Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$. 
$(3^{k+1}+7^{k+1}-2)$
$(3^k*3+7^k*7-2)$
But now I'm stuck... 
 A: Notice
\begin{align}
3^k\cdot 3+7^k\cdot 7-2&=3^k\cdot(3+4)-3^k\cdot 4+7^k\cdot 7-2\cdot 7 +2\cdot 7-2\\
&=7(3^k+7^k-2)-3^k\cdot 4+2\cdot 7-2\\
&=7(3^k+7^k-2)-3^k\cdot 4+12\\
&=7(3^k+7^k-2)-12\cdot(3^{k-1}-1)\\
\end{align}
by hypothesis $8$ divides $3^k+7^k-2$, and for $k\ge 1$ the number $3^{k-1}-1$ is even, then $8$ divides $12\cdot(3^{k-1}-1)$. Therefore  $8$ divides $3^k\cdot 3+7^k\cdot 7-2.$
A: We have $3^{k+2}=9\cdot 3^k=3^k+8\cdot 3^k$ and $7^{k+2}=7^k+(8)(6)7^k$. Make separate arguments for $c$ odd and $c$ even, using base cases $c=1$ and $c=2$ respectively. 
Or else (essentially the same idea) use strong induction.
A: By the induction hypothesis,
$$
3^k+7^k-2=8m
$$
for some integer $m$. Then $3^k=8m+2-7^k$ and so
\begin{align}
3^{k+1}+7^{k+1}-2
&=3\cdot 3^k+7^{k+1}-2\\
&=3(8m+2-7^k)+7\cdot7^{k}-2\\
&=24m+6-3\cdot7^k+7\cdot7^{k}-2\\
&=4(6m+1+7^k)
\end{align}
Can you finish up?
A: $(3^{k+1}+7^{k+1}-2)=3^k+7^k-2+2\cdot3^k+6\cdot7^k$
So, it is enough to show that $8$ divides $2\cdot3^k+6\cdot7^k=6\cdot3^{k-1}+6\cdot7^k$
Also, $3^{odd}=3$mod$(8)$ ,  $3^{even}=-1$mod$(8)$
$7^{odd}=-1$mod$(8)$, $7^{even}=1$mod$(8)$.
implies $6\cdot3^{k-1}+6\cdot7^k=0$mod$(8)$.
