Prove using induction $2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$

Question

Show that

$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$

I'm not really sure how to get started on this problem, but here is what I have done so far:

Base case $n(1)$:

$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$

But not sure where to go from here. Tips?

score 4 · Answer 1 · answered Dec 07 '14 at 18:18

4

$$2^{3n}-1 = 8^n-1 = (8-1)(8^{n-1} + 8^{n-2} + \dots + 8^2 +8+1) = 7 \cdot (\mbox{something})$$

answered Dec 07 '14 at 18:18

Crostul

1

The induction is hidden behind the formula for $x^n-y^n$, see e.g. http://math.stackexchange.com/questions/117660/proving-xn-yn-x-yxn-1-xn-2-y-x-yn-2-yn-1 – Surb Dec 07 '14 at 18:25

barak manos · Accepted Answer · 2014-12-07T18:43:32.957

4

For simplicity, please note that $2^{3n}=8^n$.

First, show that this is true for $n=1$:

Second, assume that this is true for $n$:

Third, prove that this is true for $n+1$:

edited Dec 07 '14 at 18:43

answered Dec 07 '14 at 18:37

barak manos

How did you get the $7k$ here $\frac{8\cdot7k+7}{7}$. It seems like it should be $\frac{8\cdot k+7}{7}$ – nullByteMe Dec 07 '14 at 21:15
@inquisitor: $\frac{8^n-1}{7}=k\implies8^n-1=7k$. – barak manos Dec 07 '14 at 21:29
Got it, thank you for showing me that. – nullByteMe Dec 07 '14 at 21:32
@inquisitor: You're welcome. – barak manos Dec 07 '14 at 21:32

Josh B. · Answer 3 · 2014-12-07T18:49:33.567

1

Assume $2^{3n}-1$ is divisible by 7. Look at $2^{3(n+1)}-1$ and show that it is divisible by 7.

$2^{3(n+1)}-1 = 2^{3n+3}-1 = 8*2^{3n}-1 = (7+1)*2^{3n}-1 = (7*2^{3n} + 2^{3n})-1 = 7*2^{3n} + (2^{3n}-1)$ which is divisble by 7.

edited Dec 07 '14 at 18:49

answered Dec 07 '14 at 18:40

Josh B.

I don't under your last step where you got the $7 \cdot 2^{3n}+(2^{3n}-1)$ – nullByteMe Dec 07 '14 at 18:45
2

@ inquisitor: Let $x=2^{3n}$, then $8x-1=(7x+x)-1=7x+(x-1)$. – Math-fun Dec 07 '14 at 18:48
Got it! Thanks for that – nullByteMe Dec 07 '14 at 18:49
Edited in extra steps for clarity. – Josh B. Dec 07 '14 at 18:50

score 0 · Answer 4 · answered Dec 07 '14 at 18:48

0

$2^{3n} = 8^{n}$. Let $s_{n} = 8^{n} - 1$.

$$8^{n + 1} - 1 = 8(8^{n}) - 1 = 8(8^{n} - 8) + 7 = 8(8^{n} - 1) + 7 = 8s_{n} + 7$$

answered Dec 07 '14 at 18:48

AJY

4 Answers4