Exercise 7, § 2

We accept that for every integer $a > 0$ there are infinitely many prime numbers in the set of integers $1 + ma$ ($m$ running through the set of integers $\geq 1$; this is a particular case of Dirichlet’s theorem on arithmetic progressions).

a) Let $n$ be an integer $\geq 1$, and let $s \geq 1$ be an integer such that $q = 1 + (2n + 1)!s$ is prime; show that if $p$ is a prime number such that $p - 1$ divides $2nq$ then $p - 1$ must divide $2n$ (in the opposite case one would have $p - 1 = qd$ with $d$ an integer, and $p$ would be divisible by $d + 1$).

b) Deduce from a) that for every integer $n > 0$ there exist infinitely many integers $m > n$ such that $b_{2m} - b_{2n}$ is an integer.*