= Solution
The first <characteristic polynomial> factors as
$$
\rho(w)=(w-1)(w^2-2\alpha w+1).
$$
Also $\rho(1)=0$ and $\rho'(1)=\sigma(1)=2(1-\alpha)$, so the <order conditions for a linear multistep method> hold for every $\alpha$. Convergence of the full <linear multistep method> recurrence requires, in addition, the <root condition for a multistep method>: all <polynomial roots> of $\rho$ lie in the closed unit disk and its <polynomial roots> on the unit circle are simple.
For $-1<\alpha<1$, write $\alpha=\cos\theta$ with $0<\theta<\pi$. The other <polynomial roots> are $e^{\pm i\theta}$, distinct from each other and from 1, so the <root condition for a multistep method> holds. For $|\alpha|>1$, the two quadratic <polynomial roots> are real and reciprocal, and one has modulus greater than one. At $\alpha=1$, $\rho=(w-1)^3$; at $\alpha=-1$, $\rho=(w-1)(w+1)^2$. Both endpoints violate simplicity. <Consistency of a numerical method> plus <zero-stability> therefore gives
$$
\boxed{-1<\alpha<1.}
$$
The conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha=1$ the <polynomials> contain $(w-1)^2$, and at $\alpha=-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the <common-factor cancellation defect in a multistep recurrence>, so those parameter values are not added to the interval.
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