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Substitute the linear recurrence relation for each of $p_{n+2}$ and $q_{n+2}$ in the discrete Wronskian:Iterating givesNo 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,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 $n{r}^n$, where $r=1$ or $-1$ respectively. Direct substitution verifies both, and their discrete Wronskian is $r^{2n+1}=r\ne 0$.