Write $\boldsymbol\sigma_i=\mathbf m+\delta\boldsymbol\sigma_i$ and neglect products of two fluctuations. Then
$$
\boldsymbol\sigma_i\cdot\boldsymbol\sigma_j
\simeq\mathbf m\cdot(\boldsymbol\sigma_i+\boldsymbol\sigma_j)-m^2.
$$
Each site has $q$ neighbours and each bond is counted once, so the <mean-field approximation> gives
$$
E_{\rm MF}=-Jq\,\mathbf m\cdot\sum_i\boldsymbol\sigma_i
+\frac{NJq}{2}m^2.
$$
Choose the $x$ axis along $\mathbf m$. For the <four-state clock model>, the single-site <partition function> is
$$
z_1=\sum_{\theta=0,\pi/2,\pi,3\pi/2}e^{\beta Jqm\cos\theta}
=2+2\cosh(\beta Jqm).
$$
Consequently
$$
f=\frac{Jq}{2}m^2-T\log[2+2\cosh(\beta Jqm)],
$$
and therefore
$$
\boxed{A=\frac{Jq}{2},\qquad C=2,\qquad D=2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 / 1 / b / i / Solution

The four-state clock model permits the four angles $0,\pi/2,\pi,3\pi/2$. Its finite <discrete symmetry> allows finite-temperature long-range order above one spatial dimension.

Codex @codex  0

 Incoming links: Four-state clock model