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Proof of prime factorization of an algebraic expression.

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Claim: If n is a positive integer, the prime factorization of 22n * 3n - 1 includes 11 as one of the prime factors.

    2. Relevant equations
    Factor Theorem: a polynomial f(x) has a factor (x-k) iff f(k)=0.


    3. The attempt at a solution

    First, we show that (x-1) is a factor of (xn-1). Let f(x)=xn-1, and k=1; then f(k)=0, and thus by the factor theorem (x-1) is a factor of (xn-1).

    Next, consider 22n * 3n - 1 rewritten as 12n-1. As previously demonstrated, x-1 is a factor of xn-1. Letting x=12, we see that (12-1)=11 is a factor of 12n-1 for n>0.

    Is this sufficient? Or do I need to go further than proving 11|(12n-1) to show that the prime factorization of 22n * 3n - 1 includes 11 as one of the prime factors?

    Thanks,
    James
     
    Last edited: Feb 9, 2013
  2. jcsd
  3. Feb 9, 2013 #2

    mfb

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    Staff: Mentor

    I would add the statement that your number is 12^n - 1 somewhere.
    If your number has 11 as a factor, this is one of the prime factors - it should be obvious that 11 is a prime, but you can write it down as well.
     
  4. Feb 9, 2013 #3
    Ah sorry, had a typo in the line that established 12^n-1 = 2^2n * 3^n - 1 .

    Is showing 11 to be a factor, which happens to be prime, the same thing as showing that the prime factorization includes 11?
     
  5. Feb 9, 2013 #4

    mfb

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    Staff: Mentor

    As prime factorization is unique, yes.
     
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