= Solution
Bob uses <qudit teleportation> to teleport his carrier $B_1$ to Clare through the pair $B_2C$. This transfers Bob's entanglement with Alice to Clare. Explicitly, Bob performs a <generalized Bell basis> measurement on $B_1B_2$. Contraction of the two initial pairs gives
$$
{}_{B_1B_2}\langle\psi_{rs}|
\left(|\psi_{00}\rangle_{AB_1}|\psi_{00}\rangle_{B_2C}\right)
=\frac1{n\sqrt n}\sum_j\omega^{-rj}|j\rangle_A|j+s\rangle_C
=\frac1n(I_A\otimes X_C^sZ_C^{-r})|\psi_{00}\rangle_{AC}.
$$
The probability is $1/n^2$ for each outcome. Bob sends $(r,s)$ to Clare, who applies $Z_C^rX_C^{-s}$. \b[Alice and Clare then share $|\psi_{00}\rangle_{AC}$ with certainty.] This is <entanglement swapping>; neither an additional entangled pair nor a quantum transmission during the protocol is required.
The two initial pairs are consumed, and Bob's measured carriers cease to be entangled with $AC$. Without Bob's classical record, averaging the possible <generalized Bell states> leaves $AC$ maximally mixed. Thus the conditional <entanglement swapping> does not supply faster-than-light communication.
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