= Solution
\b[They need a pre-shared entanglement resource that permits deterministic production of a Bell pair.] In particular, one shared <Bell pair> and two classical bits suffice for exact <quantum teleportation>. This condition is stronger than merely having some entanglement.
Let Alice's input be $|\psi\rangle_C=\alpha|0\rangle+\beta|1\rangle$, and let her system $A$ and Bob's system $B$ share $|\Phi^+\rangle_{AB}=(|00\rangle+|11\rangle)/\sqrt2$. Use the <Bell states> $|\Phi^\pm\rangle=(|00\rangle\pm|11\rangle)/\sqrt2$ and $|\Psi^\pm\rangle=(|01\rangle\pm|10\rangle)/\sqrt2$. Direct expansion gives
$$
|\psi\rangle_C|\Phi^+\rangle_{AB}
=\frac12\left[
|\Phi^+\rangle_{CA}|\psi\rangle_B+
|\Phi^-\rangle_{CA}Z|\psi\rangle_B+
|\Psi^+\rangle_{CA}X|\psi\rangle_B+
|\Psi^-\rangle_{CA}XZ|\psi\rangle_B
\right].
$$
Alice performs a <projective measurement> in this Bell basis and sends its two-bit outcome. Each outcome has probability $1/4$. Bob applies respectively $I,Z,X,ZX$, recovering $|\psi\rangle$ exactly, without either party knowing $\alpha,\beta$. Before he receives the outcome, his averaged state is $I/2$, so the protocol does not communicate without the classical message. Alice's measurement consumes the input and the shared entanglement.
For necessity, exact transmission of every pure qubit means the induced <quantum channel> is the identity: pure-state projectors span the operator space, and a physical channel is linear. Alice can therefore prepare a Bell pair locally and transmit one half by the proposed protocol, leaving a Bell pair shared with Bob. The protocol would itself be a deterministic <local operations and classical communication> conversion of their resource into a Bell pair. This proves the <exact teleportation resource criterion>. With no shared entanglement, it is impossible: local operations and classical communication preserve <separable quantum states>, whereas the resulting Bell pair would be entangled.
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