= Solution
Use a steady, plane, nonradiative <normal shock wave> in a <perfect gas>, and measure the normal speeds $v_{\mathrm u},v_{\mathrm d}>0$ in the <shock frame>. The labels denote upstream and downstream. Conservation of mass, normal <momentum> and total <energy> gives
$$
\rho_{\mathrm u}v_{\mathrm u}=\rho_{\mathrm d}v_{\mathrm d},\qquad
p_{\mathrm u}+\rho_{\mathrm u}v_{\mathrm u}^2=p_{\mathrm d}+\rho_{\mathrm d}v_{\mathrm d}^2,
$$
and
$$
\frac12v_{\mathrm u}^2+\frac{\gamma}{\gamma-1}\frac{p_{\mathrm u}}{\rho_{\mathrm u}}
=\frac12v_{\mathrm d}^2+\frac{\gamma}{\gamma-1}\frac{p_{\mathrm d}}{\rho_{\mathrm d}}.
$$
The enthalpy term includes the <pressure> work. For very large upstream <Mach number>, neglect $p_{\mathrm u}$ compared with $\rho_{\mathrm u}v_{\mathrm u}^2$. If $r=\rho_{\mathrm d}/\rho_{\mathrm u}$ is the <shock compression ratio>, the first two equations imply
$$
v_{\mathrm d}=\frac{v_{\mathrm u}}r,\qquad
p_{\mathrm d}=\rho_{\mathrm u}v_{\mathrm u}^2\left(1-\frac1r\right).
$$
Substitution into the <energy> equation gives
$$
1=\frac1{r^2}+\frac{2\gamma}{\gamma-1}\frac{r-1}{r^2},\qquad
(r-1)\left(r-\frac{\gamma+1}{\gamma-1}\right)=0.
$$
Discard the no-shock solution $r=1$. For a nonrelativistic monatomic gas, $\gamma=5/3$, so the <Strong-shock Rankine-Hugoniot conditions> become
$$
\boxed{\frac{\rho_{\mathrm d}}{\rho_{\mathrm u}}=\frac{v_{\mathrm u}}{v_{\mathrm d}}=4,\qquad
p_{\mathrm d}=\frac34\rho_{\mathrm u}v_{\mathrm u}^2.}
$$
The factor four requires this <adiabatic index> and negligible radiative, magnetic and escaping-particle contributions to the flux budget. For example, the same strong, nonradiative calculation with $\gamma=4/3$ gives compression seven, not four.
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