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Choose the Landau gaugeIt preserves translation invariance in the $y$-direction, so $p_y=\hslash k$ is conserved. With $p_z=0$, magnetic minimal coupling giveswhereThis is a quantum harmonic oscillator, so the Landau levels areindependent of $k$.

The periodic boundary condition in $y$ quantizes $k=2\pi r/L_y$. Requiring the guiding centre to lie in a strip of width $L_x$ gives a $k$-interval of width $|qB|L_x/\hslash$. The number of allowed integers is therefore approximatelythe degeneracy of a Landau level, up to boundary corrections.

With the magnetic minimal coupling convention for charge $e$,Here $\nabla \cdot \mathbf{A}=0$, ${\mathbf{A}}^2=B^2(x^2+y^2)/4$, andTherefore in Cartesian coordinatesIn cylindrical coordinates this isThis choice of $\mathbf{A}$ is the symmetric gauge.