Proof that $3^c + 7^c - 2$ by induction

Question

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...

Answer 1

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

Answer 2

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.

Answer 3

$(3^{k+1}+7^{k+1}-2)=3^k+7^k-2+2\cdot3^k+6\cdot7^k$

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$(8)$.

Answer 4

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?