= Solution
Let $\Phi(z)=1-\phi_1z-\cdots-\phi_pz^p$. The <causality root criterion for an autoregressive model>,
$$
\boxed{\Phi(z)\ne0\quad\text{for }|z|\leq1,}
$$
is a sufficient condition for a <weakly stationary process> solution, and gives the causal solution $X_t=\sum_{j\geq0}\psi_j\varepsilon_{t-j}$. Indeed $1/\Phi$ is analytic in a disk of radius greater than one, so its Taylor coefficients $\psi_j$ decay geometrically. The series therefore has <mean-square convergence> and defines a <linear filter of a stationary time series>; multiplying by $\Phi$ verifies the equation.
For the specified order-two model, put $\varphi=(1+\sqrt5)/2$. Since $\varphi-\varphi^{-1}=1$,
$$
\Phi(z)=1-\alpha z-\alpha^2z^2
=(1-\alpha\varphi z)(1+\alpha z/\varphi).
$$
When $\alpha\ne0$, its roots are $1/(\alpha\varphi)$ and $-\varphi/\alpha$. Both have modulus greater than one exactly when $|\alpha|<1/\varphi$. The case $\alpha=0$ is <white noise> and satisfies the same inequality. Consequently
$$
\boxed{\alpha_0=\frac{\sqrt5-1}{2},\qquad|\alpha|<\alpha_0.}
$$
This is the causal <stationarity> condition selected above. Without causality, a two-sided <autoregressive model> can also have a <noncausal stationary autoregression> when its <polynomial> has no unit-circle zeros. Thus the inequality is not a necessary condition for all conceivable <weak stationarity> solutions; the causal interpretation is the one required by the subsequent innovations and forecasting questions.
To find the <Wold representation>, expand $1/\Phi(z)=\sum_{j\geq0}\psi_jz^j$. Matching coefficients gives $\psi_0=1$, $\psi_1=\alpha$ and $\psi_j=\alpha\psi_{j-1}+\alpha^2\psi_{j-2}$. Hence
$$
\boxed{X_t=\sum_{j=0}^{\infty}F_{j+1}\alpha^j\varepsilon_{t-j},\qquad
F_{j+1}=\frac{\varphi^{j+1}-(-1/\varphi)^{j+1}}{\sqrt5}.}
$$
Here $F_1=F_2=1$ are the <Fibonacci numbers>. The coefficient formula also follows by partial fractions in the displayed factorization. Absolute summability follows from $|\alpha\varphi|<1$ and $|\alpha/\varphi|<1$. The past observation span equals the past noise span, because the causal series gives one inclusion and $\varepsilon_t=X_t-\alpha X_{t-1}-\alpha^2X_{t-2}$ gives the reverse inclusion. Thus these noises are the <linear innovations>, and there is no additional deterministic component. This is the <Fibonacci-coefficient autoregression>.
The <best linear prediction from an infinite past> is the <orthogonal projection> onto the closed span of $X_T,X_{T-1},\ldots$. Future <white noise> is orthogonal to that span. Substitution in the recursion therefore gives
$$
\boxed{\widehat X_{T+1}=\alpha X_T+\alpha^2X_{T-1},\qquad
\widehat X_{T+2}=2\alpha^2X_T+\alpha^3X_{T-1}.}
$$
The respective errors are $\varepsilon_{T+1}$ and $\alpha\varepsilon_{T+1}+\varepsilon_{T+2}$, so their <mean squared errors> are $\sigma^2$ and $(1+\alpha^2)\sigma^2$. No <normal distribution> assumption is needed for these linear forecasts.
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