Vietnamese Mathematical Olympiad

a. With $x,y\in \mathbb{R}$, $P(x,y)$ indicates the proposition containing the variable as follows $$f(x+g(y))=xf(y)+(2023-y)f(x)+g(x).$$ From $P(0,y)$, we deduce $$f(g(y))=(2023-y)f(0)+g(0)=2022(2023-y)+g(0),\forall y\in \mathbb{R}.$$ Note that the right hand side of the above equality is a polynomial of first degree in the variable $y$, so it takes all values on the set of real numbers, in other words $f$ is surjective.

We show that $g$ is injective. Consider $x_1,x_2\in \mathbb{R}$ such that $g(x_1)=g(x_2)$, then from $P(0,x_1)$ and $P(0,x_2)$, we get $$2022(2023-x_1)+g(0)=2022(2023-x_2)+g(0)\text{ or }{x}_1=x_2.$$ Thus $g$ is injective, the proof is complete.

b. Now consider the following equation $$f(g(y))=(2023-y)f(0)+g(0)=2022(2023-y)+g(0).(1)$$ From $P(g(x),y)$, we deduce $$f(g(x)+g(y))=g(x)f(y)+(2023-y)f(g(x))+g(g(x)),\forall x,y\in \mathbb{R}.$$ Combined with (1), one can get $f(g(x)+g(y))=g(x)f(y)+(2023-y)[2022(2023-x)+g(0)]+g(g(x))$. Swap $x,y$ in the above equation and then compare, we get $$\begin{array}{rl}& g(x)f(y)+g(g(x))+(2023-y)g(0)\\ =& g(y)f(x)+g(g(y))+(2023-x)g(0),\forall x,y\in \mathbb{R}.\end{array}(2)$$ Since $f$ is surjective, there exists a real number $a$ such that $f(a)=0$. Substituting $y=a$ in (2), we get $$g(g(x))=-g(0)x+g(a)f(x)+C,\forall x\in \mathbb{R}$$ where $C$ is a constant. Substitute back to (2), get $g(x)f(y)-g(0)x+g(a)f(x)+C+(2023-y)g(0)=g(y)f(x)-g(0)y+g(a)f(y)+C+(2023-x)g(0)$, for all $x,y\in \mathbb{R}$. Simplifying this to get $$g(x)f(y)+g(a)f(x)=g(y)f(x)+g(a)f(y),\forall x,y\in \mathbb{R}.$$ Substitute $y=0$ into the above equation $$2022g(x)+g(a)f(x)=g(0)f(x)+2022g(a)$$ thus $$g(x)=\frac{g(0)-g(a)}{2022}f(x)+g(a),\forall x\in \mathbb{R}.$$ If $g(a)=g(0)$ then $a=0$ or $f(a)=f(0)=0$, contradicts $f(0)=2022$. So $g(a)\ne g(0)$, from which it follows that $g$ is surjective. Then there exists some value $b$ such that $g(b)=0$. From $P(x,b)$ we get $$f(x)=xf(b)+(2023-b)f(x)+g(x),\forall x\in \mathbb{R}.$$ Substituting back to $P(x,y)$ we get $$\begin{array}{rl}f(x+g(y))& =xf(y)+(2023-y)f(x)+f(x)\\ & -xf(b)+(b-2023)f(x)\\ & =x(f(y)-f(b))+f(x)(1+b-y),\forall x,y\in \mathbb{R}.\end{array}$$ Substituting $y=1+b$ to get $$f(x+g(1+b))=x(f(1+b)-f(b)),\forall x\in \mathbb{R}$$ or $f$ is linear over $\mathbb{R}$, and the same for $g$. At this point, we just need to replace it with $P(x,y)$ to complete the solution. There are two pairs of $(f(x),g(x))$ as follows $$\{\begin{array}{l}f(x)=\frac{-1\pm \sqrt{5}}{2}x+2022,\\ g(x)=1011(-1\mp \sqrt{5})x+2\cdot 1011^2\cdot (-3\mp \sqrt{5})\end{array},\forall x\in \mathbb{R}.$$ $\square$