Given a multiset $S$ of $mn$ numbers, how hard is to find multisets $A$ and $B$, of $m$ and $n$ numbers respectively, such that $$ S = \{ xy \mid x\in A, y\in B \}~~~~\text{(multiset sense)}, $$ or prove that there are none?

For example, if I give you $S=\{6,8,15,15,20,20\}, m=2,n=3$, you will tell me $A=\{3,4\}$, $B=\{2,5,5\}$.

I used the vague word "numbers" to mean anything you like. Choose "positive integers", "integers", "real numbers", "elements of my favourite quasigroup" as you please. Personally I think "real numbers" grabs the essence of the problem, as the subtask of factoring integers is a distraction. Assume you can do exact arithmetic in constant time per basic operation.

If you are working in a ring, the task looks the same as that of arranging the $mn$ numbers in an $m\times n$ matrix so that the rank is 1.

If your numbers are "positive reals", then you can take logarithms to convert the multiplication into addition, in which case the question is a bit like this one. But I don't care about finding all factorizations, just whether there is none or more than none.

I don't any more have an application for this. I just thought it might disturb your sleep like it disturbed mine.

Given a multiset $S$ of $mn$ numbers, how hard is to find multisets $A$ and $B$, of $m$ and $n$ numbers respectively, such that $$ S = \{ xy \mid x\in A, y\in B \}~~~~\text{(multiset sense)}, $$ or prove that there are none?

For example, if I give you $S=\{6,8,15,15,20,20\}, m=2,n=3$, you will tell me $A=\{3,4\}$, $B=\{2,5,5\}$.

I used the vague word "numbers" to mean anything you like. Choose "positive integers", "integers", "real numbers", "elements of my favourite quasigroup" as you please. Personally I think "real numbers" grabs the essence of the problem, as the subtask of factoring integers is a distraction. Assume you can do exact arithmetic in constant time per basic operation.

If you are working in a ring, the task looks the same as that of arranging the $mn$ numbers in an $m\times n$ matrix so that the rank is 1.

If your numbers are "positive reals", then you can take logarithms to convert the multiplication into addition, in which case the question is a bit like this one. But I don't care about finding all factorizations, just whether there is none or more than none.

I don't any more have an application for this. I just thought it might disturb your sleep like it disturbed mine.

Given a multiset $S$ of $mn$ numbers, how hard is to find multisets $A$ and $B$, of $m$ and $n$ numbers respectively, such that $$ S = \{ xy \mid x\in A, y\in B \}~~~~\text{(multiset sense)}, $$ or prove that there are none?

For example, if I give you $S=\{6,8,15,15,20,20\}, m=2,n=3$, you will tell me $A=\{3,4\}$, $B=\{2,5,5\}$.

I used the vague word "numbers" to mean anything you like. Choose "positive integers", "integers", "real numbers", "elements of my favourite quasigroup" as you please. Personally I think "real numbers" grabs the essence of the problem, as the subtask of factoring integers is a distraction. Assume you can do exact arithmetic in constant time per basic operation.

If you are working in a ring, the task looks the same as that of arranging the $mn$ numbers in an $m\times n$ matrix so that the rank is 1.

If you numbers are "positive reals", then you can take logarithms to convert the multiplication into addition, in which case the question is a bit like this one. But I don't care about finding all factorizations, just whether there is none or more than none.

I don't any more have an application for this. I just thought it might disturb your sleep like it disturbed mine.

Given a multiset $S$ of $mn$ numbers, how hard is to find multisets $A$ and $B$, of $m$ and $n$ numbers respectively, such that $$ S = \{ xy \mid x\in A, y\in B \}~~~~\text{(multiset sense)}, $$ or prove that there are none?

For example, if I give you $S=\{6,8,15,15,20,20\}, m=2,n=3$, you will tell me $A=\{3,4\}$, $B=\{2,5,5\}$.

I used the vague word "numbers" to mean anything you like. Choose "positive integers", "integers", "real numbers", "elements of my favourite quasigroup" as you please. Personally I think "real numbers" grabs the essence of the problem, as the subtask of factoring integers is a distraction. Assume you can do exact arithmetic in constant time per basic operation.

If you are working in a ring, the task looks the same as that of arranging the $mn$ numbers in an $m\times n$ matrix so that the rank is 1.

If your numbers are "positive reals", then you can take logarithms to convert the multiplication into addition, in which case the question is a bit like this one. But I don't care about finding all factorizations, just whether there is none or more than none.

I don't any more have an application for this. I just thought it might disturb your sleep like it disturbed mine.