= Solution
Choice well-orders every infinite $X$, so Hessenberg gives $X^2\cong X$. Conversely assume this square identity for every infinite set. For arbitrary $X$, form the disjoint union $Y=X\sqcup h(X)$. A pairing $Y^2\cong Y$, together with the well-order on $h(X)$, yields by the standard tracing/comparison argument an injection either $h(X)\to X$ or $X\to h(X)$. The first is forbidden by <Hartogs theorem>, so the second well-orders $X$. This is <Tarski cardinal-square theorem>, and hence choice follows.
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