 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Neither $1$ nor $-1$ is divisible by a <prime number>, so both lie outside $U$. On the other hand, $0\in2\mathbb Z\subseteq U$. 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
$$
\boxed{\mathbb Z\setminus U=\{-1,1\}.}
$$

 Back to article page