If n is positive integer, prove that the prime factorization of $2^{2n}\times 3^n - 1$ contains $11$ as one of the prime factors

Question

I have: $2^{2n} \cdot 3^{n} - 1 = (2^2 \cdot 3)^n - 1 = 12^n - 1$. I know every positive integer is a product of primes, so that,

$$12^n - 1 = p_1 \cdot p_2 \cdot \dots \cdot p_r. $$ Also, any idea how to use math notation on this website?

Answer 1

We have to prove $11|12^n-1$. This follows easily from $$12^n\equiv 1^n=1\ (\ mod\ 11\ )$$

Answer 2

First, show that this is true for $n=1$:

$2^{2}\cdot3^{1}-1=11$

Second, assume that this is true for $n$:

$2^{2n}\cdot3^{n}-1=11k$

Third, prove that this is true for $n+1$:

$2^{2(n+1)}\cdot3^{n+1}-1=$

$\color\red{2^{2n}\cdot3^{n}-1}+11\cdot2^{2n}\cdot3^{n}=$

$\color\red{11k}+11\cdot2^{2n}\cdot3^{n}=$

$11(k+2^{2n}\cdot3^{n})$

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Please note that the assumption is used only in the part marked red.

Answer 3

By the Binomial theorem,

$$12^n-1=(11+1)^n-1\\ =11^n+n\,11^{n-1}+\frac{n(n-1)}211^{n-2}+\cdots+\frac{n(n-1)}211^2+n\,11+1-1,$$

which is a multiple of $11$.