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Algebra proof involving perfect squares

  1. Sep 10, 2010 #1
    The problem states, if c^2 = ab and (a,b) = 1, prove that a and b are perfect squares.
    ( the notation (a,b) means the GCD of a and b)

    So i have a lot of thoughts on this problem but i am getting stuck.

    1) if c^2 = ab then c(c) = ab which says c divides ab.

    2) the only real theorem available about the GCD of two numbers we have is that it can be written as a linear combination. so ax + by = 1 for some integers x and y.

    3) however this problem is in the section about prime numbers and prime factorizations so i started thinking about perfect squares prime factorization

    4 = 2^2
    9 = 3^2
    16 = 2^5 = 2^2(2^2)
    25 = 5^2
    36 = 2^2(3^2)
    49 = 7^2
    64 = 2^6 = 2^2(2^2)(2^2)
    81 = 3^4 = 3^2(3^2)
    and so on. so it seems this this should help me somehow. But i am not sure where to go with this or how i am going to use that fact that (a,b) = 1, so they are relatively prime.

    Well wait if they are relatively prime, they couldn't be 2 perfect squares like 36 and 81 or else their GCD would not be 1 it would 3^2 or 9.
    But a and b could be numbers like 25 and 16.

    Anyone can help me finish please? A lot of these ideas came up as i type this.
  2. jcsd
  3. Sep 10, 2010 #2
    Factorization seems a good way to go. I would try to find out how often a prime factor of c has to appear in the product of ab.
    Then the GCD should tell you were this factor can come from in ab.

    What's wrong with 25 and 16? They are a solution for c = 20.
  4. Sep 10, 2010 #3
    I am still a bit confused. So just as an example if a were 25 and b were 16 then thats fine because (a,b) = 1.

    so the c = 20 we have (20)^2 = 25(16) and so we have 400 = 400.
    20 = 2^2(5).

    so now i am starting to see something. c has the same prime factorization has a*b. I am having some trouble putting this all together though.

    Is there anymore help i could get?
  5. Sep 10, 2010 #4
    Do it the other way round to get a feeling what you have to do.
    E.g. try to find some a,b for c=15.
  6. Sep 10, 2010 #5
    write ab as a product of distinct prime powers, they must all be even powers if ab = c^2.

    Now if gcd(a,b)=1 then a and b are coprime, so can't contain any prime factors in common...
  7. Sep 10, 2010 #6
    Let me start with C.
    c = p1^k1(p2^k2)...pn^kn.

    c^2 = p1^2k1(p2^2k2)...pn^2kn

    and we know (a,b) = 1 are relatively prime so they do not share any of the same p sub i right?
    But i can definitely say that the prime factorization of c^2 better be the same the prime factorization of ab right?

    a = (p1^2k1)(p2^2k2)....(pr^2kr)
    b = (pr+1^2kr+1)(pr+2^2k+2).....(pn^2kn)

    aha so then a and b are perfect squares. this is right?

    does this work?
    Last edited: Sep 10, 2010
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