The DANUBE Mathematical Competition

We prove that the only function satisfying the two conditions is $f(x)=x$. We prove that $f$ is an injection. If we put $y=0$ in (i) we get $f(f(x^2)+f(0))=x^2+2f(0)$, for any $x$, or equivalently, $$f(f(a)+f(0))=a+2f(0)(1)$$ for any number $a\ge 0$. From (1), $f$ is an injection on the set of all nonnegative numbers. We fix now $y$. From the conditions (i) and (ii) we get that $f(\cdot )$ is a superior unbounded function. If $f(y_1)=f(y_2)$ then $f(f(x^2)+y_1+f(y_1))=f(f(x^2)+y_2+f(y_2))$. For large enough values of $x$ the expressions $f(x^2)+y_1+f(y_1)$ and $f(x^2)+y_2+f(y_2)$ are positive and thus $y_1=y_2$.

We prove now that $f(0)=0$. Case 1. $f(0)\le 0$. For $a=-2f(0)$, we have $f(f(-2f(a))+f(0))=0$. Thus, there exists a real number $c$ such that $f(c)=0$. Plugging $x=0$ and $y=c$ in (i) we get $f(f(0)+c)=0$. As $f$ is an injection, we get $f(0)+c=c$, that is $f(0)=0$. Case 2. $f(0)\ge 0$. In (i) we put $x=y=0$: $f(2f(0))=2f(0)$. We plug now $a=3f(0)=f(0)+f(2f(0))$. From (1) we have $f(a)=f(f(2f(0))+f(0))=2f(0)+2f(0)=4f(0)$. We add now $f(0)$ and we get $f(f(a)+f(0))=f(5f(0))=3f(0)+2f(0)=5f(0)$. By plugging now $x=0$ and $y=2f(0)$ in (i) we get $f(5f(0))=4f(0)$. Thus $f(5f(0))=5f(0)=4f(0)$, from were we obtain $f(0)=0$. From (1), for $a\ge 0$ we have $f(f(a))=a$, and from (i), for $x=0$, we have $f(y+f(y))=2f(y)$ for any real number $y$. We plug now $y=f(a)$: $f(f(a)+a)=2f(f(a))=2a$ and $y=a$: $f(a+f(a))=2f(a)$. We obtained that for any $a\ge 0$ we have $f(a)=a$. Then, (i) becomes $f(x^2+y+f(y))=x^2+2f(y)$. We fix now any real $y$. There exists a $x\in \mathbb{R}$ such that $x^2+y+f(y)>0$. Then $f(x^2+y+f(y))=x^2+y+f(y)=x^2+2f(y)$ and thus $f(y)=y$.