Are there infinitely many primes and non primes of the form $10^n+1$? Prove that there are infinitely many primes and non-primes in the numbers $10^n+1$, where $n$ is a natural number. So numbers are 101, 1001, 10001 etc.
 A: I also doubt this is homework.
Notice that if $n$ has an odd prime factor, then we can factor $10^n+1$ as we can factor $x^n+y^n$.  Therefore $10^n+1$ must be composite except when $n=2^m$ for some $m$.  The sequence $$f(n)=a^{2^n}+1$$ is a generalization of the Fermat numbers, and while we can show that each element is relatively prime since they have a similar multiplicative property, namely $$f(n)-2=a^{2^n}-1=f(n-1)f(n-2)\cdots f(1)f(0)(a-1).$$  It is unknown if there are infinitely many primes for any integer $a>1$.  Note that if $a$ is odd, the sequence has pairwise $\gcd$ $2$ rather than $1$.
Also, the problem of finding the next prime of the form $10^n+1$ after the first three, $2$, $11$ and $101$ is discussed on the following physics forum.  In short, their numerical calculations did not find another, and they took $n$ quite large.
A: I am not so sure this is homework.
One part is easy: If $\displaystyle n$ is odd (or divisible by an odd number $\displaystyle \gt 1$), then $\displaystyle 1+10^n$ is composite, using the fact that $\displaystyle x^{2n+1} + y^{2n+1}$ is divisible by $\displaystyle x+y$.
For $\displaystyle 1 + 10^n$ to be prime, $\displaystyle n$ must be a power of a $\displaystyle 2$, which makes it similar to Fermat numbers, and the question of whether there are an infinite number of Fermat primes is open. I believe the current 'expectation' based on heuristic arguments is that there are only finite number of such primes.
I would guess this would be the case with $\displaystyle 1 + 10^{2^m}$ too.
