I already have a demonstration, but need a second opinion on how to demonstrate this: Being $a,b\in\Bbb Z$, If $a^2|b^3$, then $a|b$
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2I am voting to reopen this question because $2\ne3$ but the question that duplicates this one says $a^n|b^n$ which have the same power. – Kamal Saleh Feb 28 '23 at 02:04
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2The second question that duplicates this one obviously doesn't. There is no mention of irrationality in the OP. – Kamal Saleh Feb 28 '23 at 02:08
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Welcome to Mathematics Stack Exchange. I agree with the above comments. Unlike the proposed duplicate, the claim in this question is false; e.g., $16^2\mid8^3,$ but $16\not\mid8$ – J. W. Tanner Feb 28 '23 at 02:19
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@KamalSaleh Please be patient. The dupe list was being edited and niw has exact dupes. Please retract your vote and delete the comment (so as now to mislead voters) – Bill Dubuque Feb 28 '23 at 02:23
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@J.W.Tanner Please be patient. The dupe list was being edited and niw has exact dupes. Please retract your vote and delete the comment (so as now to mislead voters) – Bill Dubuque Feb 28 '23 at 02:24
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Please check for possible typos in your exponents: with $b^2$ it is true and with $b^3$ it is false. I've added dupe links for both cases (there are many more too). – Bill Dubuque Feb 28 '23 at 02:39
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1I agree that one of the duplicates on the edited list is now a genuine duplicate, so I retracted my reopen vote – J. W. Tanner Feb 28 '23 at 03:18