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Proof about an integer being a perfect square.

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    m and n are positive integers with m,n≥2
    where [itex] m^2=kn^2 [/itex]
    3. The attempt at a solution
    we know that all prime factors of m have an even amount , their are no prime factors that
    are repeated an odd number of times. The same goes for n.
    if k is not a perfect square then it will have an odd number of prime factors.
    and then when we multiply in to n it will still be an odd number, but
    m has only an even number, so k has to the square of an integer so both sides have even number of each prime factor.
     
  2. jcsd
  3. Jan 11, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    6=2*3 has an even number of prime factors.
    But it has at least one prime factor where the exponent is odd.

    What do you want to prove? That k has to be a perfect square? You can do it that way.

    If you can use real numbers in your proof: ##a^2=k## with rational a and an integer k has a solution only if k is a perfect square.
     
  4. Jan 11, 2013 #3
    what I meant to say is that there would exists a prime factor
    that has at least an odd exponent
     
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