Team Selection Test for IMO

There is only one such function, that is $f(x)=x^2+x+1$.

Writing $x=0$ in (i) gives $f(0)=f(0{)}^2$. As $f(x)>0$, $f(0)=1$.

Plugging in $-x$ in (i) gives $f(x{)}^2-2xf(x)=f(-x{)}^2+2xf(-x)$, that is $(f(x)-f(-x))(f(x)+f(-x))=2x(f(x)+f(-x))$. Again as $f(x)>0$, we obtain $f(x)=f(-x)+2x$.

Then by (ii) we have $f(x)=2x+f(x-1)$ ().

By induction on $n$ we can show that $f(n)=n^2+n+1$ for all non-negative integer $n$.

By () applying induction on $n$ gives $f(x+n)=f(x)+2xn+n^2+n$ () for all real number $x$ and non-negative integer $n$.

Let $x=\frac{m}{n}$ where $m$ and $n$ are positive integers. Then by () we have $f((x+n{)}^2)=f(x^2+2m+n^2)=f(x^2)+2x^2(2m+n^2)+(2m+n^2{)}^2+2m+n^2$.

On the other hand by (i) and (*) we get $f((x+n{)}^2)=(f(x)+2m+n^2+n{)}^2-2(x+n)(f(x)+2xn+n^2+n)$.

The last two equations conclude that $f(x)=x^2+x+1$ for all positive rational number $x$.

Since both $f(x)$ and $x^2+x+1$ are strictly increasing on $(1,\infty )$ and there exists a rational number in any interval, we can easily show that $f(x)=x^2+x+1$ for every real number $x>1$.

Using () and (iii) we can show that $f(x)=x^2+x+1$ for all real number $x$ and it satisfies all three conditions.