Primes of the form 10^n + 1

Since $a^{n} + 1$ is factorable when $n$ is odd, numbers of the form $(10^{n+1})^{2m+1} + 1$ are factorable because $2m+1$ is always odd.

Also, numbers $10^n + 1$ are all factors of $10^{3n + 2} + 1$ because:
- $11 * 91 = 1001$
- $101 * 9901 = 1000001$
- $1001 * 999001 = 1000000001$
- $10001 * 99990001 = 1000000000001$
- etc.

This leaves us to look at numbers where $n = 1, 3, 7, 15, 31, ...$ which is equivalent to $10^{2^n} + 1$ which turns out to be the Fermat Numbers written in base 2 !

Some of these numbers have been factorized here . It is unknown whether or not any are prime besides $11$ and $101$ .