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?

First, show that this is true for $n=1$:
$12^1−7^1−4^1−1=0$
Second, assume that this is true for $n$:
$12^n−7^n−4^n−1=3k$
Third, prove that this is true for $n+1$:
$12^{n+1}−7^{n+1}−4^{n+1}−1=$
$\color\red{12^n−7^n−4^n−1}+11\cdot12^n-6\cdot7^n-3\cdot4^n=$
$\color\red{3k}+11\cdot12^n-6\cdot7^n-3\cdot4^n=$
$3k+11\cdot12\cdot12^{n-1}-6\cdot7^n-3\cdot4^n=$
$3k+132\cdot12^{n-1}-6\cdot7^n-3\cdot4^n=$
$3(k+44\cdot12^{n-1}-2\cdot7^n-1\cdot4^n)$

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