Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 1G / c

Suppose that the positive integer $n$ is a decisive number. For every prime number $p<\sqrt{n}$, the composite number $p^2$ satisfies $1<p^2<n$. It therefore cannot be coprime to $n$, so $p\mid n$.

The product of all prime numbers below $\sqrt{n}$ consequently divides $n$. Taking natural logarithms givesChanging the strict endpoint can remove at most one prime term. By part b, for every sufficiently large $n$ the right-hand side is at least, for example, $\sqrt{n}/13$. This is impossible for sufficiently large $n$, sinceThus every decisive number lies below one fixed bound, and only finitely many integers can do so.