khan.academy wrote: Problem Section #2 c). Denote by $\mathbb{Q^+}$ the set of all positive rational numbers. Determine all functions $f:\mathbb{Q^+}\to\mathbb{Q^+}$ which satisfy the following equation for all $x,y \in \mathbb{Q^+} : f(f(x)^2.y)=x^3.f(xy)$. Let $P(x,y)$ be the assertion $f(f(x)^2y)=x^3f(xy)$ $P(x,1)$ $\implies$ $f(f(x)^2)=x^3f(x)$ and so $f(x)$ is injective. $P(1,1)$ $\implies$ $f(f(1)^2)=f(1)$ and so, since injective, $f(1)=1$ 1) $f(xy)=f(x)f(y)$ $\forall x,y\in\mathbb Q^+$ $P(x,y)$ $\implies$ $f(f(x)^2y)=x^3f(xy)$ and so $y^3f(f(x)^2y)=x^3y^3f(xy)$ $P(xy,1)$ $\implies$ $f(f(xy)^2)=x^3y^3f(xy)$ and so : $y^3f(f(x)^2y)=f(f(xy)^2)$ $P(y,f(x)^2)$ $\implies$ $f(f(y)^2f(x)^2)=y^3f(f(x)^2y)$ And so $f(f(y)^2f(x)^2)=f(f(xy)^2)$ And so, using injectivity : $f(xy)=f(x)f(y)$ Q.E.D. 2) $\boxed{f(x)=\frac 1x\quad\forall x\in\mathbb Q^+}$ Using previous property, $P(x,y)$ becomes $f(f(x))^2=x^3f(x)$ And so $xf(x)$ is a perfect square and it exists a multiplicative function $g(x)$ such that $f(x)=\frac{g(x)^2}x$ Then $f(f(x))^2=x^3f(x)$ becomes $g(g(x))^4=g(x)^5$ This means that $g(x)=h(x)^4$ for some multiplicative $h(x)$ and so $h(h(x))^{16}=h(x)^5$ and so $h(x)=k(x)^4$ for some multiplicative $h(x)$ and so .... And it is easy to conclude that the only possibility is $g(x)=1$ $\forall x\in\mathbb Q^+$ Q.E.D.

They simply copied ISL 2010 A5, wow.

I think IMO SL is a bit too tough for national level in Nepal. https://artofproblemsolving.com/community/c6h418682p2362286

ZETA_in_olympiad wrote: I think IMO SL is a bit too tough for national level in Nepal. https://artofproblemsolving.com/community/c6h418682p2362286 I agree!