[proofplan]
Both identities are proved by induction on $n$ using the defining recurrences $p_n = a_n p_{n-1} + p_{n-2}$ and $q_n = a_n q_{n-1} + q_{n-2}$. For (i) the base case is a direct computation at $n = 0$ (and optionally $n = 1$), and the inductive step substitutes the recurrences into $p_n q_{n-1} - p_{n-1} q_n$ to produce a sign-flipped version of the previous determinant, yielding the pattern $(-1)^{n-1}$. Identity (ii) then follows from a single substitution of the recurrences, using (i) to simplify the resulting expression.
[/proofplan]
[step:Recall the convergent recurrences and initial data]
The convergent numerators and denominators are defined by
\begin{align*}
p_{-1} &= 1, & p_0 &= a_0, & p_n &= a_n p_{n-1} + p_{n-2} \quad (n \geq 1), \\
q_{-1} &= 0, & q_0 &= 1, & q_n &= a_n q_{n-1} + q_{n-2} \quad (n \geq 1),
\end{align*}
where $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{Z}_{\geq 1}$ for $n \geq 1$.
[/step]
[step:Prove the first determinant identity $p_n q_{n-1} - p_{n-1} q_n = (-1)^{n-1}$ by induction on $n$]
We prove the identity for all $n \geq 0$ by induction on $n$.
**Base case** $n = 0$. Using $p_0 = a_0$, $p_{-1} = 1$, $q_0 = 1$, $q_{-1} = 0$:
\begin{align*}
p_0 q_{-1} - p_{-1} q_0 &= a_0 \cdot 0 - 1 \cdot 1 = -1 = (-1)^{0 - 1},
\end{align*}
as required.
**Inductive step.** Assume the identity holds at level $n - 1 \geq 0$:
\begin{align*}
p_{n-1} q_{n-2} - p_{n-2} q_{n-1} &= (-1)^{n-2}.
\end{align*}
Substituting the recurrences $p_n = a_n p_{n-1} + p_{n-2}$ and $q_n = a_n q_{n-1} + q_{n-2}$ into $p_n q_{n-1} - p_{n-1} q_n$:
\begin{align*}
p_n q_{n-1} - p_{n-1} q_n &= (a_n p_{n-1} + p_{n-2}) q_{n-1} - p_{n-1} (a_n q_{n-1} + q_{n-2}) \\
&= a_n p_{n-1} q_{n-1} + p_{n-2} q_{n-1} - a_n p_{n-1} q_{n-1} - p_{n-1} q_{n-2} \\
&= p_{n-2} q_{n-1} - p_{n-1} q_{n-2} \\
&= -(p_{n-1} q_{n-2} - p_{n-2} q_{n-1}) \\
&= -(-1)^{n-2} = (-1)^{n-1}
\end{align*}
by the inductive hypothesis. This completes the induction.
[guided]
The strategy is to show that the cross-term $p_n q_{n-1} - p_{n-1} q_n$ alternates sign at each step. The algebra is straightforward: substitute the recurrences, let the $a_n$-terms cancel, and observe that the surviving expression is the negative of the previous cross-term.
**Base case.** For $n = 0$, we need $p_0 q_{-1} - p_{-1} q_0 = (-1)^{-1} = -1$. Using the initial data:
\begin{align*}
p_0 q_{-1} - p_{-1} q_0 &= a_0 \cdot 0 - 1 \cdot 1 = -1.
\end{align*}
**Inductive step.** Assume the identity holds at $n-1$, i.e.,
\begin{align*}
p_{n-1} q_{n-2} - p_{n-2} q_{n-1} &= (-1)^{n-2}.
\end{align*}
We compute the cross-term at $n$. Substitute $p_n = a_n p_{n-1} + p_{n-2}$ and $q_n = a_n q_{n-1} + q_{n-2}$:
\begin{align*}
p_n q_{n-1} - p_{n-1} q_n &= (a_n p_{n-1} + p_{n-2}) q_{n-1} - p_{n-1}(a_n q_{n-1} + q_{n-2}).
\end{align*}
Expanding,
\begin{align*}
&= a_n p_{n-1} q_{n-1} + p_{n-2} q_{n-1} - a_n p_{n-1} q_{n-1} - p_{n-1} q_{n-2}.
\end{align*}
The $a_n p_{n-1} q_{n-1}$ terms cancel — this is the crucial cancellation that makes the argument work; without it we would have a messy $a_n$-dependent expression. We are left with
\begin{align*}
p_{n-2} q_{n-1} - p_{n-1} q_{n-2} &= -(p_{n-1} q_{n-2} - p_{n-2} q_{n-1}) = -(-1)^{n-2} = (-1)^{n-1}
\end{align*}
by the inductive hypothesis and the sign-flip. Why is it important that the cancellation happened? Because it shows the determinant $p_n q_{n-1} - p_{n-1} q_n$ is **independent** of the specific partial quotient $a_n$: it depends only on the two preceding terms of the sequence. This is the algebraic shadow of the geometric fact that the matrix
\begin{align*}
M_n &= \begin{pmatrix} a_n & 1 \\ 1 & 0 \end{pmatrix}
\end{align*}
has determinant $-1$, and the convergent matrices are products $\prod_k M_k$, so the determinant of the product is $(-1)^{n+1}$. Our identity is this determinant expressed in coordinates.
[/guided]
[/step]
[step:Prove the second identity $p_n q_{n-2} - p_{n-2} q_n = (-1)^n a_n$]
Fix $n \geq 1$. Substituting the recurrences $p_n = a_n p_{n-1} + p_{n-2}$ and $q_n = a_n q_{n-1} + q_{n-2}$:
\begin{align*}
p_n q_{n-2} - p_{n-2} q_n &= (a_n p_{n-1} + p_{n-2}) q_{n-2} - p_{n-2}(a_n q_{n-1} + q_{n-2}) \\
&= a_n p_{n-1} q_{n-2} + p_{n-2} q_{n-2} - a_n p_{n-2} q_{n-1} - p_{n-2} q_{n-2} \\
&= a_n (p_{n-1} q_{n-2} - p_{n-2} q_{n-1}) \\
&= a_n \cdot (-1)^{n-2} = (-1)^n a_n,
\end{align*}
where the penultimate equality uses identity (i) from Step 2 applied at level $n - 1$ (which requires $n - 1 \geq 0$, i.e., $n \geq 1$, as hypothesised), and the final equality uses $(-1)^{n-2} = (-1)^n$.
[/step]
[step:Conclude]
Steps 2 and 3 establish identities (i) and (ii) respectively, so the theorem is proved.
[/step]
