Prove there exists a prime number $p > 10^{100}$ such that $p$ is the only prime number in the interval $[p − 10^{20}, p + 10^{20}]$.
I was looking through NT notes online and came across this homework question on a problem set.
I am not sure how to gain traction on this problem. So far I tried bounding a bit with Bertrand but I'm not sure that's leading anywhere.
Let $q_k$ and $q_k'$ for $k=1$ to $10^{20}$ be distinct primes, all greater than $10^{20}$. By the Chinese Remainder Theorem, the system of simultaneous congruences
$$x\equiv \begin{cases}-1\mod q_1\\ -2\mod q_2\\ \vdots\\ -10^{20}\mod q_{10^{20}}\\ 1\mod q_1'\\ 2\mod q_2'\\ \vdots\\ 10^{20}\mod q_{10^{20}}' \end{cases}$$
has a solution $n$ mod $Q$, where $Q=\prod_{k=1}^{10^{20}}q_kq_k'$. So there is an arithmetic progression, $n,n+Q,n+2Q,\ldots$, all with the property that $q_k\mid(n+mQ)+k$ and $q_k'\mid(n+mQ)-k$ for all $1\le k\le10^{20}$. Because the primes $q_k$ and $q_k'$ are all greater than $10^{20}$, none of them divide $n$, hence $\gcd(n,Q)=1$. By Dirichlet's Theorem on primes in arithmetic progressions, it follows that $n+mQ$ is prime for some (in fact infinitely many) values of $m$. Let $p=n+mQ$ be that prime.