Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that 
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.

Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (4^k)(4) -1$
$ = (12^k)(12) - (7^k)(3+4) - (4^k)(3+1) -1$

I'm not really sure about the last step, as someone just told me to do it. Am I supposed to find the right addends to use and then distribute the exponent terms until I get a multiple of the original $a_{k}$? Because I can't get it to work out evenly, and the -1 at the end gives me trouble. Also, I know that $12^n$ is a multiple of three already, but I don't know how to implement that fact to my advantage. Can I prove that $7^{n}-4^{n}-1$ is also a multiple of three and go from there?

Since $(x-12)(x-7)(x-4)(x-1)=x^4-24x^3+183x^2-496x+336$, we have that
$a_n=12^n-7^n-4^n-1$ satisfies
$$
a_n=24a_{n-1}-183a_{n-2}+496a_{n-3}-336a_{n-4}\tag{1}
$$
Since the first $4$ values
$$
a_1=0,a_2=78,a_3=1320,a_4=18078\tag{2}
$$
are all divisible by $6$, induction with $(1)$ and $(2)$ insure that $6\mid a_n$ for all $n\ge1$.
