Primes and $n^{th}$ Powers

claim: $n=1$ is the only $n$ where this works.

proof: We are looking for primes of the form $p = x^{n}-1$ where $x$ is a natural number and $n$ is a positive integer.

The case for $n=1$ was given in the problem to work.

If $n \geq 2$ , then we can factor $p=x^{n}-1$ as

$p = x^{n}-1 = (x-1) \left( \sum_{i=0}^{n-1} x^{i} \right)$

Now, for $p$ to be prime, then either $x-1=1$ or the sum is equal to $1$ .

If $x=0$ then for all $n$ , $p = x^{n}-1 = -1$ which is not prime.

If $x \geq 1$ and $n \geq 2$ , then

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

so that for $p$ to be prime, it follows that $x-1=1$ or that $x=2$ . Since this is the only possibility, it is impossible for two or more distinct $x$ to result in $p = x^{n}-1$ being prime for all $n \geq 2$ . Hence, $\boxed{1}$ is the final answer.

Let $n$ be an even number such that $n = 2s$ .

$x^{2s} - 1 = p$

$(x^s -1)(x^s + 1) = p$

Since $p$ is prime it only has two factors, $1$ and $p$ . Since $x> 0$ , $(x^s -1 ) = 1 \implies x = 2^{\frac{1}{s}}$ . Since, $1<2^{\frac{1}{s}}< 2, \forall s> 1 \implies 1<x<2$ meaning $x$ is not an integer. Therefore, $n$ has to be odd.

So if $n = 2s+1$ , then $x^{2s+1} -1 = p$ . Factoring $(x- 1)$ we get that $(x-1)\displaystyle\sum_{r=0}^{2s} x^r$ . Repeating our argument, $\displaystyle\sum_{r=0}^{2s} x^r =1 \text{ iff } s = 0$ which yields trivial $n =1$ therefore $(x - 1) =1 \rightarrow x= 2$ . Whether or not $\displaystyle\sum_{r=0}^{2s} x^r =p$ is now unimportant since $x = 2$ thus proving that $x$ can't take at least $2$ values implying the answer is $\boxed{1}$ .