<s>[INST] How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone out there will have a super smart elegant solution to this!
Base step: test when n = 1 ...
Assume true for $n = 1$
ie . $12^k - 4^k - 3^k +1 = 6M$, where $m$ is an integer
RTP: also true for $n = k+1$
ie.  $12^{k+1} - 4^{k+1} - 3^{k+1} +1 = 6N$ where $N$ is an integer
LHS: $12^{k+1} - 4^{k+1} - 3^{k+1} +1$
$= 12( 4^k + 3^k  - 1 + 6M) - 4^{k}(4) - 3^{k}(3) +1$ (from assumption)
$= 6(12M) + 12(4^k) + 12(3^k) -12 -4^{k}(4) - 3^{k}(3) +1$
Here is where I break down and go around in circles.
 [/INST] Work mod 3, and put aside induction for now. Note that  $4 \equiv_3 1$, and so for each integer $n$:
$$4^n \equiv_3 1^n = 1.$$
Thus $12^n -4^n -3^n +1 \equiv_3 (0 -1 -0 -1) = 0$.
So this quantity is divible by 3 for each positive integer $n$.
Lets now get back to mod 6 though. So $12^n -4^n -3^n +1$ is divisible by 3. But it is also even so it is divisible by 6 as well.
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