Let $0\neq \beta\in\overline{\mathbb{Z}}$ and let $n$ be a positive integer coprime to $N_{\mathbb{Q}(\beta)/\mathbb{Q}}(\beta)$. Say that $n$ is a Fermat pseudoprime to base $\beta$ if $$\beta^{n^{[\mathbb{Q}(\beta):\mathbb{Q}]}-1}\equiv 1\pmod n.$$ If $p$ is a prime which does not divide $N_{\mathbb{Q}(\beta)/\mathbb{Q}}(\beta)$ and $\mathcal{O}$ is the ring of integers of $\mathbb{Q}(\beta)$, then $\bar{\beta}\in\mathcal{O}/p\mathcal{O}$ is a unit, and $\mathcal{O}/p\mathcal{O}$ is a product of residue fields, each with degree dividing $[\mathbb{Q}(\beta):\mathbb{Q}]$, so $p$ passes the test. This justifies the use of the word 'pseudoprime'.
If $\beta\in\mathbb{Z}_{\ge 2}$, there are many ways to construct composite $n$ which are Fermat pseudoprimes to base $\beta$. For example, Cipolla's method: Let $p\nmid \beta(\beta^2-1)$ be an odd prime, then $(\beta^{2p}-1)/(\beta^2-1)$ is composite and a pseudoprime to base $\beta$.
Questions:
- For which $\beta$ do we have a way of constructing infinitely many composite base $\beta$ pseudoprimes?
- For which $\beta$ do we know there exist infinitely many composite base $\beta$ pseudoprimes?
Edit: Just read that if $n$ is a Carmichael number coprime to the discriminant of $\mathbb{Q}(\beta)$, such that the minimal polynomial of $\beta$ splits completely modulo $p$ for all $p\mid n$, then $\beta^n\equiv \beta\pmod n$. This suggests the answer to the second question is affirmative.
