# $a^n-1$ is a composite number

This is homework from my elementary theory of numbers course: I need to prove that when $n\geq 2$ and $a>2$ the equation $a^n-1 = (a-1)(1+a+\cdots+a^{n-1})$ is a composite number .

I don't know where to start

Thanks

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Do you know what "composite" means? – Chris Eagle Nov 5 '12 at 18:39
@ChrisEagle: It seems we had simultaneous identical edits. – Harald Hanche-Olsen Nov 5 '12 at 18:43
if $a>2$ and $n$ are both integers, and $n\ge 2$, then there is nothing to prove here. – user31280 Nov 5 '12 at 18:47
Hint: to prove $\rm\:b\,c\:$ is composite it suffices to show that both $\rm\,b,c\,$ are nonzero nonunits, i.e. $\ne 0,\pm1,\:$ which e.g. is true if both are $> 1.\ \$ – Bill Dubuque Nov 5 '12 at 19:01

So you need to prove that for $a>2, \ n>1, \ a^n-1$ is a composite number, i.e. $a^n-1=b\cdot c \$ for some integers $1<b,c < a^n-1$ (observe that if $1<b<a^n-1$ then necessary $1<c<a^n-1$).

You noted that $a^n-1 = (a-1)(1+a+...+a^{n-1}) .$ So if we set $b=a-1$ and $c=1+a+...+a^{n-1}$ then $a^n-1=b\cdot c. \$ For $a^n-1$ to be prime the only thing it remains to be proven is that $1<b,c < a^n-1 \ldots$

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Consider the polynomial $p(x) = x^n - 1$, since $p(1) = 0$ we have $x-1$ divides $p(x)$.

Performing the long division gives $x^{n-1} + x^{n-2} + \ldots + 1$.

In this way we have factored $p(x) = (x-1)(x^{n-1} + x^{n-2} + \ldots + 1)$ so given a number $a>2$, $p(a)$ must factor.

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But the key point is: for it to be composite it "must factor" nontrivially, i.e. both factors are $\ne \pm1, 0$ $\ \$ – Bill Dubuque Nov 5 '12 at 18:58