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

  1. #1
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    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
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  2. #2
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    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
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  3. #3
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    i'm still confused where does the a^1 come from.
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    Are you familiar with this property: $\displaystyle a^{m+n} = a^ma^n$

    Simply imagine $\displaystyle m = k$ and $\displaystyle n = 1$.
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  5. #5
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    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.
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