1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook