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Thread: Help with proof

  1. #1
    Aug 2010

    Help with proof

    For my homework one of the problems is the

    Proposition: For all integers a and n, if a divides n^2 and a<=n then a divides n.

    It either has to be proved or a counterexample given and I can't find a counterexample so I'm trying to prove it. I know that n^2 has to be positive but that isn't the case for a and n. n^2 would be equal to a times some integer so n^2 = a*x so n would be the square root of that. But in order to prove its divisible you have to prove that it is a times some integer and I cant prove the square root of a*x is an integer so I guess I just need some help with what I should do next. I haven't been doing proofs for long. Thanks for any help.
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  2. #2
    MHF Contributor
    Oct 2009
    If a can be negative, then there is an easy counterexample: a = -n^2 for some concrete n > 1.

    Suppose all numbers involved are positive. Using the Fundamental Theorem of Arithmetic, it helps to think about a number as a multiset of its prime divisors. Then m divides n iff the multiset of prime divisors of m is a subset of the corresponding set for n.

    Suppose that n = p * q for some different prime numbers p and q, then n^2 = p * p * q * q. If a divides n^2 but does not divide n, then a must have two p's or two q's, say, a = p * p. Can you think of an example of p and q such that a <= n?
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