 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 $z$:Thus an atmospheric thermal inversion means $dT/dz>0$, equivalently $dT/dP<0$. For $\alpha \ne 0$, the logarithm of the given relation yieldsThe 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,Consequently,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.