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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 ?

Bernard
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Rêveuse n
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    Hint: $x^n-1=x^{pk}-1=(x^p)^k-1^k$. – cqfd May 10 '20 at 15:29
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    I think you have the simplest way, except no need to assume $ n = pk $ where $p$ is prime. It's enough just to take any non-trivial factorisation $ n = ab $ where both $a,b > 1$. – WA Don May 10 '20 at 15:44
  • The post is for questions and answers if you don't like it or you have nothing to add just keep your opinion for yourself and do a favor for yourself and people here ,you have to respect the people here who try to find out the answer and help other people you are a rude person – Rêveuse n May 11 '20 at 17:45

1 Answers1

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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.

mrbronstedlowry
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