Prove that there exists $a_1, b_1 \in \mathbb{Z}$ such that $d = a_1 b_1$ and $a_1 \mid a$ and $b_1 \mid b$.

Question

Let $a, b, d \in\mathbb{Z}$. Suppose that $d \mid ab$. Prove that there exists $a_1, b_1 \in \mathbb{Z}$ such that $d = a_1 b_1$ and $a_1 \mid a$ and $b_1 \mid b$.

Answer 1

Consider $$a=p_1^{\alpha _1}...p_n^{\alpha _n}\quad \text{and}\quad b =q_1^{\beta _1}...q_m^{\beta _m},$$ decomposition of $a$ and $b$ in prime numbers. Then $$d\mid ab\iff d=p_1^{\gamma _1}...p_n^{\gamma _n}q_1^{\delta _1}...q_m^{\delta _m},$$ for some $\gamma _i\in \{0,...,\alpha _i\}$ and $\delta _i\in\{0,...,\beta _i\}.$

Take $$a_1=p_1^{\gamma _1}...p_n^{\gamma _n}\quad \text{and}\quad b_1=q_1^{\delta _1}...q_m^{\delta _m}.$$ The claim follows.