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!
 A: Note that by hypothesis $$8|3^n+7^n-2$$ then $8|3^{n+1}+7^{n+1}-2$ if and only if also $$8|(3^{n+1}+7^{n+1}-2)-(3^{n}+7^{n}-2)=3^n(3-1)+7^n(7-1)=\underbrace{2\cdot 3^n+2\cdot 7^n-4}_{\mathrm{divisible\ by\ 8}}+\underbrace{4+4\cdot 7^n}_{\mathrm{divisible\ by\ 8}}.$$
A: 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
$$3\cdot3^n+7\cdot7^n-2-3 \cdot \left(3^n + 7^n -2\right)=3\cdot3^n+7\cdot7^n-2-3\cdot3^n-3\cdot7^n-6\\
=7^n(7-3)+4$$
A: By direct induction following the initial step:
$$3^{n+1}+7^{n+1}-2=7(3^n+7^n-2)-\left(4\cdot3^n-12\right)$$
But the first summand in the right expression above is divisible by eight by ind. hypothesis, and the second one is
$$12(3^{n-1}-1)=12(3-1)(3^{n-2}+3^{n-3}+\ldots+3+1)$$
and this is also a multiple of eight (and observe the expression is correct for $\;n\ge 2$)
A: Let $\displaystyle f(n)=3^n+7^n-2$
$\displaystyle\implies f(m+1)-f(m)=3^{m+1}+7^{m+1}-2\left(3^m+7^m-2\right)$
$\displaystyle=2\cdot3^m+6\cdot7^m=8\cdot3^m+6(7^m-3^m)$
Now we know $\displaystyle7^m-3^m$ is divisible by $7-3=4$ (Proof)
$\displaystyle\implies f(m+1)-f(m)$ is divisible by $8$
$\displaystyle\implies  8$ will divide $f(m+1)$ if $8$ divides $f(m)$
Now show that the base case i.e. for $n=1$ holds true
A: Hints:
$$3^{2n}=9^n\equiv 1 \; (\mathrm{mod}\; 8)$$
$$7^{2n}=49^n\equiv 1 \; (\mathrm{mod}\; 8)$$
and
$$3^{2n+1}=3\cdot 9^n\equiv 3 \; (\mathrm{mod}\; 8)$$
$$7^{2n+1}=7\cdot 49^n\equiv 7 \; (\mathrm{mod}\; 8)$$
