= Solution
Apply the <exponential-symbol order criterion for a multistep method>. The exact-solution residual of the <linear multistep method> is encoded by $E(z)=\rho(e^z)-z\sigma(e^z)$; vanishing through degree $p$ is equivalent to the <order conditions for a linear multistep method> through order $p$. Here $\rho(1)=0$, $\rho'(1)=\sigma(1)=1-\alpha$, and direct <Taylor expansion> gives
$$
E(z)=-\frac{1+\alpha}{24}z^4-\frac{17+13\alpha}{360}z^5+O(z^6).
$$
For example, for $n\geq2$ the coefficient of $z^n$ can be calculated as
$$
\frac{2^n-(1+\alpha)}{n!}
-\frac{\tfrac23(1-\alpha)+2^{n-1}(5+\alpha)/12}{(n-1)!}.
$$
For $n=2,3$ this is zero, verifying the lower conditions explicitly. The constant and linear coefficients also vanish for every $\alpha$. Therefore \b[the formal order is at least three for every $\alpha$].
The fourth-degree term vanishes exactly when $\alpha=-1$. For that value the fifth-degree coefficient is $-1/90\ne0$, so
$$
\boxed{p=3\text{ if }\alpha\ne-1,\qquad p=4\text{ precisely when }\alpha=-1.}
$$
At $\alpha=1$ these are formal residual conditions only: the <characteristic polynomial> has a double root at one, so the method is not convergent. Canceling its common polynomial factor would change the recurrence and give a different order calculation. This distinction is relevant to part (b).
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