2 deleted 2 characters in body

This should be a comment as it does not answer the question, but I do not have enough points.

A sufficient, and necessary, condition for $E$ to generate $\mathbb{Q}$ is that $0 \in E$ and for every pair of numbers $(p,q)\in \mathbb Z \times \mathbb N^\ast$ such that $p \wedge q gcd(p,q) = 1$, there exists $m \in \mathbb Z^\ast$ such that $pm \in E$ and $qm \in E$. The multiplicities of such $m$ corresponding to the multiplicities of representations, as stated above $p=q=1$ gives non unique solutions, actually an infinite number of them (as E needs to be bigger then the set of prime numbers).

So apart from checking for this, a faster method would depend on how E is defined.

I'm wondering what are the smallests such sets so that every quotient $\frac p q$ admits at least n representations, or what are the smallests sets, if they exist, such that every quotient (apart from the pathological cases 0, 1, -1...) admits exactly n representations?

1

This should be a comment as it does not answer the question, but I do not have enough points.

A sufficient, and necessary, condition for $E$ to generate $\mathbb{Q}$ is that $0 \in E$ and for every pair of numbers $(p,q)\in \mathbb Z \times \mathbb N^\ast$ such that $p \wedge q = 1$, there exists $m \in \mathbb Z^\ast$ such that $pm \in E$ and $qm \in E$. The multiplicities of such $m$ corresponding to the multiplicities of representations, as stated above $p=q=1$ gives non unique solutions, actually an infinite number of them (as E needs to be bigger then the set of prime numbers).

So apart from checking for this, a faster method would depend on how E is defined.

I'm wondering what are the smallests such sets so that every quotient $\frac p q$ admits at least n representations, or what are the smallests sets, if they exist, such that every quotient (apart from the pathological cases 0, 1, -1...) admits exactly n representations?