NMO Selection Tests for JBMO

We show that the smallest element of $S$ that is greater than $1$ divides a larger element of $S$. If $p$ is of the form $m^2 + 1$, with $m \in \mathbb{N}$, we show that $p - (m - 1)^2 = 2m$ divides $p^2 - 1 = m^2$. Indeed, as $m$ is even, it follows that $2m \mid m^2$.

If $p$ cannot be written as $m^2 + 1$, $m \in \mathbb{N}$, $p$ cannot be written as $m^2 + 2m$ either (this is a composite number because $m > 1$), hence $m^2 + 1 < p < m^2 + 2m$ for some $m \in \mathbb{N}$ ($m \ge 2$). We show that $p - m^2$, which is an element of $S$ that is larger than $1$, divides another element of $S$. The condition that $p - m^2$ divides one of the numbers $p - n^2$ with $n \in \{0, 1, 2, \dots, m-1\}$ is equivalent to $p - m^2$ dividing one of the numbers $m^2, m^2 - 1^2, m^2 - 2^2, \dots, m^2 - (m-1)^2$. Being smaller than $2m$, $p - m^2$ divides one of the following $2m-1$ consecutive numbers: $1, 2, \dots, m-1$, $m$, $m+1, \dots, 2m-1$, hence it divides one of the differences $m^2 - 0^2, m^2 - 1^2, \dots, m^2 - (m-1)^2$. Moreover, it does not divide $m^2$, because it would divide $p$.