= Solution
For $|\mathbf v|<c$, define the <Lorentz factor> $\gamma=(1-|\mathbf v|^2/c^2)^{-1/2}$. A particle of <rest mass> $m$ has
$$
\boxed{\mathbf p=\gamma m\mathbf v,\qquad E=\gamma mc^2.}
$$
These expressions give the <relativistic energy-momentum relation> directly:
$$
E^2-c^2\mathbf p\cdot\mathbf p=\gamma^2m^2c^4\left(1-\frac{|\mathbf v|^2}{c^2}\right)=m^2c^4,
$$
so $\boxed{E^2=c^2\mathbf p\cdot\mathbf p+m^2c^4}$.
The <four-momentum> is
$$
\boxed{P^\mu=(E/c,\mathbf p)=mU^\mu,\qquad U^\mu=\gamma(c,\mathbf v),}
$$
where $U^\mu$ is the <four-velocity>, the proper-time derivative of $(ct,\mathbf x)$. It is a <four-vector>, so its <Minkowski metric> norm is invariant under <Lorentz transformations>. Choose <metric signature> $(+,-,-,-)$; in the particle's rest frame, $P^\mu=(mc,\mathbf0)$ and $P_\mu P^\mu=m^2c^2$. In any frame the same norm is $E^2/c^2-|\mathbf p|^2$. Equating the two yields the same <relativistic energy-momentum relation>. With the opposite metric convention both norm expressions change sign, leaving the energy relation unchanged.
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