Exercise 7, § 2

It will be admitted that for every integer $a > 0$, there exists an infinity of prime numbers in the set of integers $1 + ma$ ($m$ running through the set of integers $\geq 1$; particular case of the theorem of the arithmetic progression of Dirichlet). 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; now prove that if $p$ is a prime number such that $p - 1$ divides $2nq$, then $p - 1$ necessarily divides $2n$ (otherwise, one would have $p - 1 = qd$ with $d$ integer, and $p$ would be divisible by $d + 1$). b) Deduce from a) that for every integer $n > 0$, there exists an infinity of integers $m > n$ such that $b_{2m} - b_{2n}$ is an integer.