= Similarity of squared matrix products with one diagonalizable product
{title2=$AB\text{ diagonalizable}\Longrightarrow(AB)^2\sim(BA)^2$}
The identity $(BA)^{j+1}=B(AB)^jA$ gives $m_{BA}(t)\mid t\,m_{AB}(t)$. If $AB$ is <diagonalizable>, its <minimal polynomial> has simple roots. Hence all nonzero Jordan blocks of $BA$ have size one and its zero blocks have size at most two. Squaring kills these nilpotent blocks, making both squared products diagonalizable. Their eigenvalue multisets agree by <Characteristic polynomials of AB and BA>, so the squared products are <similar matrices>.
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