= Solution
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 integers>[coprime] to $n$, so $p\mid n$.
The product of all <prime numbers> below $\sqrt n$ consequently divides $n$. Taking <natural logarithms> gives
$$
\log n\geq\sum_{p<\sqrt n}\log p.
$$
Changing 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 $n$, since
$$
\frac{\sqrt n}{\log n}\longrightarrow\infty.
$$
\b[Thus every <decisive number> lies below one fixed bound, and only finitely many <integers> can do so.]
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