South African Mathematical Olympiad Third Round

By setting $n = 1$, we obtain $mf(1) = 0$ for all $m$, so $f(1) = 0$. Next we notice that since every positive integer can be written as a product of prime numbers, the function $f$ is determined completely by its value on the primes. If there exists $m \ge 2$ such that $f(m) = 0$, then $f(m) = f(1)$ and we are done. Thus assume for all $m \ge 2$ that $f(m) \ge 1$.
Thus
$$\begin{aligned} f(2024) &= f(8 \cdot 11 \cdot 23) = 23f(8 \cdot 11) + 88f(23) = 23(11f(8) + 8f(11)) + 88f(23) \\ &= 253f(8) + 184f(11) + 88f(23). \end{aligned}$$
To study $f(8)$, we instead derive a general formula for perfect powers:
$$f(a^2) = af(a) + af(a) = 2af(a),$$
$$f(a^3) = a^2f(a) + af(a^2) = 3a^2f(a).$$
We assume as inductive hypothesis that
$$f(a^n) = na^{n-1}f(a),$$
and prove it for $n+1$ by
$$f(a^{n+1}) = f(a^n \cdot a) = af(a^n) + a^n f(a) = a(na^{n-1})f(a) + a^n f(a) = (n+1)a^n f(a).$$
Thus
$$10120 = f(2024) = 253(3 \cdot 2^2)f(2) + 184f(11) + 88f(23) = 3036f(2) + 184f(11) + 88f(23). \quad (1)$$
Considering this equation modulo 11, we conclude that $f(11)$ is divisible by 11, and by the assumption that $f(11) \ge 1$, we conclude that $f(11) \ge 11$. Similarly, working modulo 23 tells us that $f(23)$ is divisible by 23 and thus $f(23) \ge 23$. Finally, working modulo 8 tells us that $4f(2)$ is divisible by 8, so $f(2) \ge 2$. Thus the right hand-side of (1) is greater than or equal to
$$3036 \cdot 2 + 184 \cdot 11 + 88 \cdot 23 = 10120.$$
Since we have equality, we must conclude that
$$f(2) = 2, \quad f(11) = 11 \quad \text{and} \quad f(23) = 23.$$
Our only hope to find two integers $m$ and $n$ for which $f(m) = f(n)$ is to use the information just obtained. Thus, we set
$$m = 2^a 11^b 23^c \quad n = 2^d 11^e 23^f.$$
Intuitively we hope to find values of d, e, f that are close to a, b, c (e.g. differ by 1) and then solve for values a, b, c that happen to work. We notice that $2 \cdot 11 = 22$ is close to 23, so let us study
$$\begin{align*} f(22^a 23^c) &= 23^c f(22^a) + 22^a f(23^c) \\ &= 23^c (a \cdot 22^{a-1} \cdot f(22)) + 22^a (c \cdot 23^{c-1} \cdot f(23)) \\ &= 23^c (a \cdot 22^{a-1} \cdot (2f(11) + 11f(2))) + 22^a (c \cdot 23^{c-1} \cdot 23) \\ &= 23^c (a \cdot 22^{a-1} \cdot 44) + 22^a (c \cdot 23^{c-1} \cdot 23) \\ &= (2a + c) 22^a 23^c \end{align*}$$
Now we compare $f(22^a 23^{b+1})$ with $f(22^{a+1} 23^b)$ and hope to find values of a, b for which they are equal.
$$\begin{aligned} f(22^a 23^{b+1}) &= (2a + b + 1) 22^a 23^{b+1}, \\ f(22^{a+1} 23^b) &= (2(a + 1) + b) 22^{a+1} 23^b. \end{aligned}$$
Thus if we can find values of a, b such that
$$2a + b + 1 = 22 \quad \text{and} \quad 2a + b + 2 = 23$$
we would be done. Luckily there are many such pairs, for example
$(\mathbf{a}, \mathbf{b}) \in \{(1, 19), (2, 17), (3, 15), (4, 13), (5, 11), (6, 9), (7, 7), (8, 5), (9, 3), (10, 1)\}$.