A set $S\subset \mathbb{Z}$ is Diophantine if there is an integer polynomial $P(n, \bar{m})$ such that$$n\in S \iff (\exists \bar{m} \in \mathbb{Z}^{k})(P(n,\bar{m})=0).$$A set $S\subset \mathbb{Z}$ is bounded Diophantine if there is an integer $N$ and an integer polynomial $P(n, \bar{m})$ such that for any $n\in \mathbb{Z}$ there is at most $N$ vectors $\bar{m}\in\mathbb{Z}^k$ such that $P(n, \bar{m})=0$ and$$n\in S \iff (\exists \bar{m} \in \mathbb{Z}^{k})(P(n,\bar{m})=0).$$A set $S\subset \mathbb{Z}$ is uniquely Diophantine if there is an integer polynomial $P(n, \bar{m})$ such that for any $n\in \mathbb{Z}$ there is at most one $\bar{m}\in\mathbb{Z}^k$ such that $P(n, \bar{m})=0$ and$$n\in S \iff (\exists \bar{m} \in \mathbb{Z}^{k})(P(n,\bar{m})=0).$$It's a theorem that Diophantine sets are precisely the recursively enumerable ones.
What can be said about the bounded Diophantine and the uniquely Diophantine sets? Are they the same class of sets? Are non-negative integers bounded Diophantine?
