= Solution
The displacement equation is $\rho\mathbf u_{tt}=(\lambda+\mu)\nabla(\nabla\cdot\mathbf u)+\mu\Delta\mathbf u$. With $\mathbf u=\nabla\phi+\nabla\times\boldsymbol\psi$ and divergence-free vector-potential gauge, divergence and curl give <wave equations> with
$$
\boxed{c_P^2=(\lambda+2\mu)/\rho,\qquad c_S^2=\mu/\rho.}
$$
Thus $\phi_{tt}=c_P^2\Delta\phi$ and $\boldsymbol\psi_{tt}=c_S^2\Delta\boldsymbol\psi$, up to gauge constants. The <P wave> displacement is parallel to its propagation direction; <S wave> displacement is perpendicular, with two transverse polarizations.
Take the plane of incidence to be $xy$. Let $k$ be the shared tangential wave number, $h=(\omega^2/c_P^2-k^2)^{1/2}$ and $s=(\omega^2/c_S^2-k^2)^{1/2}$. Incident P, reflected P and reflected SV waves have respectively wave vectors $(k,h)$, $(k,-h)$ and $(k,-s)$. Thus the reflected P angle equals $\theta$ and $\sin\theta_S=(c_S/c_P)\sin\theta$. No SH wave is excited by this planar P incidence.
Use scalar potentials $\mathbf u=(\phi_x+\psi_y,\phi_y-\psi_x)$, with incident potential amplitude one, reflected P amplitude $R$, and reflected SV amplitude $Q$. At the surface their phases are common. Vanishing $xy$ and $yy$ <tractions> gives, with $D=s^2-k^2$,
$$
2kh(1-R)+DQ=0,\qquad D(1+R)+2ksQ=0.
$$
Eliminating $Q$ gives
$$
\boxed{R=\frac{4k^2hs-D^2}{4k^2hs+D^2}.}
$$
There is no reflected P wave precisely when $4k^2hs=D^2$. If $\beta=c_S^2/c_P^2$ and $\sigma=k^2c_S^2/\omega^2=\beta\sin^2\theta$, then $h=(\omega/c_S)\sqrt{\beta-\sigma}$, $s=(\omega/c_S)\sqrt{1-\sigma}$ and $D=(\omega^2/c_S^2)(1-2\sigma)$. Substitution proves
$$
\boxed{4\sigma\sqrt{1-\sigma}\sqrt{\beta-\sigma}=(1-2\sigma)^2.}
$$
The condition applies only to incidence angles in the physical range $0\leq\sigma\leq\beta$; it need not have a solution for every elastic-modulus ratio.
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