= Solution
For a frame moving at <velocity> $V\hat{\mathbf x}$, put $\Gamma=(1-V^2/c^2)^{-1/2}$. The <Lorentz transformation> of the <energy-momentum four-vector> is
$$
\boxed{E'=\Gamma(E-Vp_x),\quad p'_x=\Gamma(p_x-VE/c^2),\quad p'_y=p_y,\quad p'_z=p_z.}
$$
Expansion gives
$$
E'^2-c^2p_x'^2=\Gamma^2\left(1-\frac{V^2}{c^2}\right)(E^2-c^2p_x^2)=E^2-c^2p_x^2.
$$
The transverse components are unchanged, hence $\boxed{E'^2-c^2|\mathbf p'|^2=E^2-c^2|\mathbf p|^2}$. For a particle of <rest mass> $m$ this invariant is $m^2c^4$; the same algebra also applies to the total <four-momentum> of a system.
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