prove (a^n)-1 is divisible by a-1

• Nov 13th 2008, 03:08 PM
JwEagle
prove (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 PM
o_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 PM
JwEagle
i'm still confused where does the a^1 come from.
• Nov 13th 2008, 03:26 PM
o_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 PM
whipflip15
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.