Solutions to Problems 1331 - 1340

Q1331 Given any positive integers $m$ and $n$ prove that every divisor of $mn$ can be expressed as a product of a divisor of $m$ and a divisor of $n$.

ANS: Let $d$ be a divisor of $mn$, which we denote by $d|mn$, and let $d_1$ be the greatest common divisor of $d$ and $m$, which we write by $d_1=\mbox{g.c.d.} (d,m)$. We can now write
with $k, d_2 , m_1\in\mathbf{Z}^+$ and $\mbox{g.c.d}(m_1,d_2)=1$. It now follows that $kd_1d_2=m_1d_1n$ so that $kd_2=m_1n$ and as $\mbox{g.c.d.}(m_1,d_2)=1$ then $d_2|n$  and thus $d=d_1d_2$ with $d_1|m$ and  $d_2|n$.