 Third-level degeneracy in a rectangular quantum box

For $b>a$ in a two-dimensional infinite square well, the lowest two states are $(1,1)$ and $(1,2)$, both nondegenerate. The next energy is the smaller of $E_{21}$ and $E_{13}$: every other candidate lies above at least one of these. Requiring equality gives $3/a^2=8/b^2$, or $b/a=\sqrt{8/3}$. In units ${\pi }^2{\hslash }^2/(2M{a}^2)$ the lowest distinct energies are then $11/8$, $5/2$ and $35/8$, with degeneracies one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.