SAUDI ARABIAN IMO Booklet 2023

Put $x=0,y=-1$: $0=-2f(0)-f(0)$ thus $f(0)=0$.

Put $y=1$: $2f(x)=f(2x)$, for all $x$ so the given condition can rewrite as $$(y+1)f(x)f(y-1)=yf(xy)-f(x),\forall x,y$$ Continue to put $y=0$ then $f(x)f(-1)=-f(x)$. If $f(-1)\ne -1$ then $f(x)=0$ for all $x$, which is satisfied the given condition. Now suppose that there exist some $x_0$ such that $f(x_0)\ne 0$, this implies that $f(-1)=-1$ and $f(1)=1$.

In (), put $x=1$ then $(y+1)f(y-1)=yf(y)-1$ for all $y$. Replace to the LHS of (), one can get $$(yf(y)-1)f(x)=yf(xy)-f(x)$$ so $f(xy)=f(x)f(y)$, $\forall x,y\ne 0$ which implies that $f$ is multiplication function. Also put $y=-1$ in () then $f(x)=-f(-x)$ which implies that $f(x)$ is odd.

Suppose that $a$ is some number such that $f(a)=0$ then by substituting $x=a$ into (), we get $0=yf(ay)-0$ so $f(ay)=0,\forall y\ne 0$. If $a\ne 0$ then $ay$ can take any values on $\mathbb{R}$ so $f(x)\equiv 0$, contradiction. Thus $a=0$ is the unique value such that $f(a)=0$. Since $(y+1)f(y-1)=yf(y)-1$, change $y\to -y$ and using the property of odd function, one can get $$(-y+1)f(-y-1)=yf(y)-1,\forall y\in \mathbb{R}.$$ Thus for all $y\ne \pm 1$ we have $$(y+1)f(y-1)=(y-1)f(y+1)\iff \frac{f(y+1)}{f(y-1)}=\frac{y+1}{y-1}$$ for all $y\ne \pm 1$. Due to the multiplication property of $f$, rewrite the last equation as $$f\left(\frac{y+1}{y-1}\right)=\frac{y+1}{y-1},\forall y\in \mathbb{R}\setminus \{\pm 1\}.$$ Note that $t=\frac{y+1}{y-1}$ can take any value on $\mathbb{R}\setminus \{1\}$ so $f(t)=t,\forall t\ne 0,1$. But $f(1)=1,f(0)=0$ so we have $f(x)=x,\forall x\in \mathbb{R}$. Hence, there are two functions that satisfying the given condition: $f(x)=0,\forall x\in \mathbb{R}$ and $f(x)=x,\forall x\in \mathbb{R}$.