, 2023

First notice that $f(1)$ divides $1$, and hence $f(1) = 1$. For all primes $p$, we have
$$\begin{align*} p \cdot f(p^{2n}) &= \text{lcm}(p \cdot f(p^{2n}), 1) = \text{lcm}(p \cdot f(p^{2n}), f(1^2)) \\ &= p \cdot f(p^n \cdot 1) = p \cdot f(p^n), \end{align*}$$
which shows that $f(p^n) = f(p^{2n})$ for all $n$. Now we show by induction on $m$ that $f(p^m) = f(p)$. We know that $f(p^2) = f(p)$. Assume that $f(p^m) = f(p)$. Now
$$\begin{gather*} p f(p \cdot p^m) = \text{lcm}(p \cdot f(p^2), f(p^{2m})) = \text{lcm}(p \cdot f(p), f(p^m)) = p \cdot f(p), \\ \text{and hence } f(p^{m+1}) = f(p). \end{gather*}$$
Since $f(p)$ divides $p$, we know that $f(p) = p^{\alpha_p}$ for $\alpha_p \in \{0, 1\}$. For each prime $p_i$, let $f(p_i) = p_i^{\alpha_{p_i}}$ with $\alpha_{p_i} \in \{0, 1\}$. For two primes $p_1 \neq p_2$, we have
$$\begin{align*} p_1^n f(p_1^n \cdot p_2^m) &= \text{lcm}(p_1^n \cdot f(p_1^{2n}), f(p_2^{2m})) = \text{lcm}(p_1^n \cdot p_1^{\alpha_{p_1}}, p_2^{\alpha_{p_2}}) \\ &= p_1^n \cdot p_1^{\alpha_{p_1}} \cdot p_2^{\alpha_{p_2}}, \end{align*}$$
and hence $f(p_1^n \cdot p_2^m) = p_1^{\alpha_{p_1}} \cdot p_2^{\alpha_{p_2}}$ for all non-negative integers $n$ and $m$. By induction on $r$, it follows that
$$f(p_1^{n_1} p_2^{n_2} \cdots p_r^{n_r}) = p_1^{\alpha_{p_1}} p_2^{\alpha_{p_2}} \cdots p_r^{\alpha_{p_r}}.$$