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Relatively prime proof involving a^n and b^n

  1. Jan 24, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Show that if a, b, n, m are Natural Numbers such that a and b are relatively prime, then a^n and b^n are relatively prime.

    2. Relevant equations
    Relatively prime means 1 = am + bn where a and b are relatively prime. gcd(a,b) = 1

    We have a couple corollaries that may be beneficial:
    1. Suppose a, n are positive integers. Then a and a^n have the same prime factors.
    2. Let a, n be positive integers. a^(1/n) is either an integer or it's irrational.

    3. The attempt at a solution

    By definition of relatively prime, 1 = ax + by where x,y exist as integers.
    By the fundamental theorem o arithmetic:
    a = p1*p2*...*pm where this is the unique prime factorization.
    b = p1*p2*...*pd
    m,d are natural numbers.

    By the corollarly, a^n = p1^n*p2^n*...*pm^n
    b^n = p1^n*p2^n*...*pd^n

    So we start with
    1 = ax + by
    1 = (p1*p2*...*pm)*x + (p1*p2*...*pd)y


    ****All I have left to do is find the way to raise this to the power of n. If I can raise it such that
    1^n = (ax)^n + (by)^n the proof is done, easily, with what I have set up.

    Question: Is this step needed? Can I just use the corollary to say that "by the corollary" since a and a^n have the same prime factorization and b and b^n have the same prime factorization, then we see a^n and b^n must be relatively prime?
     
  2. jcsd
  3. Jan 24, 2016 #2

    Samy_A

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    Yes to the last question. I was wondering why this looks so complicated.
    Can there be a prime number that divides both ##a^n## and ##b^n##, when a and b are relatively prime?
     
  4. Jan 24, 2016 #3

    RJLiberator

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    Since a and a^n and b and b^n have the same prime factorizations, and a,b are relatively prime, then a^n,b^n are relatively prime.

    No.
    Since they are relatively prime, that means 1 = ax+bm and by this proof, 1 = a^n*x+b^n*m
     
  5. Jan 24, 2016 #4

    Samy_A

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    Correct.
    Hmm, A little sloppy.
    The "x" and "m" in the RHS don't have to be the same as the "x" and "m" in the LHS.

    Example: 5*2-3*3=1, but 5²*2-3²*3≠1. (But -5²*5+3²*14=1)
    It doesn't matter, you already solved the exercise.
     
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