= Solution
Alice holds the input <qubit> $Q$ in $|\psi\rangle_Q=a|0\rangle+b|1\rangle$ and her half $A$ of a shared <Bell pair> $|\phi_+\rangle_{AB}$; Bob holds $B$. Alice applies a <controlled-NOT gate> $Q\to A$, then a <Hadamard gate> on $Q$. In the system order $Q,A,B$, the initial <quantum state> is
$$
\frac1{\sqrt2}\left(a|000\rangle+a|011\rangle+b|100\rangle+b|111\rangle\right).
$$
After these two <quantum gates>, direct expansion yields the <quantum teleportation> identity
$$
\frac12\Bigl[
|00\rangle_{QA}(a|0\rangle+b|1\rangle)_B
+|01\rangle_{QA}(b|0\rangle+a|1\rangle)_B
+|10\rangle_{QA}(a|0\rangle-b|1\rangle)_B
+|11\rangle_{QA}(-b|0\rangle+a|1\rangle)_B
\Bigr]
=\frac12\sum_{m,n=0}^1|mn\rangle_{QA}X_B^nZ_B^m|\psi\rangle_B.
$$
Alice performs <computational-basis measurements> on $Q,A$, and sends the two classical <bits> $m,n$ to Bob. Each outcome has <probability> $1/4$, independently of the input <pure state>. Bob applies the indicated <Pauli gates>: first $X^n$, then $Z^m$, so his total correction is $Z^mX^n$. In every branch,
$$
\boxed{Z^mX^nX^nZ^m|\psi\rangle=|\psi\rangle.}
$$
This proves exact <quantum teleportation> using one <Bell pair> and two classical <bits>. Alice's original <qubit> has been measured; the protocol transfers the <quantum state> without leaving her another copy.
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