i need help on trying to prove this (a^n)-1 is divisible by a-1 for every natural number n and every integer a > 1

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- Nov 13th 2008, 03:08 PMJwEagleprove (a^n)-1 is divisible by a-1
i need help on trying to prove this (a^n)-1 is divisible by a-1 for every natural number n and every integer a > 1

- Nov 13th 2008, 03:13 PMo_O
Induction works. Skipping all the formalities ...

Inductive hypothesis:**Assume**$\displaystyle {\color{blue}a^k - 1}$ is divisible by $\displaystyle a - 1$. It remains to show that $\displaystyle a^{k+1} - 1$ is also divisible by $\displaystyle a-1$.

But note that:

$\displaystyle \begin{aligned}a^{k+1} - 1 & = a^ka^1 - 1 \\ & = a^ka {\color{red} - a + a} - 1 \qquad (\text{We're adding '0'}) \\ & = a({\color{blue}a^k - 1}) + (a-1)\end{aligned} $

And the conclusion follows - Nov 13th 2008, 03:22 PMJwEagle
i'm still confused where does the a^1 come from.

- Nov 13th 2008, 03:26 PMo_O
Are you familiar with this property: $\displaystyle a^{m+n} = a^ma^n$

Simply imagine $\displaystyle m = k$ and $\displaystyle n = 1$. - Nov 13th 2008, 04:09 PMwhipflip15
Would it help if is said

$\displaystyle x^n-1=(x-1)(x^{n-1}+x^{n-2}\dots x+1)$

for every natural number n.