SAUDI ARABIAN IMO Booklet 2023

Let $x=y=0$ in the problem, we have $f(0)=f(0)+f(0{)}^2$ so $f(0)=0$.

Continue to replace $(x,y)=(1,-1)$ into the problem, we have $f(-1)=f(1)\cdot f(-1)$ deduce $f(1)=1$ or $f(-1)=0$. However, if $f(-1)=0$, substituting $y=-1$ in the problem, then $f(-x)=0$ for all real numbers $x$, not satisfied. Thus, $f(1)=1$.

Continue to replace $x=y=-1$, we have $f(1)=f(0)+f(-1{)}^2$ inferred $f(-1)=1$ or $f(-1)=-1$. But if $f(-1)=1$, substitute $y=-\frac{1}{2}$ then $f(\frac{x}{2})=-f(x)f(-\frac{1}{2})$, $\forall x\in \mathbb{R}$; continue to substitute $x=-1$ then $f(-\frac{1}{2})=0$ so $f(\frac{x}{2})=0$, $\forall x\in \mathbb{R}$, also not satisfied. Hence, $f(-1)=-1$.

Finally, substitute $y\to -1$ in the given, then $f(-x)=-f(x)$ for all real numbers $x$, that is, $f$ is an odd function.

Change $y$ by $-y$, $y-1$ and using the property $f$ as an odd function, then $$f(x)f(y)=f(2xy-x)-f(xy-x)=f(xy)+f(x)f(y-1)-f(xy-x).(\ast )$$ Thus for all positive integers $y$ then $$\begin{array}{rl}f(xy)-f(x)f(y)& =f(x(y-1))-f(x)f(y-1)\\ & =f(x(y-2))-f(x)f(y-2)\\ & =\cdots =f(0)-f(x)f(0)=0.\end{array}$$ From this it follows that $f(xy)=f(x)f(y)$ for all $x\in \mathbb{R},y\in {\mathbb{Z}}^{+}$. In $(\ast )$, let $(x,y)=(1,2)$ then $f(3)-f(1)=f(2{)}^2$. Let $y=1$ in the given equation, then $$f(2x+1)=f(x+1)+f(x),\forall x\in \mathbb{R}.$$ Continue to replace $x=1$ then $f(2)=f(3)-f(1)=f(2{)}^2$ so $f(2)=2$ or $f(2)=0$. Notice that if $f(2)=0$, then due to the multiplication, we have $f(2x)=f(2)f(x)=0$, which is also not satisfied. Therefore, $f(2)=2$ and $f(2x)=2f(x)$, $\forall x\in \mathbb{R}$ and $f(2x+1)=f(x)+f(x+1)$.

Here, by induction, we can prove $f(n)=n$ for all positive integers $n$. Replace $y\to \frac{y}{x}$ in the given, then $$f(2y+x)=f(x+y)+f(x)f\left(\frac{y}{x}\right),\forall x\ne 0,y\in \mathbb{R}.(1)$$ Change the role $x,y$ of the above expression then $$f(2x+y)=f(x+y)+f(y)f\left(\frac{x}{y}\right),\forall y\ne 0,x\in \mathbb{R}.(2)$$ Replace $x\to 2x$ in (1) then $$f(2x+2y)=f(2x+y)+f(2x)f\left(\frac{y}{2x}\right)=f(x+y)+f(y)f\left(\frac{x}{y}\right)+f(2x)f\left(\frac{y}{2x}\right),\forall x,y\ne 0.$$ On the other hand, $f(2x)=2f(x)$ so $f(x+y)=f(x)f\left(\frac{y}{x}\right)+f(y)f\left(\frac{x}{y}\right)$, $\forall x,y\ne 0$. Adding the sides of (1) and (2), combined with this last equality, we get $$f(2x+y)+f(2y+x)=2f(x+y)+f(x+y)+f(y)f\left(\frac{x}{y}\right)+f(x)f\left(\frac{y}{x}\right)=3f(x+y),\forall x,y\ne 0.$$ To handle the condition $x,y\ne 0$, replace $(x,y)\to (\frac{2b-a}{3},\frac{2a-b}{3})$ with $a,b\in \mathbb{R}$ then $$f(a)+f(b)=f(a+b);\forall a\ne 2b\text{ vaˋ }b\ne 2a.$$ For every pair of numbers $a,b\in \mathbb{R}$, we choose $c$ such that $c>4|a|+4|b|$ then make sure $c\ne 2a,c\ne 2b$. We have $$f(c)+f(a+b)=f(a+b+c)=f(c+a)+f(b)=f(c)+f(a)+f(b).$$ From here we infer that $f$ is an additive function on $\mathbb{R}$ so $f(xy)=f(2xy+x)-f(xy+x)=f(x)f(y)$, $\forall x,y\in \mathbb{R}$. Therefore, $f$ adds, multiplies, and is non-constant, so $f(x)=x$. It is easy to check that they satisfy the given condition. So $f(x)=x$ with $x\in \mathbb{R}$. $\square$