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If $n=qm+r, \ 0 \leq r <m$, then show that the remainder upon dividing $a^n-1$ by $a^m-1$ is $a^r-1$.

Answer:

If $q$ is a positive integer, then $a^{m}$ divides $\large a^{qm}-1$. For,

$a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ \cdots +a^{m(q-1})$.

But I can not finish the problem.

Help me.

MAS
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2 Answers2

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Hint: If $n=qm+r$, then $a^n-1 = a^r(a^{qm} - 1) + (a^r - 1)$.

Clive Newstead
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    Is $a^{qm}-1=(a^m-1)(1+a^m+a^{2m}+ \cdots +a^{m(q-1})$ true? I mean how to show that $a^m-1$ divides $a^{qm}-1$. Is it true? – MAS Jan 18 '19 at 19:24
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    @M.A.SARKAR: Yes that is true. It immediately shows that $a^m-1$ divides $a^{qm}-1$, since you've expressed $a^{qm}-1$ as $a^m-1$ multiplied by an integer. It follows from the fact that $$x^n-1 = (x-1)(1+x+x^2+\cdots+x^{n-1})$$ with $x=a^m$ and $n=q$. – Clive Newstead Jan 18 '19 at 19:26
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$\bmod\, \color{#c00}{a^{\large m}\!-\!1}\!:\,\ a^{\large r+mq}\!= a^{\large r} (\color{#c00}{a^{\large m}})^{\large q}\!\equiv a^{\large r} \color{#c00}1^{q}\!\equiv a^{\large r}$

i.e. if $\ a^{\large m}\equiv 1$ then $\, a^{\large n}\!\equiv a^{\large n\bmod m},\,$ i.e. $ $ expts on $\,a\,$ can be reduced $\!\bmod m$

Bill Dubuque
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