How to prove that $8^{18} - 1$ is divisible by $7$

Question

How to prove that: $$ 8^{18}-1\equiv0\pmod7 $$ In the simplest way?

Use http://math.stackexchange.com/questions/188657/why-an-bn-is-divisible-by-a-b — lab bhattacharjee, Oct 19 '14 at 17:40

Aaron Maroja · Answer 1 · 2014-10-19T18:05:45.720

8

Hint: $$ 8 \equiv 1 \mod 7\ \ \text{and}\ \ a^n \equiv b^n \mod m$$ Given any $a,b \in \mathbb{Z}, \ \ m > 1$ and $n \geq 1$ such that $a \equiv b \mod m$.

edited Oct 19 '14 at 18:05

answered Oct 19 '14 at 17:46

Aaron Maroja

17,571

2

You mean: given any $a$ and $b$ such that $a\equiv b\mod m$. – Did Oct 19 '14 at 18:04
That's right, I typed fast. – Aaron Maroja Oct 19 '14 at 18:06

score 8 · Accepted Answer · edited Oct 19 '14 at 18:06

8

Yet another one: $a^{18}-b^{18}=(a-b)\left(a^{17}+a^{16}b+\cdots+ab^{16}+b^{17}\right) \ ,$ hence $$8^{18}-1=(8-1)\left(8^{17}+8^{16}+\cdots+8+1\right),\quad\text{which is a multiple of $7$.}$$

edited Oct 19 '14 at 18:06

beep-boop

11,595

answered Oct 19 '14 at 17:51

Did

279,727

Reptile · Answer 3 · 2014-10-19T18:00:29.550

0

Ok, I found overall formula: $$ a^n \equiv b^n \pmod {a-b} $$ hence: $$ 8^{18} \equiv 1^{18} \pmod {8-1} $$

$$ 8^{18} - 1 \equiv 0 \pmod {8-1} $$

edited Oct 19 '14 at 18:00

answered Oct 19 '14 at 17:55

Reptile

41

2

Which is a way of not solving the problem. – Did Oct 19 '14 at 17:56

score 0 · Answer 4 · answered Oct 19 '14 at 18:09

0

Alternatively, you can just note that $8^{18}-1=7\cdot2573485501354569$.

answered Oct 19 '14 at 18:09

user2097

2,425

Shouldn't that be a comment? – Oct 21 '14 at 10:35
@DigitalBrain I believe that this really answers the question. – user2097 Oct 21 '14 at 16:58

How to prove that $8^{18} - 1$ is divisible by $7$

4 Answers4