= Solution
Put $w=V_{11}<0$. At $M=(\pi/a,\pi/a)$, the free energy is $2E_X$. The four cyclic combinations in the supplied <secular equation> give the <four-plane-wave splitting at a square-zone corner>
$$
\boxed{E_{M,1}=2E_X+w+2u,\qquad E_{M,2}=E_{M,3}=2E_X-w,\qquad E_{M,4}=2E_X+w-2u.}
$$
For example, the normalized coefficient vectors in the source's cyclic basis are $(1,1,1,1)/2$, $(1,0,-1,0)/\sqrt2$, $(0,1,0,-1)/\sqrt2$, and $(1,-1,1,-1)/2$. Because $u<w<0$, these are ordered as written: a lower singlet, a central doublet and an upper singlet. The sketch puts the $X$ and $M$ energies on one common scale; its representative negative couplings are stated in the caption.
\Image[/past-exam-of-the-general-relativity-and-cosmology-and-condensed-matter-physics-paper-of-part-b-of-the-physics-course-of-the-university-of-oxford/2011-square-lattice-energy-levels.png]
{title=Common energy scale at the square-zone edge and corner, with illustrative couplings V10 = −1 eV and V11 = −0.2 eV}
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