= Solution
Use base-two logarithms for <Von Neumann entropy>, so the units are <bits>, and adopt $0\log_2 0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed <Von Neumann entropies> are finite. The <density operator> of the joint <pure state> is $\rho_{AB}=|\Psi_{AB}\rangle\langle\Psi_{AB}|$. It is a rank-one <orthogonal projector>, whose only nonzero <eigenvalue> is one. Consequently
$$
\boxed{S(A,B)=-\operatorname{Tr}(\rho_{AB}\log_2\rho_{AB})=0.}
$$
The <reduced density matrices> are $\rho_A=\operatorname{Tr}_B\rho_{AB}$ and $\rho_B=\operatorname{Tr}_A\rho_{AB}$. The <quantum conditional entropy> is defined by the entropy difference
$$
S(B|A):=S(\rho_{AB})-S(\rho_A).
$$
For the <pure state> under consideration this becomes $S(B|A)=-S(\rho_A)$. To determine when it is negative, write the <Schmidt decomposition>
$$
|\Psi_{AB}\rangle=\sum_{j=1}^r\sqrt{\lambda_j}|a_j\rangle|b_j\rangle,
\qquad \lambda_j>0,\qquad\sum_j\lambda_j=1.
$$
The <eigenvalues> of $\rho_A$ are the $\lambda_j$, together with possible zeros. Each contribution $-\lambda_j\log_2\lambda_j$ is nonnegative, and their sum vanishes exactly when one <eigenvalue> equals one. Equivalently the <Schmidt rank> is one and $|\Psi_{AB}\rangle$ is a <product state>. If the <pure state> is <entangled>, then $r\geq2$, every nonzero <eigenvalue> is strictly below one, and the <Von Neumann entropy> is strictly positive. Thus the <pure-state negative conditional entropy criterion> gives
$$
\boxed{S(B|A)<0\quad\Longleftrightarrow\quad|\Psi_{AB}\rangle\text{ is entangled}.}
$$
The purity assumption matters: for mixed <density operators>, negative <quantum conditional entropy> is not necessary for <entanglement>.
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