Show that $x$ is square free iff for any $y,z$ positive integers $x=yz \Rightarrow \mathrm{hcf}(y,z) = 1$

Question

Show that x is square free if and only if

$$x = yz\Rightarrow\mathrm{hcf}(y,z) = 1$$

where x and y are positive integers. I have tried using coprime factorisation leading to

$$1 = jy + kz$$

But cant get any further

Help appreciated

Answer 1

$x$ is not squarefree: Then $x=a^2b=a\cdot ab$ for some $a,b\in\mathbb Z, a\geq 2$, but $\mathrm{hcf}(a,ab)=a\neq 1$.

$x$ is squarefree: $\mathrm{hcf}(y,z)\mid y, z$, this implies $(\mathrm{hcf}(y,z))^2\mid yz=x$, so $\mathrm{hcf}(y,z)=1$.