= Solution
Take the equilibrium <mass density> to be constant and absorb any conservative equilibrium force into the <pressure>. The linearized <ideal magnetohydrodynamic induction equation>, using the <solenoidal vector field> conditions, is
$$
\partial_t\mathbf b=(\mathbf B\cdot\nabla)\mathbf u-(\mathbf u\cdot\nabla)\mathbf B.
$$
For $\mathbf B=Jr\mathbf e_\phi$, differentiation of the cylindrical basis is important. Acting on a vector whose cylindrical components have azimuthal dependence $e^{im\phi}$ gives $(\mathbf B\cdot\nabla)\mathbf u=imJ\mathbf u+J\mathbf e_z\times\mathbf u$. Also $(\mathbf u\cdot\nabla)\mathbf B=J\mathbf e_z\times\mathbf u$. These basis contributions cancel in induction, giving \b[$i\omega\mathbf b=imJ\mathbf u$].
The perturbation of the <Lorentz force> is
$$
(\nabla\times\mathbf b)\times\mathbf B+(\nabla\times\mathbf B)\times\mathbf b=(\mathbf B\cdot\nabla)\mathbf b+(\mathbf b\cdot\nabla)\mathbf B-\nabla(\mathbf B\cdot\mathbf b).
$$
Both of the first two terms now contribute a basis-rotation term. With $\widetilde p=p_1/\rho+\mathbf B\cdot\mathbf b/(\mu_0\rho)$, the linearized <ideal magnetohydrodynamic momentum equation> is therefore
$$
\boxed{i\omega\mathbf u=-\nabla\widetilde p+\frac1{\mu_0\rho}(imJ\mathbf b+2J\mathbf e_z\times\mathbf b),\qquad i\omega\mathbf b=imJ\mathbf u.}
$$
The magnetic part of $\widetilde p$ is the perturbation of <magnetic pressure>, not an extra restoring force to retain after projecting out <pressure>.
Put $D=\omega^2-m^2\widetilde J^2$, where $\widetilde J=J/\sqrt{\mu_0\rho}$. For $mJ\ne0$, substitute $\mathbf u=\omega\mathbf b/(mJ)$ into the momentum equation and multiply by $mJ$. Taking a curl removes the modified <pressure>. Since
$$
\nabla\times(\mathbf e_z\times\mathbf b)=\mathbf e_z\nabla\cdot\mathbf b-\partial_z\mathbf b=-ik\mathbf b,
$$
we obtain the <uniform-current toroidal-field curl reduction>
$$
D\nabla\times\mathbf b=-2mk\widetilde J^2\mathbf b.
$$
A second curl, using $\nabla\times\nabla\times\mathbf b=-\nabla^2\mathbf b$, gives
$$
\boxed{\nabla^2\mathbf b=-\frac{4m^2k^2\widetilde J^4}{(\omega^2-m^2\widetilde J^2)^2}\mathbf b.}
$$
The divided expression assumes $D\ne0$. If $mkJ\ne0$, the undivided curl equation shows that $D=0$ would force $\mathbf b=0$, so this restriction loses no such magnetic mode. The $k=0$ degenerate limit can instead be taken in the undivided equations; it has no growing branch of the kind considered below. Axisymmetric $m=0$ disturbances are likewise excluded by the <velocity> elimination, and induction gives no nonzero-frequency magnetic disturbance in this particular equilibrium.
For a vector Laplacian eigenmode with $\alpha^2+k^2>0$, elimination gives
$$
D^2=\frac{4m^2k^2\widetilde J^4}{\alpha^2+k^2},\qquad \boxed{\omega^2_\pm=\widetilde J^2\left(m^2\pm\frac{2|mk|}{\sqrt{\alpha^2+k^2}}\right).}
$$
Thus \b[both squared frequencies are real]. The lower branch satisfies $\omega_-^2\geq\widetilde J^2(|m|^2-2|m|)$, so it is positive for $|m|>2$. For $|m|=2$ it is positive when $\alpha\ne0$, but the printed strict-positivity statement needs an endpoint qualification: \b[$|m|=2$, $\alpha=0$ permits a neutral mode unless boundary conditions exclude it].
This is a genuine counterexample to the unqualified endpoint, not merely a loose algebraic bound. The <neutral quadrupolar perturbation of a uniform-current toroidal field>
$$
\mathbf b=Cr(\mathbf e_r+i\mathbf e_\phi)e^{2i\phi+ikz},\qquad \mathbf u=0,\qquad \widetilde p=0,\qquad \omega=0
$$
is regular at the axis, is solenoidal, and has $\nabla\times\mathbf b=k\mathbf b$, hence $\nabla^2\mathbf b=-k^2\mathbf b$. Also $\mathbf e_z\times\mathbf b=-i\mathbf b$, so the non-gradient part of the linearized <Lorentz force> vanishes for $m=2$. The gas-pressure perturbation $p_1=-\mathbf B\cdot\mathbf b/\mu_0$ cancels its <magnetic pressure> gradient, as required by $\widetilde p=0$. Taking its real part gives a real disturbance. Decay at radial infinity or an appropriate homogeneous radial boundary condition would exclude it, restoring strict positivity for the allowed eigenmodes; no such condition is printed.
For $|m|=1$ the lower branch is unstable precisely when
$$
\boxed{m=\pm1,\qquad k\ne0,\qquad \alpha^2<3k^2.}
$$
At equality it is marginal. With the time convention $e^{i\omega t}$, the growing member has $\omega=-is$ and
$$
s=|\widetilde J|\sqrt{\frac{2|k|}{\sqrt{\alpha^2+k^2}}-1}.
$$
The other branch remains oscillatory. Actual admissibility of a chosen $\alpha$ depends on the radial domain and its <boundary conditions>.
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