Let a,b,c $\in\mathbb{Z}$ such that a|k, b|k and gcd (a,b)=1. Prove that ab|k.

Question

Let a,b,c $\in\mathbb{Z}$ such that a|k, b|k and gcd (a,b)=1. Prove that ab|k.

Any hints would be helpful.

Answer 1

k divides a and b implies k = ua, k =vb, gcd(a,b)=1 implies da+eb=1, thus k(da+eb)=k= (vb)(da)+(ua)(eb) =ab(vd+ue)

Answer 2

Since $a|k$ and $b|k$, then $a n_1 = k = b n_2$ for some $n_i \in \mathbb{Z}$. Then $a n_1 = b n_2$, but since $(a,b)=1$ then $a|n_2$, so $n_2 = a n_3$. Therefore $k = b (a n_3)$, so $ab |k$.