<s>[INST] Divisibility induction proof: $8\mid 7^n+3^n-2$ I'm stuck on the following proof by induction: $$8\mid3^n +7^n -2$$
And this is how far I've gotten:
$$\begin{aligned}3&\cdot3^n+7\cdot7^n-2\\3&(3^n+7^n-2)+7^n(7-3)-2\end{aligned}$$
Any help on where to go after this would be great!
 [/INST] You did a mistake:
$$3\cdot3^n+7\cdot7^n-2=3(3^n+7^n-2)+7^n(7-3)+4$$
Bow by $P(n)$ you know that $8|3(3^n+7^n-2)$. Thus, to complete the problem, you have to prove that
$$8|7^n\cdot 4+4$$
you can do this either by induction or by simply observing that $7^n+1$ is even, thus $4(7^n+1)$ is a multiple of $8$.
Added $P(n+1)$ contains $3\cdot3^n+7\cdot7^n-2$. Our goal is to create $3(3^n+7^n-2)$.
So, we start by adding and subtracting the missing terms:
$$3\cdot3^n+7\cdot7^n-2 =3 \cdot3^n +\left( 3 \cdot 7^n +3 \cdot (-2)- 3 \cdot 7^n -3 \cdot (-2) \right)+7\cdot7^n-2\\
=3 \left(\cdot3^n + 7^n -2\right)- 3 \cdot 7^n +6 +7\cdot7^n-2\\
=3 \left(\cdot3^n + 7^n -2\right)+7\cdot7^n- 3 \cdot 7^n +6 -2$$
An alternate way to figure this is the following:
You have
$$A=3\cdot3^n+7\cdot7^n-2$$
and need
$$B=3 \cdot \left(3^n + 7^n -2\right)$$
Then you can write 
$$A=B+(A-B) \,.$$
Thus, what you need to add to $3 \cdot \left(3^n + 7^n -2\right)$ is
$$3\cdot3^n+7\cdot7^n-2-3 \cdot \left(3^n + 7^n -2\right)=3\cdot3^n+7\cdot7^n-2-3\cdot3^n-3\cdot7^n-6\\
=7^n(7-3)+4$$
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