What exactly is $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$ supposed to be? I need help with understanding the space $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$. I am unable to really comprehend what this is actually supposed to mean.
To begin, I found the following definition. $R/I$ is a set of equivalence classes of elements of $R$, where $a,b\in R$ are in the same equivalence class, say $[a]$, if the following holds: $$a,b\in [a]\in R/I~~~ \Leftrightarrow ~~~ (a-b)\in I.$$
So in this specific case, $$a,b\in [a]\in \mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2~~~ \Leftrightarrow ~~~ (a-b)\in (\mathbb{Q}_p^*)^2.$$
But here I fail to see which property some $a\in\mathbb{Q}_p^*$ has to satisfy, such that $a\in(\mathbb{Q}_p^*)^2$ , specifically, how this is dependent on $p$. I would think that a square in $\mathbb{Q}_2$ is also a square in $\mathbb{Q}_3$ for instance, as I do not understand how the property of being a square can be impacted by some valuation on the set.
I hope my question has become somewhat clear and I would be glad, if someone could offer me a very explicit definition for when $$a=p^\alpha\frac{u}{v}\in(\mathbb{Q}_p^*)^2$$ and by extention when $$a,b\in [a]\in \mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$$
Thank you very much!
 A: 
I would think that a square in $\mathbb{Q}_2$ is also a square in $\mathbb{Q}_3$ for instance, as I do not understand how the property of being a square can be impacted by some valuation on the set.

Remember that $\mathbb{Q}_p$ is not just $\mathbb{Q}$ endowed with the $p$-adic norm; rather, it is the completion of $\mathbb{Q}$ with respect to that norm. As we vary $p$ we're not looking at different norms on the same ring, but rather genuinely different rings. So there's no mystery to the definition of $\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2$, it's exactly what it seems to be - but it is after all dependent on $p$.
An important point here is that even the set of rationals which are squares in $\mathbb{Q}_p$ (that is, $(\mathbb{Q}_p^*)^2\cap\mathbb{Q}$) depends on $p$: if $\alpha\in\mathbb{Q}$ is a square in $\mathbb{Q}_{p_1}$, its square roots in the sense of $\mathbb{Q}_{p_1}$ might not be in $\mathbb{Q}$ and so $\alpha$ might not be a square in the sense of some other $\mathbb{Q}_{p_2}$.
See also Hesel's lemma and various MSE questions (e.g. 1, 2).
