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Proving a and b are squares of an integer

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    If a, b and c are positive integers with (a,b) = 1 and c2 = ab, show that each of a and b is the square of an integer.


    I've been starring at this problem for a while now and I've got not clues. I don't see the relation between a and b being relatively prime and the rest of the problem. Any hints would be appreciated
     
  2. jcsd
  3. Aug 25, 2012 #2
    Can you prove that c does not divide either a or b?

    And I think it should be required that a, b, c > 1.
     
  4. Aug 25, 2012 #3

    uart

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    Science Advisor

    All the prime factors of a square (eg c^2) have an even multiplicity. Relative prime "a" and "b" means that they have no prime factors in common. Does that give you a hint?
     
  5. Aug 25, 2012 #4
    Yes I think so.

    Suppose c divides a and b.
    Then a=c.m and b=c.n it follows that c2=2.m.n and a=b=c. Obviously this cant be since (a,b)=1

    Suppose c divides a but not b.
    Then c2=c.m.b and c=m.b. Therefore b|c and c|a implying b|a which clearly can't be since (a,b)=1. This also implies that c can't divide b and not a.

    Is that a sufficient proof?

    Thanks for the hint but I'm still stuck.
    Does it follow from c=(a.b)/c and since c doesn't divide a or b that c must divide the product and some how use that fact.
     
  6. Aug 25, 2012 #5

    jgens

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    Gold Member

    If you can use the fact that the integers are a unique factorization domain, then the proof is pretty much trivial, as uart suggested. Otherwise, consider both the greatest divisor of c that also divides a and the greatest divisor of c that also divides b.
     
  7. Aug 25, 2012 #6
    If c does not divide a and b separately, then c must equal AB, where A divides a and B divides b. What does that imply?
     
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