If $x|yz$ and $\gcd(x, y)$ = 1 then $x|z, x, y, z \in \mathbb{Z}$

Question

If $x|yz$ and gcd$(x, y) = 1$ then $x|z $ $ , x, y, z \in \mathbb{Z}$

My thoughts:

$yz= dx$ and $mx +ny = \gcd(x, y)= 1$ where $d, m, n \in \mathbb{Z} \implies z =dx/y \text{ and } y = (mx-1)/n$

but I cannot prove it.

Answer 1

As you stated, since $\gcd(x,y) = 1$, by Bézout's identity, there exist integers $m$ and $n$ such that

$$mx + ny = 1 \tag{1}\label{eq1A}$$

Rather than trying to divide anything like you did with $yz = dx \implies z = \frac{dx}{y}$, it's easier to multiply both sides of \eqref{eq1A} by $z$ to get

$$mxz + nyz = z \tag{2}\label{eq2A}$$

Note $x \mid xz$, so $x \mid mxz$. Also, $x \mid yz$, so $x \mid nyz$. Thus, $x$ divides their sum, i.e., $x \mid z$.