= Solution
Part (b) proves one implication. Conversely, assume in ZF that
$$
\kappa^2=\kappa
$$
for every infinite cardinal $\kappa$. Let $X$ be any set and let $\gamma=h(X)$ be its Hartogs ordinal. Finite $X$ is already well-orderable, so suppose $X$ is infinite and form the <disjoint union>
$$
A=X\sqcup\gamma.
$$
By hypothesis, $A\times A$ is bijective with $A$. Since $\gamma\times X$ injects into $A\times A$, we obtain
$$
|\gamma|\,|X|\leq|\gamma|+|X|.
$$
Apply part (c), with $K=\gamma$ and $L=X$. It gives either an injection $\gamma\to X$ or a surjection $s:\gamma\to X$. The first alternative contradicts <Hartogs theorem>. In the second, every fiber $s^{-1}(\{x\})$ is a nonempty set of ordinals and therefore has a least member. The map
$$
x\longmapsto\min s^{-1}(\{x\})
$$
is an injection of $X$ into the ordinal $\gamma$ and pulls its well-order back to $X$. Thus every set is well-orderable, so the <well-ordering theorem> gives the <axiom of choice>. This proves the <Tarski cardinal-square theorem>.
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