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How do you prove that n must divide m if n is square-free and n divides m squared?
Wiki User
∙ 15y ago
Consider m, and its set of prime factors. e.g. m = 130 = 2 x 5 x 13 then m2 = (2 x 5 x 13)(2 x 5 x 13) n is a factor of m2. It must therefore be formed by the multiplication of a subset of the prime factors of m2. If this subset is contained in just one pair of brackets (see above) then it is contained in the prime factorisation of m, and so n is definitely a factor of m. [For instance, 2 x 5 = 10 is a factor of 1302 and 130. (2 x 5 x 13)(2 x 5 x 13)] If this subset spans both pairs of brackets, then n is a factor of m2 but not of m. If this is the case, however, n contains a repeated factor, and is not 'square-free'. [For instance, 2 x 2 x 5 = 20 is a factor of 1302 but not 130. (2 x 5 x 13)(2 x 5 x 13).But being divisible by 4, it is not square free.] To summarise: for a number to be divisible by m2 but not by m, it is necessarythat it has a repeated prime factor. For n to divide m, it is therefore sufficient (but not necessary) that n divides m2 and is square free.
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∙ 15y ago
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