= Solution
Write $c_s^2=p/\rho$ for the constant <isothermal sound speed> squared and $v_a^2=B^2/(\mu_0\rho)$ for the constant <Alfvén speed> squared. The horizontal <magnetic field> has no vertical <magnetic tension>; <magnetostatic equilibrium> is
$$
\frac{d\Pi}{dz}=-g\rho,\qquad \Pi=p+\frac{B^2}{2\mu_0}=\left(c_s^2+\frac{v_a^2}{2}\right)\rho.
$$
Consequently the <magnetized isothermal atmosphere> has
$$
\boxed{H=\frac{c_s^2+v_a^2/2}{g},\qquad
\rho=\rho_0e^{-z/H},\qquad p=c_s^2\rho_0e^{-z/H},\qquad
B=B_0e^{-z/(2H)},\quad B_0^2=\mu_0\rho_0v_a^2.}
$$
The sign of $B_0$ specifies the orientation of the <magnetic field>. The <adiabatic sound speed> is $v_s^2=\gamma p/\rho=\gamma c_s^2$, also constant.
For the proposed <normal mode>, let $A=\exp(i\mathbf k\cdot\mathbf x+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\boldsymbol\xi$ for its constant complex amplitude. The <divergence> amplitude is $\Delta=ik_x\xi_x+ik_y\xi_y+(ik_z+a)\xi_z$. Using $\rho'/\rho=-1/H$ and $B'/B=-a$, the <Eulerian and Lagrangian fluid perturbations> give
$$
\frac{\delta\rho}{\rho A}=-\Delta+\frac{\xi_z}{H},\qquad
\frac{\delta\mathbf B}{BA}=ik_y\boldsymbol\xi-\mathbf e_y\Delta+a\mathbf e_y\xi_z.
$$
Since $\Pi'/\rho=-g$, the <magnetohydrodynamic total pressure> amplitude is
$$
\psi=\frac{\delta\Pi}{\rho A}=-(v_s^2+v_a^2)\Delta+v_a^2ik_y\xi_y+g\xi_z.
$$
The vertical derivative of $\delta\Pi=\rho A\psi$ is $\rho A(ik_z-a)\psi$. For the <magnetic tension> perturbation, the terms proportional to $aik_y\xi_z\mathbf e_y$ from $\delta\mathbf B\cdot\nabla\mathbf B$ and $\mathbf B\cdot\nabla\delta\mathbf B$ cancel, leaving
$$
\frac{\delta\mathbf B\cdot\nabla\mathbf B+\mathbf B\cdot\nabla\delta\mathbf B}{\mu_0\rho A}
=-v_a^2 k_y^2\boldsymbol\xi-v_a^2ik_y\Delta\,\mathbf e_y.
$$
Thus the <linearization> really reduces to the constant-coefficient <eigenvalue problem>
$$
\begin{aligned}
-\omega^2\xi_x&=-ik_x\psi-k_y^2v_a^2\xi_x,\\
-\omega^2\xi_y&=-ik_y(\psi+v_a^2\Delta)-k_y^2v_a^2\xi_y,\\
-\omega^2\xi_z&=g(\Delta-\xi_z/H)-(ik_z-a)\psi-k_y^2v_a^2\xi_z.
\end{aligned}
$$
For a static <ideal magnetohydrodynamics> equilibrium with conservative <gravitational acceleration>, the <magnetohydrodynamic energy principle> gives a <Hermitian operator> for the restoring force in the <mass density> weighted <inner product>, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential <mass density> weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then $\omega^2\boldsymbol\xi=\mathsf K\boldsymbol\xi$, where
$$
\mathsf K=
\begin{pmatrix}
qk_x^2+bk_y^2&sk_xk_y&qk_xk_z+ik_x(g-qa)\\
sk_xk_y&sk_y^2&sk_yk_z+ik_y(g-sa)\\
qk_xk_z-ik_x(g-qa)&sk_yk_z-ik_y(g-sa)&q(k_z^2+a^2)+bk_y^2
\end{pmatrix}.
$$
This is a <Hermitian matrix> for real $\mathbf k$ and real positive $s,b$. Therefore \b[every squared frequency $\omega^2$ is real]; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized <plane waves>, not finite-total-energy global modes.
At $\omega^2=0$ and $k_y\ne0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta=g\xi_z$. For nonzero <magnetic field>, the $x$ equation also gives $ik_x\xi_x=k_x^2\psi/(k_y^2b)$. Substituting $\psi=-b[\Delta-ik_y\xi_y]+g\xi_z-s\Delta$ and then $s\Delta=g\xi_z$ yields
$$
\boxed{s\Delta=g\xi_z,\qquad -(k_x^2+k_y^2)\psi=k_y^2b(ik_z+a)\xi_z.}
$$
For a nontrivial mode $\xi_z\ne0$; otherwise these equations and $k_yb\ne0$ force every component to vanish. Substitution in the vertical equation gives
$$
\boxed{b k_y^2\left(|\mathbf k|^2+\frac{1}{4H^2}\right)+\frac gH\left(1-\frac{gH}{s}\right)(k_x^2+k_y^2)=0.}
$$
For example, the <determinant> provides an independent check:
$$
\det\mathsf K=s b k_y^2\left[b k_y^2(|\mathbf k|^2+a^2)+\left(\frac gH-\frac{g^2}{s}\right)(k_x^2+k_y^2)\right].
$$
To prove the claimed “if and only if”, rather than merely locating a zero <eigenvalue>, use the <magnetic buoyancy energy criterion>
$$
\begin{aligned}
\boldsymbol\xi^\dagger\mathsf K\boldsymbol\xi={}&s\left|\Delta-\frac g{s}\xi_z\right|^2
+b\left|ik_x\xi_x+(ik_z+a)\xi_z\right|^2\\
&+bk_y^2(|\xi_x|^2+|\xi_z|^2)
+\left(\frac gH-\frac{g^2}{s}\right)|\xi_z|^2.
\end{aligned}
$$
If $gH\leq s$, every term is nonnegative, so all <eigenvalues> are nonnegative for every <wavevector>, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne0$, and $\xi_z=1$, $\xi_x=-a/(ik_x)$, $\xi_y=g/(sik_y)$. The first two squares vanish; the remaining value is
$$
bk_y^2\left(1+\frac{a^2}{k_x^2}\right)+\frac gH-\frac{g^2}{s},
$$
which is negative for sufficiently small nonzero $k_y$. The <Rayleigh quotient> of this displacement is therefore negative, proving a negative <eigenvalue> and an unstable <normal mode>. This uses precisely the assumption that <boundary conditions> do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.
Finally the <plasma beta> is $\beta=p/[B^2/(2\mu_0)]=2c_s^2/v_a^2$. Hence
$$
\boxed{\text{instability}\ \Longleftrightarrow\ gH>v_s^2
\ \Longleftrightarrow\ \frac{v_a^2}{2}>(\gamma-1)c_s^2
\ \Longleftrightarrow\ \beta(\gamma-1)<1.}
$$
The <magnetic field> is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma>1$ it is stable. <Parker instability> occurs because <magnetic pressure> supports part of the atmosphere's weight while the gas can drain along gently bent <magnetic field lines>. The resulting buoyant displacement can release <Newtonian gravitational potential energy>; choosing long wavelengths along the <magnetic field> makes the stabilizing <magnetic tension> small. The completed-square calculation quantifies this competition, including the <adiabatic process> restoring force.
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