Taiwan IMO 2J, 2020

Easy to see that $f(n) \leq f(1)n$ by substituting $a = 1$. By substituting $a = nb - f(b)$ in the inequality for any large enough $n$, we have
$$nb \mid (nb - f(b))^2 + b f(nb - f(b))$$
and hence $b \mid f(b)^2$. In particular, for every prime $p$, $f(p) = k_p p$ for some integer $0 < k_p \leq f(1)$. Therefore, there must be an integer $k$, such that $f(p) = kp$ for infinitely many prime $p$. Thus, for infinitely many $p$,
$$a + kp \mid (a^2 + p f(a)) - a(a + kp) = p f(a) - p k a$$
thus $a + kp \mid f(a) - k a$. Since $p$ can be infinitely large, we must have $f(a) = k a$.

Substitute $b = 1$ and rearrange to find that
$$\frac{f(a) + f(1)^2}{a + f(1)} = f(1) - a + \frac{a^2 + f(a)}{a + f(1)}$$
is a positive integer and since $f(a) \leq a f(1)$, follows that $\frac{f(a) + f(1)^2}{a + f(1)} \leq f(1)$, hence for some positive integer $k$, $\frac{f(a) + f(1)^2}{a + f(1)} = k$, i.e., $f(n) = k n + f(1)(k - f(1))$ for infinitely many $n$. Fixing an arbitrary $a$, rearrange the given inequality, we have
$$\frac{a^2 + n f(a)}{a + k n + f(1)(k - f(1))}$$
is an integer for infinitely many $n$. Since
$$\frac{a^2 + n f(a)}{a + k n + f(1)(k - f(1))} \rightarrow \frac{f(a)}{k}$$
as $n \to \infty$, we have
$$\frac{a^2 + n f(a)}{a + k n + f(1)(k - f(1))} = \frac{f(a)}{k}$$
for infinitely many $n$. Thus
$$\frac{f(a)}{k}(a + f(1)(k - f(1))) = a^2. \qquad (1)$$
Let $X = f(1)(k - f(1))$, we have $a + X \mid a^2 + (X + a)(X - a) = X^2$ holds for arbitrary $a$. Thus $X = 0$, that is, $k = f(1)$. By equation (1), $f(a) = k a$. $\square$