Proof of prime factorization of an algebraic expression.

1. Feb 9, 2013

jcoughlin

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. Feb 9, 2013

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.

3. Feb 9, 2013

jcoughlin

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?

4. Feb 9, 2013

Staff: Mentor

As prime factorization is unique, yes.