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Math Help - Primes and Squares again

  1. #1
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    Primes and Squares again

    Q: Show that if a, b are positive integers such at (a,b)=1 and ab is a square, then a and b are also squares.

    Pf: (by Contradiction)
    Suppose a and b are not squares, then a=h*k and b=p*q where h,k,p,q are all unique primes. (Is this the right approach to this proof?)
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  2. #2
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    Hi,

    Hint: A number is a perfect square iff all exponents in its prime decomposition are even.

    Use this on ab. Since (a,b)=1, it is clear how prime decompositions of a and of b must look like.
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  3. #3
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    Quote Originally Posted by dlbsd View Post
    Q: Show that if a, b are positive integers such at (a,b)=1 and ab is a square, then a and b are also squares.

    Pf: (by Contradiction)
    Suppose a and b are not squares, then a=h*k and b=p*q where h,k,p,q are all unique primes. (Is this the right approach to this proof?)

    Let ab=x^2
    Let p be a prime, such that p|a
    Now p can't divide b as (a,b)=1
    So, p|x
    \Rightarrow p^2|x^2
    \Rightarrow p^2|ab
    \Rightarrow p^2|a

    Thus power of prime in a(and by similar logic in b) are all even. Thus either is a perfect sqaure as well.

    This is wrong ! Please see below from post by BingK
    Last edited by aman_cc; October 3rd 2009 at 07:31 AM.
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Thus power of prime in a(and by similar logic in b) are all even. Thus either is a perfect sqaure as well.
    Could you explain this to me?
    So p^2|a which means there is an integer h such that a=p^2 * h

    So the power of the prime is 2 but what about h?
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  5. #5
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    Quote Originally Posted by dlbsd View Post
    Could you explain this to me?
    So p^2|a which means there is an integer h such that a=p^2 * h

    So the power of the prime is 2 but what about h?

    This is wrong, please see post below from BingK


    The proof is valid for any p. So in your specific question -
    case 1: h is prime - then be same logic you can show h^2|a. But as that is not the case hence h can't be prime.
    so h is composite - then also if q is a prime such that q|h then you can show that q^2|a. Infact h itself is a perfect sqaure - as it should be

    The idea of the proof is that for "any" prime , p, which divides a, p^2 also divides a. So in the prime- factorization of a all the powers will be even.
    Logic for a can be replicated for b. Thus the statement will be true for ab as well. If all powers of prime are even then the number is a perfect square. Using this fact we conclude ab is a perfect square.

    Hope it helps.
    Last edited by aman_cc; October 3rd 2009 at 07:39 AM.
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  6. #6
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    I think there's a bit of a problem with your logic that if p^2|a then all the powers in the prime factorization of a will be even. If you think about it p^2|p^3, and 3 is not even ...

    But if you use Taluivren's hint, you can take the prime decomposition of ab, and all the powers have to be even. Since (a,b)=1, we can rearrange the prime decomposition of ab so it is written as the prime decomposition of a multiplied by the prime decomposition of b. Then, we look at the respective prime decompositions and notice that the powers are all even. Again, by the hint, we have that a and b are also perfect squares.
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  7. #7
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    Quote Originally Posted by Bingk View Post
    I think there's a bit of a problem with your logic that if p^2|a then all the powers in the prime factorization of a will be even. If you think about it p^2|p^3, and 3 is not even ...

    But if you use Taluivren's hint, you can take the prime decomposition of ab, and all the powers have to be even. Since (a,b)=1, we can rearrange the prime decomposition of ab so it is written as the prime decomposition of a multiplied by the prime decomposition of b. Then, we look at the respective prime decompositions and notice that the powers are all even. Again, by the hint, we have that a and b are also perfect squares.
    @Bingk - I get what you are saying. Yes - it is incorrect. Thanks for pointing that out.
    Last edited by aman_cc; October 3rd 2009 at 07:29 AM.
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