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Math Help - Square free problem 2

  1. #1
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    Square free problem 2

    Prove that every integer n > 1 is the product of a square-free integer and a perfect square.

    proof.

    Should I start by saying n = a product of primes by the fundamental theorem? But I can't seem to get anywhere after that.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Prove that every integer n > 1 is the product of a square-free integer and a perfect square.

    proof.

    Should I start by saying n = a product of primes by the fundamental theorem? But I can't seem to get anywhere after that.
    Let us start easy. Suppose p,q,r are prime and n=p^2q^3 r. Then we can write n = (p^2 q^2) \cdot (qr) now p^2 q^2 = (pq)^2 is a square and qr is square free. Another example, n=p^2 q^4 s^3 r^5 = (p^2 q^4 s^2 r^4)(sr) = (pq^2sr^2)^2 (sr). Get the idea?
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  3. #3
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    I don't understand that you can set <br />
n=p^2q^3 r<br />
, what allows an integer to be a product of two or more of the same prime number?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    I don't understand that you can set <br />
n=p^2q^3 r<br />
, what allows an integer to be a product of two or more of the same prime number?
    Because I am trying to show you an example. I am saying suppose that n has that specific type of factorization, then we can do what I did above. Those were just examples. There is a general rule that we use, that rule is what you have to find and apply it to the general case.
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