= Solution
With $\rho_0=a$, $\rho_1=\theta$, $\rho_2=1$, the constant order condition gives $a=-1-\theta$. The next three <order conditions for a linear multistep method> give
$$
\theta+2=\sigma_0+\sigma_1+\sigma_2,\qquad \theta+4=2\sigma_1+4\sigma_2,\qquad \theta+8=3\sigma_1+12\sigma_2.
$$
Solving this linear system yields \b[the unique coefficients of order at least three]:
$$
\boxed{a=-1-\theta,\qquad \sigma_0=\frac{5\theta+4}{12},\quad \sigma_1=\frac{2(\theta+2)}3,\quad \sigma_2=\frac{4-\theta}{12}.}
$$
The characteristic polynomial controlling <zero-stability> factors as
$$
\rho(\zeta)=\zeta^2+\theta\zeta-1-\theta=(\zeta-1)(\zeta+1+\theta).
$$
The other root is $-1-\theta$. Requiring its modulus to be at most one gives $-2\leq\theta\leq0$, but $\theta=-2$ produces a repeated root at one and violates the <root condition for a multistep method>. At $\theta=0$, the roots $1,-1$ are both simple, so that endpoint is allowed. By the <Dahlquist equivalence theorem>, \b[the method converges precisely for]
$$
\boxed{-2<\theta\leq0.}
$$
The fourth defect coefficient is $C_4=\theta$, so the method has order exactly three for nonzero $\theta$. At $\theta=0$ it is the two-step Simpson method, of order four; thus the extra-order endpoint has correctly remained in the answer.
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