Show that $n$ is prime.

Question

Let $x$ and $n$ be positive integers such that $\displaystyle \sum_{i=0}^{n-1} x^i$ is a prime number

Thus, show that $n$ is also prime

Answer 1

Two cases:

If $x>1$

$$\sum_{i=0}^{n-1} x^i=\frac{x^n-1}{x-1}$$

If $n$ is not a prime, then $n=pq$ with $p, q>1$, thus $x^n-1=(x^p)^q-1$ is divisible by $x^p-1$ (see below [1]), which is greater than $1$, and greater than $x-1$, hence $\frac{x^n-1}{x-1}$ has some divisor, and is not a prime. Contradiction, so $n$ is indeed a prime.

If $x=1$

$$\sum_{i=0}^{n-1} x^i=n$$

So it's immediate that $n$ is a prime.

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[1] $$\frac{(x^p)^q-1}{x^p-1}=\sum_{k=0}^{q-1} x^p$$

and the sum is an integer, so the quotient is exact, and $x^p-1$ divides $(x^p)^q-1$.