Baltic Way 2023 Shortlist

First notice that $f(1)$ divides $1$, and hence $f(1)=1$. For all primes $p$, we have $$\begin{array}{rl}p\cdot f(p^{2n})& =\text{lcm}(p\cdot f(p^{2n}),1)=\text{lcm}(p\cdot f(p^{2n}),f(1^2))\\ & =p\cdot f(p^n\cdot 1)=p\cdot f(p^n),\end{array}$$ 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 $$\begin{array}{c}pf(p\cdot p^m)=\text{lcm}(p\cdot f(p^2),f(p^{2m}))=\text{lcm}(p\cdot f(p),f(p^m))=p\cdot f(p),\\ \text{and hence }f(p^{m+1})=f(p).\end{array}$$ 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 $p_1\ne p_2$, we have $$\begin{array}{rl}{p}_1^{n}f(p_1^n\cdot p_2^m)& =\text{lcm}(p_1^n\cdot f(p_1^{2n}),f(p_2^{2m}))=\text{lcm}(p_1^n\cdot p_1^{{\alpha }_{p_1}},p_2^{{\alpha }_{p_2}})\\ & =p_1^n\cdot p_1^{{\alpha }_{p_1}}\cdot p_2^{{\alpha }_{p_2}},\end{array}$$ 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 $$f(p_1^{n_1}{p}_2^{n_2}\cdots p_r^{n_r})=p_1^{{\alpha }_{p_1}}{p}_2^{{\alpha }_{p_2}}\cdots p_r^{{\alpha }_{p_r}}.$$