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
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
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
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.