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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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    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.\ \ 2012-11-05

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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 (observe that if $1 then necessary $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

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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 $\ne \pm1, 0$ $\ \ $2012-11-05