<s>[INST] 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... 
 [/INST] $(3^{k+1}+7^{k+1}-2)=3^k+7^k-2+2\cdot3^k+6\cdot7^k$
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$(8)$.
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