Exercise 0.2.3 (If $n$ is composite, there are integers $a,b$ such that $n\mid ab$ but $n\nmid a,\, n\nmid b$)

Proof. To give a counterexample, $6\mid 3\times 4$, but $6\nmid 3$ and $6\nmid 4$.

We show that this is the same if $n$ is any composite number, so that $n=\mathit{𝑢𝑣}$, where $u>1$ and $v>1$. Put $a=u$, and $b=v$. Then $n\mid n=\mathit{𝑢𝑣}=\mathit{𝑎𝑏}$, but $1<a<n$ and $1<b<n$, thus $n\nmid a$ and $n\nmid b$ (if $n\mid a$, then $n=\mathit{𝑘𝑎}$, where $k\ne 0$ since $n>1$, so $k\ge 1$ and $n=\mathit{𝑘𝑎}\ge a$, which contradicts $a<n$).

For every composite number $n$, there are integers $a$ and $b$ such that $n$ divides $\mathit{𝑎𝑏}$ but $n$ does not divide either $a$ or $b$. □