Commuting squares of ordinals ($\alpha \beta =\beta \alpha ⟺{\alpha }^2{\beta }^2={\beta }^2{\alpha }^2$)

Two ordinals commute under ordinal multiplication if and only if their squares commute. One implication follows from associativity. For the converse, squaring is strictly increasing: $A<B$ implies $A^2\le BA<B^2$. If $\alpha \beta <\beta \alpha$, monotonicity givesa contradiction to commuting squares. Swap the factors to exclude the reverse inequality; zero factors cause no exception.