= Solution
For each hidden variable $\lambda$, a <deterministic local hidden-variable model> assigns values $a_i(\lambda),b_j(\lambda)\in\{0,\ldots,d-1\}$ to all settings. The responses depend only on the local setting; their averaging distribution is setting independent, as required by <measurement independence>.
Let $T(\lambda)$ be the sum of the nonnegative modular representatives in the <chained modular Bell inequality>. Before reduction modulo $d$, the chain telescopes:
$$
\sum_{i=1}^N(a_i-b_i)
+\sum_{i=1}^{N-1}(b_i-a_{i+1})
+(b_N-a_1-1)=-1.
$$
Reduction changes this integer only by multiples of $d$. Since all reduced terms are nonnegative,
$$
T(\lambda)\geq0,\qquad
T(\lambda)\equiv-1\equiv d-1\pmod d.
$$
The smallest possible nonnegative value in this residue class is $d-1$. Averaging over the hidden variable proves
$$
\boxed{I_{N,d}=\mathbb E_\lambda T(\lambda)\geq d-1.}
$$
Adding local random seeds to $\lambda$ gives the same result for stochastic <local hidden-variable theories>.
For the <coplanar qubit realization of the chained Bell inequality>, share
$$
|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}.
$$
At angle $\theta$ in the $xz$ plane, use the <projective measurement> of the observable
$$
M(\theta)=\cos\theta\,\sigma_z+\sin\theta\,\sigma_x,
\qquad
P_a(\theta)=\frac{I+(-1)^aM(\theta)}2,\quad a=0,1.
$$
Here $\sigma_x,\sigma_z$ are <Pauli matrices>, and the classical <bit> $a$ records eigenvalue $(-1)^a$. Since $\langle\sigma_z\otimes\sigma_z\rangle=\langle\sigma_x\otimes\sigma_x\rangle=1$ and both crossed <expected values> vanish in this <Bell state>,
$$
\langle M(\theta)\otimes M(\phi)\rangle=\cos(\theta-\phi).
$$
The one-party <expected values> vanish, so the complete joint <probabilities> are
$$
p(a,b\mid\theta,\phi)
=\frac14\bigl(1+(-1)^{a+b}\cos(\theta-\phi)\bigr).
$$
For binary outcomes, $[a-b]$ is one exactly when $a\neq b$, while $[b-a-1]$ is one exactly when $a=b$. Consequently
$$
\mathbb E([a-b]_2)=\frac{1-\cos(\theta-\phi)}2,\qquad
\mathbb E([b-a-1]_2)=\frac{1+\cos(\theta-\phi)}2.
$$
For $N\geq2$, set $\delta=\pi/(2N)$ and choose Alice's angle $\theta_i=2(i-1)\delta$ and Bob's angle $\phi_i=(2i-1)\delta$. Every adjacent pair $A_i,B_i$ or $B_i,A_{i+1}$ differs by $\delta$, giving expectation $\sin^2(\delta/2)$. The closing pair has angle difference $(2N-1)\delta=\pi-\delta$; its agreement <probability> is also $\sin^2(\delta/2)$. Thus every one of the $2N$ terms contributes equally:
$$
\boxed{I_{N,2}^{\mathrm{quantum}}=2N\sin^2\left(\frac{\pi}{4N}\right)<1.}
$$
Indeed $\sin x<x$ for $x>0$, so this value is below $\pi^2/(8N)\leq\pi^2/16<1$. For an explicit two-setting instance, Alice measures $\sigma_z,\sigma_x$ and Bob measures $(\sigma_z+\sigma_x)/\sqrt2,(-\sigma_z+\sigma_x)/\sqrt2$. The four relevant <expected values> sum to
$$
\boxed{I_{2,2}^{\mathrm{quantum}}=4\sin^2(\pi/8)=2-\sqrt2,}
$$
violating the local bound one. At $N=1$ there is no violation; the two terms always sum to one. As $N\to\infty$, the displayed quantum value is asymptotic to $\pi^2/(8N)$ and approaches the algebraic lower bound zero.
The violation rules out a <local hidden-variable theory> with setting-independent preassigned local responses reproducing these quantum correlations. It does not permit faster-than-light communication: the joint formula gives each local outcome <probability> $1/2$, independent of the remote setting, as required by the <no-signalling theorem>. The contradiction concerns the joint correlations and the Bell assumptions, not the ability to send a signal.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-59-chained-bell.png]
{title=Coplanar chained Bell measurements and their quantum violation of the local bound}
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