Let $P(x)\in\mathbb{Z}[x]$ be monic and irreducible over $\mathbb{Q}[x]$, and let $\theta$ be a root of $P(x)$. Let $K = \{a + b\theta\} \subseteq \mathbb{Z}[\theta]$. When is it the case that there are infinitely many $\alpha\in K$ s.t. $\text{Nm}(\alpha) = 1$?
closed as offtopic by Venkataramana, Andreas Blass, Karl Schwede, Stefan Kohl, S. Carnahan♦ Sep 27 '14 at 2:50
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4$\begingroup$ The notation $K$ is a bit unfortunate (since it will connote a field to many readers) for something that in general won't even be a ring. $\endgroup$ – GNiklasch Sep 26 '14 at 16:59

1$\begingroup$ If $\deg(P) \geq 3$ there will only be finitely many by Thue's theorem. If $\deg(P) = 2$, it will depend on whether or not $\mathbb{Q}[\theta]$ is imaginary quadratic or real quadratic. $\endgroup$ – Jeremy Rouse Sep 26 '14 at 17:02
If and only if $\mathbb{Q}(\theta)$ is a real quadratic field.
In the imaginary quadratic case, and when $P$ has degree $1$, there are only finitely many units in the ring of integers of the field. When the degree of $P$ is at least $3$, the norm condition amounts to a Thue equation. In the real quadratic case, some power of the fundamental unit will be congruent to $1$ modulo the conductor ideal of the order and will thus be an element of the order along with its powers.