r = e \omega r B \ \Rightarrow \ \omega = \frac{e B}{M},$$ with the cyclotron frequency $\omega$ being independent of the radius $r$. Obviously, for this system all classical orbits are closed. The same is true for a particle moving in a $1/r$ or $r^2$ potential. In those cases, the fact that all bound classical orbits are closed is related to the conservation of the Runge-Lenz vector which generates a hidden accidental dynamical symmetry.

Let us now investigate the question of accidental symmetry for the particle in the constant magnetic field. The Lagrange function then takes the form $$L = \frac{M}{2} \vec v^2 - e \vec A(\vec x) \cdot \vec v  =
\frac{M}{2} \left(\dot x^2 + \dot y^2\right) - e B x \dot y,$$ and the corresponding conjugate momenta are $$p_x = \frac{{\partial}L}{{\partial}\dot x} = M \dot x = M v_x, \
p_y = \frac{{\partial}L}{{\partial}\dot y} = M \dot y - e B x = M v_x - e B x.$$ First of all, in the gauge that we picked, $y$ is a cyclic coordinate and hence the canonically conjugate momentum $p_y$ is conserved as a consequence of translation invariance in the $y$-direction. Despite the fact that the system 