Proving that $3^n+7^n+2$ is divisible by 12 for all $n\in\mathbb{N}$. Can someone help me prove this? :( I have tried it multiple times but still cannot get to the answer.
Prove by mathematical induction for $n$ an element of all positive integers that   $3^n+7^n+2$ is divisible by 12.


A general rule of thumb for induction is to try to get
$3^{n+1} + 7^{n+1}+2 = somethingtodo with(3^n + 7^n+2) = something to do with(multiple of 12) = multiple of 12$
So $3^{n+1} + 7^{n+1} +2 = $
$3*3^n + 7*7^n + 2=$
$3*3^n + 3*7n + 4*7^n + 2= $
$3(3^n + 7^n) + 4*7^n + 2 = $
$3(3^n + 7^n + 2) - 6 + 4*7^n + 2 = $
$3(multiple of 12) + 4*7^n -4  $
so now it's a matter of proving $4*7^n -4$ is a multiple of $12$.
$3(multiple of 12) + 4*7^n -4 = $
$3(multiple of 12) + 4(7^n - 1)=$
$3(multiple of 12) + 4*(7-1)(7^{n-1} + .... + 7 + 1)=$
$3(multiple of 12) + 4*6(7^{n-1} + .... + 7 + 1)=$
$3(multiple of 12) + 24*(7^{n-1} + .... + 7 + 1)=$
$3(multiple of 12) + (multiple of 12)*(7^{n-1} + .... + 7 + 1)=$
$multiple of 12$
....
So to formally put this together.
Base case: 
$n = 1$.  $3^1 + 7^1 + 2 =12$ is a multiple of $12$.
Inductive case: 
Assume $3^n + 7^n + 2 = 12K$ is a multiple of $12$..
Then $3^{n+1}+7^{n+1} + 2 =$
$3*3^n + 3*7^n+ 6  + 4*7^n-4 =$
$3(3^n + 7^n + 2) + 4(7^n - 1)=$
$3(12K) + 4(7-1)(7^{n-1} + ... + 1)$  
Let $M = (7^{n-1} + ... + 1)$. It's worth noting that if $n = 1$ then $7^{n-1} + ..... + 1 = 1$.  It is important in a proof by induction to not make any assumptions that are not provable for the base $n=1$ case.  In this case that $7^{n-1} + .... +1$ actually exists.)
$3(12K) + 4(7-1)(7^{n-1} + ... + 1)=$
$3(12K) + 4*6*M = $
$12[3K + 2M]$
is a multiple of $12$.
Conclusion: $3^n + 7^n + 2$ is a multiple of $12$ for all natural $n$.
