Primes in consecutive integers

Let $\pi(n)$ denote the number of [positive] prime numbers $\leq n$ . Then we are interested in whether the function $f(n) = \pi(n+1000) - \pi(n)$ ever equals $100$ .

To show this, we may confirm

• $f(0) = \pi(1000) = 168$ and $f(1002! + 1) = 0$
• $f(n+1) - f(n) \in \{-1, 0, 1\}$ , so $f$ satisfies an "intermediate value theorem" on the integers.

Together, this implies there is some integer $N$ with $0 < N < 1002!+1$ such that $f(N) = 100$ , which means that the integers $N+1, N+2, N+3, \ldots, N+1000$ contain exactly 100 primes.