On page 55 of the 3rd edition of Ribenboim's ``The New Book of Prime Number Records'', he defines two associated sequences, $U_n(P,Q)=\left( \alpha^n-\beta^n \right)/\left( \alpha-\beta \right)$ and $V_n(P,Q)=\alpha^n+\beta^n$, where $P$ and $Q$ are nonzero integers and $\alpha$, $\beta$ are the roots of $X^2-PX+Q$.
On page 64, he notes in equation (IV.24) that $\gcd \left( U_n, Q \right)=\gcd \left( V_n, Q \right)=1$ if $\gcd(P,Q)=1$.
Putting $D=P^2-4Q$ (as he does on page 54), we have $$ V_n+\sqrt{D} \,U_n=2\alpha^n. $$
Suppose that $D$ is not a perfect square. If we replace $2$ on the right-hand side of this relationship with another non-zero algebraic integer, $\gamma$, in ${\mathbb Q} \left( \sqrt{D} \right)$, then we can also form two associated binary recurrence sequences in a similar way: $$ V_n+\sqrt{D} \,U_n=\gamma\alpha^n. $$
My questions are the following.
- What can we say about $\gcd \left( U_n, N(\alpha) \right)$ and $\gcd \left( V_n, N(\alpha) \right)$ in this case?
Here $N(\alpha)$ is the norm, $N_{{\mathbb Q}(\sqrt{D})/{\mathbb Q}}(\alpha)$. So this question is an analog of Ribenboim's equation (IV.24).
What can we say about $\gcd \left( U_n, N(\gamma) \right)$ and $\gcd \left( V_n, N(\gamma) \right)$, or even $\gcd \left( U_n^2, N(\gamma) \right)$ and $\gcd \left( V_n^2, N(\gamma) \right)$ in this case?
Lastly, what can be said about $\gcd \left( U_n, V_n \right)$ in this case?
This is an analog of Ribenboim's equation (IV.25).
For all of these questions, it is ideally the best possible results that I am looking for.
For example, if $\alpha$ is a non-torsion unit in the quadratic number field and $\gamma=A+B\sqrt{D}$ for non-zero integers $A$ and $B$, then it appears that $\gcd \left( U_n^2, N(\gamma) \right)\mid 4\gcd \left( A^2, DB^2 \right)$, $\gcd \left( V_n^2, N(\gamma) \right) \mid 4\gcd \left( A^2, B^2 \right)$ and $\gcd \left( U_n, V_n \right) \mid \left( 2 \gcd(A,B) \right)$.
References to any such results in the literature, as well as suggestions for how to prove such sharp results, would be very welcome.