ $P(u,x)$ $\implies$ $f(u+g(x))=uf(x)+(a+1-x)f(u)+g(u)$ $P(g(x),a+1)$ $\implies$ $f(g(x)+u)=g(g(x))+vg(x)$ Subtracting, this becomes $g(g(x))=uf(x)+(a+1-x)f(u)+g(u)-vg(x)$ $P(g(x),y)$ $\implies$ $f(g(x)+g(y))=g(x)f(y)+(a+1-y)f(g(x))+g(g(x))$ $=g(x)f(y)+(a+1-y)((a+1-x)a+f(u))+uf(x)+(a+1-x)f(u)+g(u)-vg(x)$ Swapping there $x,y$ and subtracting, we get : $(g(x)-u)(f(y)-v)=(g(y)-u)(f(x)-v)$ And so $g(x)=\alpha f(x)+\beta$ for some $\alpha\ne 0$ (since injective) and some $\beta$ Note that this implies that $g(x)$ is also surjective and $f(x)$ is also injective. Let then $t$ such that $g(t)=0$ : $P(x,t)$ $\implies$ $f(x)=xf(t)+(a+1-t)f(x)+\alpha f(x)+\beta$ Which is $(a+\alpha-t)f(x)=-xf(t)-\beta$ If $a+\alpha-t\ne 0$, this implies $f(x)=bx+a$ (since $f(0)=a$) and so $g(x)=dx+e$ for some $b,d\ne 0,c,e$ Plugging this in original equation, we get $t=a\frac{-1\pm\sqrt 5}2$ and $\alpha=a\frac{-3\pm\sqrt 5}2$ and so unfortunately $a+\alpha=t$, and so this does not fit. So $a+\alpha-t=0$, which implies $f(t)=0$ and $\beta=0$ Plugging $g(x)=\alpha f(x)$ in $P(x,y)$, this becomes $f(x+\alpha f(y))=xf(y)+(a+1+\alpha-y)f(x)$ And so $f(x+\alpha f(y)+\alpha f(z))=(x+\alpha f(z))f(y)+(a+1+\alpha-y)f(x+\alpha f(z))$ $=(x+\alpha f(z))f(y)+(a+1+\alpha-y)(xf(z)+(a+1+\alpha-z)f(x))$ Swapping $y,z$ and subtracting, we get, setting $x=1$ : $(a+\alpha-y)f(z)=(a+\alpha-z)f(y)$ And so $f(x)=bx+a$ (since $f(0)=a$) and so $g(x)=\alpha(bx+a)$ Plugging this in original equation, we get two solutions : $\boxed{\text{S1 : }f(x)=\frac{-1-\sqrt 5}2x+2022\text{ and }g(x)=2022\frac{-1+\sqrt 5}2x+2022^2\frac{-3+\sqrt 5}2\quad\forall x}$, which indeed fits $\boxed{\text{S2 : }f(x)=\frac{-1+\sqrt 5}2x+2022\text{ and }g(x)=2022\frac{-1-\sqrt 5}2x+2022^2\frac{-3-\sqrt 5}2\quad\forall x}$, which indeed fits

how you can find that $$\boxed{ f(x)=\frac{a(x-a)}{a-2022} \textrm{ and } g(x)=a(x-a), \, \forall \, x \in \mathbb{R}; \textrm{ where } a=1011\left ( \pm \sqrt{5}-1 \right ).}$$

ZAO wrote: how you can find that $$\boxed{ f(x)=\frac{a(x-a)}{a-2022} \textrm{ and } g(x)=a(x-a), \, \forall \, x \in \mathbb{R}; \textrm{ where } a=1011\left ( \pm \sqrt{5}-1 \right ).}$$ I dont undestand your question. My proof of this result is given all along my previous post. What is the first part you dont understand ?

No it is laikhanhhoang_3011's solution I don't understand his conclusion

ZAO wrote: No it is laikhanhhoang_3011's solution I don't understand his conclusion Notice that $f$ is surjective and $f(a+(a-2022)f(y))=af(y)$, i can say $$f(a+(a-2022)y)=ay \Rightarrow f(y)=\frac{a(y-a)}{a-2022}.$$Then you can do some smail calculation to get $g(x).$