= Solution
Put $D=\sqrt{n^2-2n+5}=\sqrt{(n-1)^2+4}$. For $0<n<1$, $D>n+1$, since $D^2-(n+1)^2=4(1-n)>0$. Remembering $\beta<0$, the two lowest trapezium <molecular orbital> energies are
$$
\epsilon_1=\alpha+\frac\beta2(n+1+D)\quad(A_1),\qquad
\epsilon_2=\alpha+\frac\beta2(D-n-1)\quad(B_1).
$$
The remaining levels lie at $2\alpha-\epsilon_2$ and $2\alpha-\epsilon_1$. Filling the lowest spatial orbitals with opposite-spin pairs gives
$$
\boxed{\begin{array}{c|cc}
&\text{square}&\text{trapezium}\\\hline
2\ \text{electrons}&2\alpha+4\beta&2\alpha+\beta(n+1+D)\\
4\ \text{electrons}&4\alpha+4\beta&4\alpha+2\beta D
\end{array}}
$$
For the square four-electron case, two <electrons> occupy the lowest orbital and two occupy the degenerate $E_u$ set. Their summed one-electron <energy> is fixed here, but this <Hückel method> calculation alone does not decide the spin or correlation within that open shell.
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