= Solution
Take $P_0>0$, real $\alpha$, and the real branch $T\ge T_0$, with physical <temperature> $T>0$. In <hydrostatic equilibrium> the <pressure> decreases with height $z$:
$$
\frac{dP}{dz}=-\rho g<0.
$$
Thus an <atmospheric thermal inversion> means $dT/dz>0$, equivalently $dT/dP<0$. For $\alpha\ne0$, the <logarithm> of the given relation yields
$$
T(P)=T_0+\frac{\log^2(P/P_0)}{\alpha^2},\qquad
\frac{\log(P/P_0)}{\alpha}=\sqrt{T-T_0}\ge0.
$$
The sign restriction must be retained after squaring. In particular, the inverted branch has $0<P\le P_0$, whereas the non-inverted branch has $P\ge P_0$. By <differentiation>, on the interior of either branch,
$$
\frac{dT}{d\log P}=\frac{2\log(P/P_0)}{\alpha^2}
=\frac{2\sqrt{T-T_0}}{\alpha},\qquad
\frac{dT}{dz}=-\frac{2\rho g\sqrt{T-T_0}}{\alpha P}.
$$
Consequently,
$$
\boxed{\alpha<0:\ \text{thermal inversion};\qquad
\alpha>0:\ \text{temperature decreases outward}.}
$$
At $T=T_0$, the inverse <temperature gradient> vanishes at the branch endpoint, rather than producing a finite isothermal interval. If $\alpha=0$, the relation becomes $P=P_0$ independently of $T$; it is not a nontrivial vertical <atmospheric pressure-temperature profile> under <hydrostatic equilibrium>. These distinctions prevent the two sides of the squared parabola from being interpreted as one permissible atmosphere.
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