<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] 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 $3^k\cdot 3+7^k\cdot 7-2.$
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