Saudi Arabian IMO Booklet

Put $c=1$ in the given condition, we have $$f(ab+a+b)=f(a)f(b)+f(a)f(1)+f(b)f(1);\forall a,b\in {\mathbb{Q}}^{+},(1)$$ Put $b=3$ into (1), we have $$f(4a+3)=f(a)f(3)+f(a)f(1)+f(3)f(1);\forall a\in {\mathbb{Q}}^{+}.$$ Put $b=1$ into (1), we have $$f(2a+1)=2f(a)f(1)+f(1{)}^2;\forall a\in {\mathbb{Q}}^{+}.$$ Thus $$f(4a+3)=2f(2a+1)f(1)+f(1{)}^2=4f(1{)}^{2}f(a)+2f(1{)}^3+f(1{)}^2;\forall a\in {\mathbb{Q}}^{+}.$$ From these, we can conclude that $$[f(3)+f(1)]f(a)+f(3)f(1)=4f(1{)}^{2}f(a)+2f(1{)}^3+f(1{)}^2;\forall a\in {\mathbb{Q}}^{+}.$$ If $f(3)+f(1)\ne 4f(1{)}^2$ then $f$ is constant. Thus $f(3)+f(1)=4f(1{)}^2$, otherwise $f$ will be constant. So we must have $$f(3)+f(1)=4f(1{)}^2\text{ and }f(3)f(1)=2f(1{)}^3+f(1{)}^2.$$ Thus $f(3),f(1)$ are solutions of the quadratic equation $t^2-2f(1)t+2f(1{)}^3+f(1{)}^2=0$, thus $$f(1{)}^2-4f(1{)}^2+2f(1{)}^3+f(1{)}^2=0\iff f(1{)}^2(f(1)-1)=0.$$ $$f(ab+a+b)=f(a)f(b)+f(a)+f(b);\forall a,b\in {\mathbb{Q}}^{+}(2)$$ Continue to put $b=1$ and $b=3$, we get $$f(4a+3)=4f(a)+3\text{ and }f(2a+1)=2f(a)+1.$$ Put $a=b=c=\frac{1}{3}$ into the given condition, $f(\frac{1}{3})=3f(\frac{1}{3}{)}^2$ so $f(\frac{1}{3})=\frac{1}{3}$. Put $a=2$ and $b=\frac{1}{3}$ into (2), we have $f(3)=f(2)f(\frac{1}{3})+f(\frac{1}{3})+f(2)$, thus $f(2)=2$. Put $b=c=2$ into the given condition, $f(4a+4)=4f(a)+4$; $\forall a\in {\mathbb{Q}}^{+}$ thus $$4f(a)+4=f(4a+4)=f\left(4\left(a+\frac{1}{4}\right)+3\right)=4f\left(a+\frac{1}{4}\right)+3.$$ From these, we can conclude that $f(a+\frac{1}{4})=f(a)+\frac{1}{4}$, thus $$f(4a+4)=4f(a)+4;\forall a\in {\mathbb{Q}}^{+}.$$ Hence, by induction, one can show that $f(x+n)=f(x)+n$ for all positive integer $n$ and positive real number $x$; thus $f(n)=n,\forall n\in {\mathbb{Z}}^{+}$. Finally, put $b\to n$ and $a\to \frac{m}{n+1}$ for some $m,n\in {\mathbb{Z}}^{+}$ into (2), we get $$f(m+n)=f(n)f\left(\frac{m}{n+1}\right)+f\left(\frac{m}{n+1}\right)+f(n)\to f\left(\frac{m}{n+1}\right)=\frac{m}{n+1}.$$ Thus $f(x)=x$ for all $x\in {\mathbb{Q}}^{+}$. It is easy to check this function satisfies the condition. $\square$