= Solution
Put $\ell=p_\phi$. The <Hamiltonian> has no explicit time or azimuthal dependence, so $\dot H=\{H,H\}=0$ and $\dot\ell=-H_\phi=0$. Its equations of motion are
$$
\dot R=p_R,\qquad\dot\phi=\frac\ell{R^2},\qquad
\dot p_R=\frac{\ell^2}{R^3}+\frac A{R^2},\qquad\dot\ell=0.
$$
Write $C=p_R\ell\sin\phi+(\ell^2/R+A)\cos\phi$. Differentiating along these equations gives
$$
\begin{aligned}
\dot C={}&\left(\frac{\ell^3}{R^3}+\frac{A\ell}{R^2}\right)\sin\phi
+\frac{p_R\ell^2}{R^2}\cos\phi
-\frac{p_R\ell^2}{R^2}\cos\phi\\
&-\left(\frac{\ell^3}{R^3}+\frac{A\ell}{R^2}\right)\sin\phi=0.
\end{aligned}
$$
Equivalently, the same cancellation evaluates $\{C,H\}$. Hence $\boxed{H,\ p_\phi,\ C\text{ are all constants of motion}.}$ The extra conserved component is associated with the Runge-Lenz symmetry of this inverse-distance central potential, rather than with conservation of radial momentum.
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