How do you prove that (2^n * 3^(2n)) -1 is always divisible by 17?

Answer 1

(2^n * 3^(2n)) - 1
= (2^n . 9^n) - 1
= 18^n - 1
≡ 1^n - 1 = 0 (mod 17).

If you're not familiar with modular arithmetic, we can use the binomial theorem:
18^n - 1
= (17 + 1)^n - 1
= 17^n + ... + C(n, k) 17^k 1^(n-k) + ... + n (17) + 1 - 1
= (stuff divisible by 17) + 1 - 1
= (stuff divisible by 17).

More formally we could write (stuff divisible by 17) as 17.[17^(n-1) + ... + C(n, k) 17^(k-1) + ... + n] and then note that since it has 17 as a factor, it is divisible by 17.

Answer 2

(2^n * 3^(2n)) -1 = 2^n * 3^n * 3^n = (2*3*3)^n = 18^n - 1

(18^n - 1) / 17 = (18^n - 1) / (18 - 1)

Which is 18^(n-1) + 18^(n-2) + 18^(n-3) ... + 18^0, which is an integer.