How Many Primes Exist Here?

$10001, 100010001, 1000100010001, ...$

$= \frac{99999999}{9999}, \frac{999999999999}{9999}, \frac{9999999999999999}{9999}, ...$

$= \frac{10^{8} - 1}{10^4 - 1}, \frac{10^{12} - 1}{10^4 - 1}, \frac{10^{16} - 1}{10^4 - 1}, ...$

$= \frac{10^{4n} - 1}{10^4 - 1}$ for $n \geq 2$

$= \frac{(10^{n} - 1)(10^{n} + 1)(10^{2n} + 1)}{10^4 - 1}$ for $n \geq 2$

By inspection, $n = 2$ and $n = 3$ are not prime because $10001 = 73 \cdot 137$ and $100010001 = 3 \cdot 33336667$ .

For $n \geq 4$ , there will be at least two factors of the numerator (namely $(10^{n} + 1)$ and $(10^{2n} + 1)$ ) that will be larger than the denominator $10^4 - 1$ which will result in a number with at least two factors greater than $1$ ; in other words, a number that will never be prime, either.

There are therefore $\boxed{0}$ prime numbers in the given sequence.

We're looking at the number $10^{4n-4} + 10^{4n-8} + \cdots + 10^4 + 1 = \frac{10^{4n}-1}{10^4-1}$ for $n \ge 2.$

If $n$ is even, then $10^{4n} -1$ is divisible by $10^8-1,$ so $\frac{10^{4n}-1}{10^4-1}$ is divisible by $10^4+1.$ There are two cases: when $n=2$ our number is $10^4+1 = 73 \cdot 137,$ and when $n \ge 4$ is even $10^4+1$ is a nontrivial proper factor.

If $n$ is odd, then I claim that $\frac{10^n-1}9$ is a factor; or, equivalently, $\frac{(10^{4n}-1)(10-1)}{(10^4-1)(10^n-1)}$ is an integer. But this is a consequence of the fact that $\frac{(x^{4n}-1)(x-1)}{(x^4-1)(x^n-1)}$ is a polynomial with integer coefficients; this is easy to see because $\text{gcd}(x^4-1,x^n-1) = x-1,$ so $\frac{x^4-1}{x-1}$ and $\frac{x^n-1}{x-1}$ are relatively prime polynomials which both divide $\frac{x^{4n}-1}{x-1},$ so their product divides it.

So, to sum up, when $n = 2,$ the number factors by inspection. When $n \ge 4$ is even, $10^4+1$ is a nontrivial proper factor. When $n \ge 3$ is odd, $\frac{10^n-1}9$ is a nontrivial proper factor.