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Every integer can be written as a sum of a square and square free integer

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data

    show that every positive integer can be written as the product of a square and a square-free integer (an integer that is not divisible by any perfect squares other than one


    3. The attempt at a solution

    well i can see by example that this works: i.e 60=22*3*5


    Basically we want to show that any integer n>0 can be written as

    n=p1a1p2a2 ...pkak

    if ai is even, then its a perfect square
    if ai is odd, then piai-1 is a perfect square and pi is square free.

    I dont know where to go from here... or if what i have said is correct

    any help is appreciated
     
  2. jcsd
  3. Oct 14, 2008 #2

    Dick

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    The p's are the prime factorization, right? Then you are done. Multiply the perfect squares to get one factor and multiply the square free parts to get the other factor.
     
  4. Oct 14, 2008 #3
    Oh really?

    so i am completley done with this problem?

    but how do i deal with numbers like 6 whose prime factorization is 2*3
     
  5. Oct 14, 2008 #4

    Dick

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    2 is 2^1. So that falls into your odd case. 2 is square free and 2^(1-1)=1, which is a perfect square, but not a very interesting one.
     
  6. Oct 14, 2008 #5
    ok thanks feel free to lock.
     
  7. Oct 14, 2008 #6

    HallsofIvy

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    1= 12 is itself a square. 6= 1*6: a square times a square free number.
     
    Last edited: Oct 15, 2008
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