Basic Induction Can someone give me a hint on how to solve this?
Prove that $3^{(3n+4)} + 7^{(2n+1)}$ is divisible by 11 for all natural numbers n.
So far I've got
Base P(1): $3^{(3(1)+4)} + 7^{(2(1)+1)} = 2530$, which is divisible by 11 thus true
P(k): $3^{(3k+4)} + 7^{(2k+1)} = 11a$ for which $a$  $\in \mathbb{N}$ 
P(k+1): $3^{(3k+7)} + 7^{(2k+3)} = 11b$ for which $b$  $\in \mathbb{N}$ 
But i'm not sure how to continue from here in terms of breaking it down and proving the divisibility because of having 2 different indices. Appreciate any advice. Thank you.
 A: Your $P(k+1)$ is what you hope to be true, but you should avoid writing it as though it is fact.
What we do know is that $P(k)$ is true for some $k\geq 1$, and we hope to show that $P(k+1)$ follows.
You have $3^{(3k+4)} + 7^{(2k+1)} = 11a$ for some $a\in\mathbb{N}$
So, $3^{(3(k+1)+4)}+7^{(2(k+1)+1)} = 3^{3k+7} + 7^{2k+3} = 3^3\cdot 3^{3k+4} + 7^2\cdot 7^{2k+1}$
Now, notice that $3^3 = 27 = 2\cdot 11 + 5$
and that $7^2 = 49 = 4\cdot 11 + 5$
The result should follow soon after.

$$=(2\cdot 11 + 5)3^{3k+4} + (4\cdot 11+5)7^{2k+1} \\= 11(2\cdot 3^{3k+4}+4\cdot 7^{2k+1}) + 5\cdot (3^{3k+4}+7^{2k+1})\\=\dots$$

A: HINT : 
$$\begin{align}3^{3k+7}+7^{2k+3}&=27\cdot 3^{3k+4}+49\cdot 7^{2k+1}\\&=27\cdot 3^{3k+4}+(27+22)\cdot 7^{2k+1}\end{align}$$
A: P(k+1):$3^{3k+7} + 7^{2k+3} = 27 \cdot 3^{3k+4} + 49\cdot 7^{2k+1}\tag{1}$ 
From P(k): $3^{3k+4} = \color{blue}{11a - 7^{2k+1}}$
So $(1)$ becomes 
$27(\color{blue}{11a - 7^{2k+1}})+49\cdot 7^{2k+1}=27\cdot 11a -27\cdot 7^{2k+1}+ 49\cdot 7^{2k+1}=27\cdot 11a +2\cdot 11\cdot 7^{2k+1}=\color{red}{11}(27a+2\cdot 7^{2k+1})$ 
Hence $3^{3n+4} + 7^{2n+1}$ is a multiple of $\color{red}{11} \space\forall \space n\in \mathbb{N}$ 
QED
A: Modulo $11$ we have $3^{3n+4}+7^{2n+1}\equiv 3^{3n}.3^4+7^{2n}.7\equiv 3^{3n}.81+7^{2n}.7\equiv 3^{3n}.4+7^{2n}.(-4)\equiv 4(3^{3n}-7^{2n})\equiv 4( 27^n-49^n)\equiv 4(5^n-5^n)=0.$
