Let $p$ be odd prime.Why $a,b∈\mathbb Z_p^×-{(\mathbb Z_p^×)}^2$ implies $a/b∈{(\mathbb Z_p^×)}^2$?

Question

Let $p$ be odd prime.Why $a,b∈\mathbb Z_p^×-{(\mathbb Z_p^×)}^2$ implies $a/b∈{(\mathbb Z_p^×)}^2$ ?

For example, $2, -1∈\mathbb Z_3^×-{(\mathbb Z_3^×)}^2$ and $-2∈{(\mathbb Z_3^×)}^2$.

I heard the titled question is true, but I have trouble how to confirm this in general. I uses this titled fact to classify quadratic extension of $\mathbb Q_p$. Thank you in advance.

Answer 1

Once you know that $$\mathbb Z_p^\times/(\mathbb Z_p^\times)^2\cong \mathbb Z/2\mathbb Z,$$ then you know that you can write $$\mathbb Z_p^\times = (\mathbb Z_p^\times)^2\sqcup x(\mathbb Z_p^\times)^2$$ for some (well, any) non-square $x\in \mathbb Z_p^\times$. Can you see how this helps you show that $a/b$ is a square?