Let $K$ be a number field (finite extension of $\mathbb{Q}$) and $\mathcal{O}_K$ its ring of integers (elements of $K$ which are integral over $\mathbb{Z}$). It's an exercise in "Undecidability in number theory" (Koenigsmann, 2013, exercise 3.3, arXiv) to show that the set of non-zero elements in $\mathcal{O}_K$ is diophantine. More specifically even, it should hold that for $x \in \mathcal{O}_K$,
$$ x \neq 0 \Leftrightarrow \exists y, z : xz = (2y-1)(3y-1). $$
The direction from right to left is easy, as $(2y-1)(3y-1)$ can never be equal to $0$ ($2$ and $3$ are not invertible in $\mathcal{O}_K$), but I struggle to prove the other direction, even when $\mathcal{O}_K = \mathbb{Z}$. Both hints and full solutions are welcome.
In $\Bbb Z$ we need to show that for any nonzero number $n$, the congruence $(2y-1)(3y-1)\pmod n$ is soluble (we may as well suppose $n>0$). If either $2$ or $3$ is invertible modulo $n$, this is clear. But that's enough since by the Chinese Remainder theorem, we may reduce to the case where $n$ is a prime power, and a prime power can't be divisible both by $2$ and by $3$.
I think this works in ${\mathcal O}_K$ too. A finite quotient of ${\mathcal O}_K$ will be a direct product of rings of prime power order etc.