, 2023

Solution:
Answer: $f(x)=\frac{1}{x}$.

It is straightforward to check that $f(x)=\frac{1}{x}$ works. We then focus on proving that there are no other solutions. Let $P(x, y)$ denote the given functional equation. First note that for all $x, y \in \mathbb{Q}^{+}$, $f(x)>f(x+y)$, so $f$ is strictly decreasing and hence injective. Now, for all $x, y \in \mathbb{Q}^{+}$,
$$\begin{aligned} P(x, y) & \Longrightarrow f(x)=f(x+y)+f\left(x+x^{2} f(y)\right) \\ P\left(x, x^{2} f(y)\right) & \Longrightarrow f(x)=f\left(x+x^{2} f(y)\right)+f\left(x+x^{2} f\left(x^{2} f(y)\right)\right) \end{aligned}$$
Equating these two equations gives
$$\begin{aligned} f\left(x+x^{2} f\left(x^{2} f(y)\right)\right) & =f(x+y) \\ x+x^{2} f\left(x^{2} f(y)\right) & =x+y \\ f\left(x^{2} f(y)\right) & =\frac{y}{x^{2}} \end{aligned}$$
for all $x, y \in \mathbb{Q}^{+}$. In particular, plugging in $x=1$ gives $f(f(y))=y$, and replacing $y$ with $f(y)$ gives $f\left(x^{2} y\right)=\frac{f(y)}{x^{2}}$ for all $x, y \in \mathbb{Q}^{+}$. There are two ways to finish.

Finish 1: The above implies that $f\left(x^{2}\right)=\frac{f(1)}{x^{2}}$ for all $x \in \mathbb{Q}^{+}$. Since $\left\{x^{2}: x \in \mathbb{Q}^{+}\right\}$ is dense in $\mathbb{Q}^{+}$ and $f$ is decreasing we find that $f(x)=c / x$ for some constant $c$. Plugging back in now gives $c=1$.

Finish 2: For all $x \in \mathbb{Q}^{+}$,
$$\begin{aligned} P\left(x, \frac{9 x}{16}\right) \Longrightarrow f(x) & =f\left(\frac{25 x}{16}\right)+f\left(x+x^{2} f\left(\frac{9 x}{16}\right)\right) . \\ & =\frac{16 f(x)}{25}+f\left(x+\frac{16 x^{2} f(x)}{9}\right) \end{aligned}$$
Hence, we have
$$\begin{aligned} f\left(\frac{25 x}{9}\right)=\frac{9 f(x)}{25} & =f\left(x+\frac{16 x^{2} f(x)}{9}\right) \\ \frac{25 x}{9} & =x+\frac{16 x^{2} f(x)}{9} \Longrightarrow f(x)=\frac{1}{x} \end{aligned}$$
as desired.