= Solution
For disturbances varying only along the imposed <magnetic field>, set $k=0$ and $x=m^2>0$. The steady <horizontal-field magnetoconvection dispersion relation> becomes
$$
\boxed{R=\frac{(m^2+\pi^2)^3}{m^2}+Q(m^2+\pi^2).}
$$
Expand it in powers of $x$:
$$
R(x)=\pi^2Q+(Q+3\pi^2)x+x^2+3\pi^4+\frac{\pi^6}{x}.
$$
Its derivative and second derivative are
$$
R'(x)=Q+3\pi^2+2x-\frac{\pi^6}{x^2},\qquad
R''(x)=2+\frac{2\pi^6}{x^3}>0.
$$
Because $R$ diverges as $x\to0$ and $x\to\infty$, the unique minimum is determined by
$$
x^2(Q+3\pi^2+2x)=\pi^6.
$$
For large <Chandrasekhar number>, this equation first gives $x\sim\pi^3Q^{-1/2}$ and hence the requested <strong-field steady magnetoconvection varying along the field> selection law:
$$
\boxed{m_c^4\sim\frac{\pi^6}{Q}.}
$$
For the threshold through order unity, the leading $Qx+\pi^6/x$ contribution at its optimum is $2\pi^3\sqrt Q$. More precisely, the stationary equation gives
$$
x=\pi^3Q^{-1/2}\left[1-\frac{3\pi^2}{2Q}+O(Q^{-3/2})\right].
$$
The correction to $Qx+\pi^6/x$ has no first-order contribution because that expression is already stationary at its leading minimizer. Also $3\pi^2x=O(Q^{-1/2})$ and $x^2=O(Q^{-1})$. Therefore
$$
\boxed{R_{\min}(Q)=\pi^2Q+2\pi^3\sqrt Q+3\pi^4+O(Q^{-1/2}).}
$$
The constant term $3\pi^4$ must be retained. This restricted family bends <magnetic field lines>, so its large stationary threshold is compatible with the unrestricted field-aligned minimum in the preceding part.
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