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: Hint:
If $ f(n)=3^n+7^n+2,$
find $$f(m+1)-7f(m)=?$$
Clearly, $f(m+1)\equiv f(m)\pmod{12}$ for $m\ge1$
Now establish the base case i.e., for $f(1)$
A: Here's another hint for the core part of your inductive proof (I am sure there are more elegant ways, but the following seems to work fine):
$$
\begin{align}
3^{k+1}+7^{k+1}+2&= 7(3^k+7^k+2)-12-4\cdot3^k & \text{(rewrite)}\\[0.5em]
&= 7(12\ell)-12-12\cdot3^{k-1} & \text{(by ind. hyp.)}\\[0.5em]
&= 12(7\ell)-12(1+3^{k-1}) & \text{(factor out 12)}\\[0.5em]
&= 12(7\ell-1-3^{k-1}) & \text{(factor out 12 again)}\\[0.5em]
&= 12\eta. & \text{(desired result)}
\end{align}
$$
A: With induction:
When $n=1$: $3^1+7^1+2=12$ is divisible by $12$. Therefore, we assume that $(3^k+7^k+2)$ is divisible by $12$ for some $k≥1$ (thus, $3^k+7^k+2=12a$ where $a$ is a function of $3$ and $7$).
Then, $$3^{k+1}+7^{k+1}+2=3×3^k+7×7^k+2$$
$$=3×3^k+7^k(3+4)+6-6+2$$
$$=3×3^k+3×7^k+6+4×7^k-4$$
$$=3(3^k+7^k+2)+4(7^k-1)$$
$$=3×(12a)+4(7-1)(7^{k-1}+7^{k-2}+⋯+2+1)$$
$$=36a+24(7^{k-1}+7^{k-2}+⋯+2+1)$$
$$=12b$$ where $b$ is also a function of $3$ and $7$.
Thus, $(3^{k+1}+7^{k+1}+2)$ is also divisible by $12$.
A: 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 $n$.
A: Let $x_n = 3^n+7^n+2$ and $y_n=x_n-2= 3^n+7^n$.
Then $y_{n+2}= 10 y_{n+1} -21 y_n$ and so $x_{n+2}= 10 x_{n+1} -21 x_n +24$.
Since $x_1=12$ and $x_2=60$ are multiples of $12$, so is $x_n$ by induction.
