Let $x,n ∈ \mathbb{N}$ such that :
$1+x^1+x^2+x^3+\dots +x^{n-1}$ is prime .
Proof that $n$ is prime .
I actually suppose that $n$ isn't prime absurdly so $n=p×k$ such as $p$ is prime, I also use the $a^n-b^n$ formula .
Does anyone has another answer and can help ?
If $x=1$, then clearly $n$ is prime. Let us consider $x>1$.
Suppose, towards a contradiction, that $n$ is composite. Then we may write $n=ab$ where $a,b\ge 2$. Thus
$$\begin{align*}1+x+\dots+x^{n-1}&=\frac{x^n-1}{x-1}\\ &=\frac{x^{ab}-1}{x-1}\\ &=\frac{(x^{b(a-1)}+x^{b(a-2)}+\dots+1)(x^b-1)}{(x-1)}\\ &=\frac{(x^{b(a-1)}+x^{b(a-2)}+\dots+1)(x^{b-1}+x^{b-2}\dots+1)(x-1)}{(x-1)}\\ &=(x^{b(a-1)}+x^{b(a-2)}+\dots+1)(x^{b-1}+x^{b-2}\dots+1)\end{align*}$$
which, provided $x>1$, is clearly composite. So we conclude that $n$ is in fact prime.