Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 6 / a (Prime divisors)

The set $S_n$ is nonempty because $n\in S_n$. By the well-ordering principle for the natural numbers, it has a least element $p$. In particular $p>1$ and $p$ divides $n$.

If $p$ were not a prime number, it would have a positive integer divisor $d$ with $1<d<p$. Since $d$ divides $p$ and $p$ divides $n$, divisibility is transitive and $d$ divides $n$. This would put $d$ in $S_n$, contradicting the minimality of $p$. Hence $\boxed{p\text{ is prime}}$.

Neither $1$ nor $-1$ is divisible by a prime number, so both lie outside $U$. On the other hand, $0\in 2\mathbb{Z}\subseteq U$. If an integer $z$ has $|z|>1$, part (a)(i) supplies a prime number $p$ dividing the positive integer $|z|$; the same $p$ divides $z$, so $z\in p\mathbb{Z}\subseteq U$. These cases exhaust $\mathbb{Z}$, proving