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 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? 

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 $11$.
