= Solution
For a general <Lagrangian>,
$$
p=\frac{\partial L}{\partial\dot x},
\qquad
H=p\cdot\dot x-L.
$$
Along an Euler-Lagrange trajectory,
$$
\frac{dH}{dt}=-\frac{\partial L}{\partial t},
$$
so energy is conserved when $L$ has no explicit time dependence.
For the given <Lagrangian>, write $v=|\dot x|$. At points where $v\ne0$,
$$
\boxed{p=mc\frac{\dot x}{v}}.
$$
The Euler-Lagrange equation is therefore
$$
\boxed{
\frac d{dt}\left(mc\frac{\dot x}{|\dot x|}\right)
=-\nabla V(x)}.
$$
In expanded form its left-hand side is
$$
mc\left(
\frac{\ddot x}{v}
-\frac{\dot x(\dot x\cdot\ddot x)}{v^3}
\right).
$$
The <momentum> has fixed magnitude,
$$
\boxed{p\cdot p=m^2c^2},
$$
so this is constant without needing to solve the equation of motion. The <Hamiltonian> is
$$
\boxed{H=p\cdot\dot x-L=V(x)}.
$$
The <singular Legendre transform of a degree-one velocity Lagrangian> applies here: $p$ determines only the direction of $\dot x$, not its magnitude, and the <velocity> Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the <Lagrangian> from this <Hamiltonian>.
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