IMO functional equations

To get the ball rolling, here is a partial proof, assuming that the function $f$ is continuous at $1$ .

Putting $x=y=1$ gives $f(1)\,ff(1) + ff(1) \;= \; f(1)[ff(1) + ff(1)] \;=\; 2f(1)\,ff(1)$ and so $f(1) = 1$ . Putting $x=1$ gives $ff(y) + f(y) \; = \; f(y)[1 + ff(y^2)]$ for $y > 0$ . Putting $y=1$ gives $x f(x^2) + ff(x) \; = \; f(x)[ff(x^2) + 1]$ for $x > 0$ . But this implies that $xf(x^2) + ff(x) \; = \; ff(x) + f(x)$ for all $x > 0$ , so that $xf(x^2) = f(x)$ for all $x > 0$ . Thus $g(x) = xf(x)$ is such that $g(x^2) = g(x)$ for all $x > 0$ .

Assuming that $f$ is continuous at $1$ , we deduce that $g$ is continuous at $1$ . Since $g(x) \; = \; g\big(\sqrt[2^n]{x}\big) \hspace{2cm} n \ge 1$ we see, letting $n \to \infty$ , that $g(x) = g(1) = 1$ for all $x > 0$ . Thus $f(x) = x^{-1}$ for all $x > 0$ , making the answer $\boxed{2017}$ (again).