39th Balkan Mathematical Olympiad | BMO 2022 - Problem 3 | Solution - 6 May 2022


Problem 3 Find all functions $f: (0, \infty) \to (0, \infty)$ such that $\begin{align} f(y(f(x))^3 + x) = x^3f(y) + f(x) \end{align}$ for all $x, y>0$ . Solution We claim that the only solution is $f(x)=x$ for all positive $x$ . It obviously works so we will prove that no other function works. Claim 1: $f$ is increasing Proof Assume that $x_0>x_1$ . Subtituting $x=x_1, y= \frac{x_0-x_1}{f(x_1)^3}$ we get $f(x_0)=x^3f(\frac{x_0-x_1}{f(x_1)^3})+f(x_1)>f(x_1)$ Claim 2: $\lim_{y \to 0} (f(y)) = 0$ Proof $f$ is increasing and therefeore continuous almost everywhere, so there is some $x_0$ in which it is continuous. Subtitute $x=x_0,y \to 0$ . We get that $\lim_{y \to 0} (f(x_0)+x_0^3f(y)) = \lim_{y \to 0} (f(y(f(x_0))^3+x_0))=f(x_0)$ , and therefore $\lim_{y \to 0} (f(y)) = 0$ . Claim 3: $f$ is continuous Proof Subtitute $y \to 0$ : $\lim_{y \to 0} (f(y(f(x))^3+x))= \lim_{y \to 0} (x^3f(y)+f(x))=f(x)$ ,so $f$ is continuous from above. We will prove that $f$ is also continuous from below. We will fix some $x$ and take small enough $x_0$ . Then there exists exactly one value of $y$ for which $x=y(f(x_0))^3+x_0$ . Notice that $y(f(x_0))^3+x_0$ is monotone in $x_0$ and therefore for each value of $x_0$ we will get a different $y$ . If for all small enough $x_0$ $f$ won't be continuous at $y$ then it won't be continuous in an uncountable number of points, and that is a contradiction to Claim 1. So there exists some $x_0$ for which $f$ is continuous in the unique $y_0$ satisfiyng $x=y(f(x_0))^3+x_0$ . subtitute $x=x_0$ and $y \to y_0^-$ and get: $\lim_{y \to y_0^-} (f(yf(x_0)^3+x_0)) = \lim_{y \to y_0^-} (f(x)+x^3f(y))=f(x)+x^3f(y_0)$ , and therefore $f$ is also continuous from below in $x$ . Claim 4: $f$ is surjective Proof $f$ is continuous and $\lim_{y \to 0} (f(y)) = 0$ so we only need to prove that $f$ is unbounded, which we get by subtituting a large $x$ in the equation. Claim 5: $f(1)=1$ Proof assume that $f(1)<1$ . So there exist some $y_0$ for which $y_0=y_0*f(1)^3+1$ . Subtitute $(1,y_0)$ in the equation: $f(y_0)=f(y_0)+f(1)$ , contradiction. Assume now that $f(1)>1$ . From the surjectivity there exist some $x_0$ such that $f(x_0)=1$ . $f$ is increasing, and therefore $x_0<1$ . subtitute $x=x_0$ in the equation: $f(y) < f(y+x_0)=x_0^3f(y)+1$ , which is a contradiction for large enough $f(y)$ (we proved that $f$ is surjective so there exist large enough values of $f(y)$ ). Therefore, $f(1)=1$ . Claim 6: $f(x)=x$ for all positive $x$ Proof $f$ is continuous so it is enough to prove that $f(x)=x$ for all rational $x$ . subtitute $x=1$ : $f(y+1)=f(y)+1$ , so from induction $f(x)=x$ for all positive integers $x$ . take some rational $y_0$ . There exist some positive integer $x_0$ such that $y_0x_0^3$ is an integer. By subtituting $(x_0,y_0)$ we get that $y_0x_0^3+x_0=f(y_0x_0^3+x_0)=x^3f(y_0)+x_0$ and therefore $f(y_0)=y_0$ .



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