Suppose $p$ is an odd prime.The number of quadratic extensions of $\mathbb{Q}_p$ is $3$. They are $\mathbb{Q}_p(\sqrt a)$, $\mathbb{Q}_p(\sqrt{ap})$, $\mathbb{Q}_p(\sqrt p)$, where $a$ is chopped as square free unit of $\mathbb Z_p$.
But , why $\mathbb Z_p$ has square free unit element ? And why $\mathbb{Q}_p(\sqrt a)$ does not depend on the choice of $a$ ?
Thank you in advance.