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Here's another howler some people commit: If m, n are integers such that m divides n^2 then m divides n. It's true sometimes, for example if m is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers p, q, r with p squarefree such that $ m = p q^2 $ and $ n = p q r $ The usual counterexample is that 8 divides 4^2 but not 4 ;-) |
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1 | [made Community Wiki] | ||
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Here's another howler some people commit: If m, n are integers such that m divides n^2 then m divides n. It's true sometimes, for example if m is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers p, q, r with p squarefree such that $ m = p q^2 $ and $ n = p q r $ |
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