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Via inductive proof show that $3^{2n}-1$ is divisible by 8 for all natural numbers n

The reason why that I was confused in this problem was because my steps has gotten me nowhere useful as shown below:

(I've omitted the words for the inductive proof for the sake of simplicity)

=$3^{2(k+1)}-1$

$=9-3^{2k}-1$

$=8-9*3^k$

But this has gotten me nowhere and I was wondering if there was any other way to prove this with inductive proof

John Rawls
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  • You probably shouldn't omit the words. :-) As things stand your 'proof' looks very confusing, as I have no idea what '$9-3^{2k}-1$' is supposed to represent. – Steven Stadnicki May 21 '17 at 23:04
  • How did you get $3^{2(k+1)} -1= 9 - 3^{2k}-1$? You should have gotten $3^{2(k+1)} -1= 3^{2k + 2}-1 = 3^{2k}3^2 - 1 = 9*3^{2k} - 1$. – fleablood May 21 '17 at 23:08
  • ... and from there $3^{2(k+1)} -1 = 93^{2k} - 1 = 83^{2k} + (3^{2k} - 1)$. As we are assuming $8|3^{2k} -1$ and we know that $8| 83^{2k}$ we can conclude that if $8|3^{2k} -1$ then $8|3^{2k}-1 + 83^{2k} = 3^{2(k+1) } -1$. – fleablood May 21 '17 at 23:11
  • @StevenStadnicki I think it is very clear that $9- 3^{2k} -1$ represents the number nine minus 3 to the 2k power minus 8 (ex. if $k = 2$ is is the number $9- 81 - 1 = -73$. The only problem is that it is wrong. $3^{2(k+1)} -1 \ne 9 - 3^{2k} -1$ and $k = 2$ would have $3^{2(k+1)} -1 = 3^6 - 1=728 \ne -73.$ – fleablood May 21 '17 at 23:15

2 Answers2

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Since the title asks for a proof by induction I will provide one. For a non-inductive proof see user8012403's answer.

For base case, $n = 0$, we simply note $3^{2\cdot0}-1 = 0$ which is a multiple of 8.

For the inductive step, suppose that $8|3^{2n}-1$. Then note that $$3^{2n+2}-1 = 9\cdot3^{2n}-1=(3^{2n}-1)+(8\cdot3^{2n}).$$

Both terms on the right are multiples of 8 and thus their sum is too; hence we're done.

fractal1729
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Let $P(n)$ be the proposition.

When $n=1$, we have $3^2 -1=8$, thus $P(1)$ is true. Assume that $P(k)$ is true for some $k\in \mathbb{Z}^{+}$, i.e. $\exists k\in \mathbb{Z}^{+}, 3^{2k}-1\equiv 0 \mod 8\iff 3^{2k}\equiv 1 \mod 8$

now, when $n=k+1$,

$3^{2k+2}-1=9*3^{2k}-1$

$\equiv 9*1-1 \mod 8$

$\equiv 0 \mod 8$

By the principle of mathematical induction, $P(n)$ is true for $\forall n\in \mathbb{Z}^{+}$

Thanks Steven.

BAI
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