66th Czech and Slovak Mathematical Olympiad

We will show that the only solution is a function $f$ such that $$f(m)=\{\begin{array}{ll}p,& \text{if }m\text{ is a non-trivial power of a prime }p,\text{ i.e. }m=p^k,k\ge 1,\\ 1,& \text{otherwise.}\end{array}$$ Number $1$ has the unique divisor $1$, hence plugging $m=1$ in the given equality we get $f(1)=1$. Let $m=p$ be a prime. Then $$f(1)\cdot f(p)=p\text{i.e.}f(p)=p.$$ For $m=p^2$ we get $$f(1)\cdot f(p)\cdot f(p^2)=p^2\text{i.e.}f(p^2)=p$$ and in general for $k>1$ and $m=p^k$ $$f(1)\cdot f(p)\cdot f(p^2)\cdots f(p^k)=p^k.$$ By induction, $f(p^k)=p$ for all positive integers $k$. Now let us consider positive integer $m$ with at least two distinct prime factors whose factorization is $m=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_k^{{\alpha }_k}$, where $k\ge 2$ and ${\alpha }_i\ge 1$ for every $i\in \{1,2,\dots ,k\}$. The divisors of $m$ include the powers $$p_1,p_1^2,\dots ,p_1^{{\alpha }_1},p_2,p_2^2,\dots ,p_2^{{\alpha }_2},\dots ,p_k,p_k^2,\dots ,p_k^{{\alpha }_k},$$ of its prime factors but the product of the corresponding function values $$\begin{array}{rl}& f(p_1)f(p_1^2)\dots f(p_1^{{\alpha }_1})f(p_2)f(p_2^2)\dots f(p_2^{{\alpha }_2})\dots f(p_k)f(p_k^2)\dots f(p_k^{{\alpha }_k})\\ & =\underset{{\alpha }_1}{\underbrace{p_1{p}_1\dots p_1}}\underset{{\alpha }_2}{\underbrace{p_2{p}_2\dots p_2}}\dots \underset{{\alpha }_k}{\underbrace{p_k{p}_k\dots p_k}}=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\dots p_k^{{\alpha }_k}=m\end{array}$$ yields $m$. This means that the product of all the other function values (none of which is a non-trivial power of a prime), including $f(m)$, is equal to $1$ so all the other function values, including $f(m)$, are equal to $1$. This specifies $f$ uniquely and also shows that it has the desired properties.