<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] 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?
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