= Solution
Assume $0<\epsilon\ll1$ and $X,Y,\alpha=O(1)$. For $n=2$, use the <method of multiple scales> with $T=\epsilon t$ and <complex amplitudes>
$$
x_0=A(T)e^{it}+\overline A(T)e^{-it},\qquad y_0=B(T)e^{2it}+\overline B(T)e^{-2it}.
$$
At first order the forcing equations are
$$
(\partial_{t_0}^2+1)x_1=-2\partial_{t_0}\partial_Tx_0-\alpha x_0-x_0y_0,\qquad
(\partial_{t_0}^2+4)y_1=-2\partial_{t_0}\partial_Ty_0-x_0y_0.
$$
The product has <frequencies> $\pm1,\pm3$. Its $e^{it}$ coefficient is $\overline A B$, while it has no $e^{2it}$ coefficient. Removing the <secular terms> gives
$$
2iA_T+\alpha A+B\overline A=0,\qquad B_T=0.
$$
The leading initial conditions give $A(0)=X/2$ and $B=Y/2$. Put $b=Y/2$, $A=p+iq$. The <primary parametric resonance of a two-to-one oscillator pair> is governed by
$$
p_T=-\frac{\alpha-b}{2}q,\qquad q_T=\frac{\alpha+b}{2}p,\qquad
\lambda^2=\frac{b^2-\alpha^2}{4}.
$$
For $b^2>\alpha^2$, choose $\lambda>0$. Solving with $p(0)=X/2,q(0)=0$ yields
$$
\boxed{x(t)=X\left[\cosh(\lambda\epsilon t)\cos t-\frac{\alpha+b}{2\lambda}\sinh(\lambda\epsilon t)\sin t\right]+O(\epsilon),\quad
y(t)=Y\cos2t+O(\epsilon)}.
$$
If $b^2<\alpha^2$, let $\nu=\sqrt{\alpha^2-b^2}/2$; replace $\cosh(\lambda T)$ by $\cos(\nu T)$ and $\sinh(\lambda T)/\lambda$ by $\sin(\nu T)/\nu$. On the boundary $b^2=\alpha^2$, take the continuous limit:
$$
x(t)=X\left[\cos t-\frac{\alpha+b}{2}\epsilon t\sin t\right]+O(\epsilon).
$$
Thus \b[exponential growth occurs when $|Y|>2|\alpha|$ and $X\ne0$], on the scale $t=O(\epsilon^{-1})$. Equality is a marginal case that can produce linear slow growth: for $b=\alpha\ne0$ the displayed initial phase excites it, whereas for $b=-\alpha$ it does not. If $X=0$, $x$ stays identically zero and the other oscillator is exactly uncoupled. The leading $y$ <amplitude> is constant; its backreaction is a higher-order effect on this time scale.
These are leading approximations for bounded <slow time> $T$, hence $t=O(\epsilon^{-1})$, while the <amplitudes> stay $O(1)$. First-order corrections can enforce the initial velocities exactly; the leading formula alone has an $O(\epsilon)$ initial-velocity mismatch. In the unstable case the expansion must not be extrapolated through arbitrarily large amplification, where weak coupling and omitted terms cease to be uniform.
For $n=1$, both leading oscillations have <frequencies> $\pm1$, so their quadratic product contains only <frequencies> $0,\pm2$. There is no first-order resonant coupling: the <solvability condition in the method of multiple scales> gives $A_T=i\alpha A/2$ and $B_T=0$, a <frequency> shift without first-order <amplitude> growth. The constant and second-harmonic corrections can feed back into the fundamental at second order. Consequently the first possible growth scale from this <second-order resonance from quadratic oscillator coupling> is
$$
\boxed{T_2=\epsilon^2t=O(1),\qquad t=O(\epsilon^{-2})}.
$$
This identifies the possible slow scale, not a guaranteed instability for every parameter or initial state. In particular, the first-order $\alpha$ <frequency> shift may detune the second-order resonance unless appropriately tuned; a second-order <amplitude> analysis would decide actual growth.
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