= Solution
The useful <autoregressive model> is for the squares, not for signed observations. Put $Y_t=X_t^2$ and
$$
u_t=(\varepsilon_t^2-1)(\alpha_0+\alpha_2Y_{t-2}).
$$
Then
$$
\boxed{Y_t=\alpha_0+\alpha_2Y_{t-2}+u_t.}
$$
The errors form a <martingale difference sequence> relative to the noise history, because the current standardized noise is independent of the past. When the <fourth moment> is finite they have finite <variance> and are uncorrelated across distinct times, although their <conditional variance> depends on the regressor. In centered form, $Y_t-m_2=\alpha_2(Y_{t-2}-m_2)+u_t$.
For <ordinary least squares>, use the $n-2$ response-regressor pairs $y_t=X_t^2$, $r_t=X_{t-2}^2$, $t=3,\ldots,n$. With their separate means $\bar y$ and $\bar r$, minimize $\sum_{t=3}^n(y_t-a-br_t)^2$. If the regressor sum of squares is positive, the estimators are
$$
\boxed{\widehat\alpha_2=\frac{\sum_{t=3}^n(r_t-\bar r)(y_t-\bar y)}{\sum_{t=3}^n(r_t-\bar r)^2},\qquad
\widehat\alpha_0=\bar y-\widehat\alpha_2\bar r.}
$$
The two means use the matched pairs; replacing them indiscriminately by a single full-sample mean is not the exact least-squares formula. The noise-series representation supplies an <ergodic stationary process>. If $3\alpha_2^2<1$, finite regressor <second moments> and the error's zero <conditional expectation> justify the usual population regression and <statistical consistency> argument. The observations still define a finite-sample least-squares fit outside that moment range, but the ordinary finite-<variance> justification must not be claimed there. If parameter constraints are required, minimize the same criterion subject to $a>0$ and $0<b<1$, rather than assert that unconstrained estimates automatically satisfy them.
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