National Competition

Setting $y=x$ and $y=1$ yields $$f(x+1)=1+f(x)+\alpha x(1)$$ and $$f(x+1)=\frac{f(x)}{f(1)}+f(1)+\alpha x(2)$$ respectively. Equating (1) and (2) implies $$f(x)\left(1-\frac{1}{f(1)}\right)=f(1)-1.$$ As $f$ cannot be constant due to (1), we obtain $f(1)=1$. By induction, we get $$f(x)=\frac{\alpha }{2}x(x-1)+x\text{for all }x\in {\mathbb{Z}}^{+}.(3)$$ In particular, this implies $f(2)=\alpha +2$ and $f(4)=6\alpha +4$. Setting $x=4$ and $y=2$ in the functional equation yields $${\alpha }^2-2\alpha =0.$$ Thus we must have $\alpha =2$ in order to obtain solutions. From now on, we only consider this case. From (3), we obtain $f(x)=x^2$ for $x\in {\mathbb{Z}}^{+}$. By induction, we obtain that for $x\in {\mathbb{Q}}^{+}$ and $n\in {\mathbb{Z}}^{+}$, from the relation $f(x+n)=(x+n{)}^2$ it follows that $f(x)=x^2$. Let now $\frac{a}{b}\in {\mathbb{Q}}^{+}$ with $a,b\in {\mathbb{Z}}^{+}$. We set $x=a$ and $y=b$ and obtain $$f\left(\frac{a}{b}+b\right)=\frac{a^2}{b^2}+b^2+2a={\left(\frac{a}{b}+b\right)}^2.$$ The above remark implies that $f(\frac{a}{b})=(\frac{a}{b}{)}^2$. It is easily verified that $f(x)=x^2$ is indeed a solution. Thus there is no solution for $\alpha \ne 2$ and the solution $f(x)=x^2$ for $\alpha =2$.