SAUDI ARABIAN IMO Booklet 2023

The answer is No. Put $k=\frac{2022}{2023}$ and let $x=0$ then $$f(y+af(y))=k\cdot \frac{y}{a}+f(0)$$ so $f$ is surjective over $\mathbb{Q}$. Put $x=-af(y)$, then $f(y)=k\cdot \frac{y}{a}+f(-af(y))$ it is easy to see that $f$ is injective. So $f$ is bijective. Put $y=0$, then $f(x+af(0))=f(x)$ so $x+af(0)=x$, resulting in $f(0)=0$. Replace $x=0$ then $$f(y+af(y))=k\cdot \frac{y}{a}$$ so $y+af(y)=f^{-1}(k\cdot \frac{y}{a})$, where $f^{-1}$ is the inverse of $f$. From this, it follows that $y+af(y)$ is also surjective over $\mathbb{Q}$. Rewrite the problem as $$f(x+y+af(y))=f(y+af(y))+f(x)$$ and replace $y+af(y)=t\in \mathbb{Q}$ then $$f(x+t)=f(t)+f(x),\forall x,t\in \mathbb{Q}.$$ Therefore, $f$ is additive on $\mathbb{Q}$, thus there exist $c\in \mathbb{Q}$ such that $f(x)=cx,\forall x\in \mathbb{Q}$. Replace to the original equation to get $$c(x+y+acy)=k\cdot \frac{y}{a}+cx$$ or $$\left(c+a{c}^2-k\cdot \frac{1}{a}\right)y=0,\forall y\in \mathbb{Q}.$$ From that we have $(ac{)}^2+ac-k=0$, obviously $\Delta =1+4k$ is not the square of the rational number so the equation of variable $t=ac$ has no rational solution. Therefore, there does not exist $a\in \mathbb{Q}$ for a function $f:\mathbb{Q}\to \mathbb{Q}$ that satisfies.