= Solution
Substitute the <linear recurrence relation> for each of $p_{n+2}$ and $q_{n+2}$ in the <discrete Wronskian>:
$$
\begin{aligned}
W_{n+1}
&=p_{n+1}[-b(n)q_{n+1}-c(n)q_n]
-[-b(n)p_{n+1}-c(n)p_n]q_{n+1}\\
&=c(n)(p_nq_{n+1}-p_{n+1}q_n)=c(n)W_n.
\end{aligned}
$$
Iterating gives
$$
\boxed{W_{n+1}=W_1\prod_{m=1}^nc(m).}
$$
No division was used, so vanishing $c(m)$ cause no exception.
For constant coefficients, seek a solution $x_n=r^n$. The <characteristic equation of a linear recurrence> is $r^2+\alpha r+1=0$. Choose $\theta\in[0,\pi]$ with $\cos\theta=-\alpha/2$. Its roots are $r=e^{i\theta}$ and $e^{-i\theta}$, since $r+r^{-1}=2\cos\theta$. For the two resulting solutions,
$$
\boxed{W_n=e^{-i\theta}-e^{i\theta}=-2i\sin\theta.}
$$
It is independent of $n$, exactly as the product identity predicts when $c(n)=1$.
When $-2<\alpha<2$ the roots are distinct and the <discrete Wronskian> is nonzero. At $\alpha=-2$ or $2$, the two displayed solutions coincide and $W_n=0$; the identity still holds. If an independent pair is wanted at these endpoints, take $r^n$ and $nr^n$, where $r=1$ or $-1$ respectively. Direct substitution verifies both, and their <discrete Wronskian> is $r^{2n+1}=r\ne0$.
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