How to show that $\mathbb{Q}_3(\sqrt {-1}) \cong \mathbb{Q}_3(\sqrt 2) \cong \mathbb{Q}_3(\sqrt 5)$? How to show that $\mathbb{Q}_3(\sqrt {-1}) \cong \mathbb{Q}_3(\sqrt 2) \cong \mathbb{Q}_3(\sqrt 5)$ ?
There is a lemma which says that $\mathbb{Q}_p(\sqrt a) \cong \mathbb{Q}_p(\sqrt b)$ iff $a/b$ is a square in $\mathbb{Q}_p$.
If I use this lemma, then $2/5$ must be square in $Q_3$ if $\mathbb{Q}_3(\sqrt 2) \cong \mathbb{Q}_3(\sqrt 5)$.
But how to show $2/5$ is square in $\mathbb{Q}_3$ ?
Any simple way to understand rather than using Hensel lemma.
 A: (Perhaps fruitful is to reframe the question as: How can I show that something is not a square in $\mathbb{Q}_3$.  I came to this by picking some random multiples of $1/5$ and trying to decide whether they were squares in $\mathbb{Q}_3$.)
(I've never seen an argument of this form before, so I may have inadvertently overlooked one or more subtleties.  I really should read more on $p$-adics.)
Suppose $x^2 =_{\mathbb{Q}_3} \dfrac{a}{5}$ for $a \in \mathbb{Z}$ (i.e., $a \in \mathbb{Q}_3$ having a finite $3$-adic expansion) and $x \in \mathbb{Q}_3$.  From the form of Hensel's lemma, we wish to study the polynomial $f(x) = 5x^2 - a$.  Hensel's lemma allows us to extend a prefix of a solution, $y \cong x \pmod{3^k}$, to $f(x) = 0$ as long as $f'(y) \neq 0$.  Since 
$$   f'(y) = 10 y = 3^2 y + y =_{\mathbb{Q}_3} 0  $$ requires $y \cong 0 \pmod{9}$ (because the first two $3$-adic digits of $y$ are unaffected by adding $3^2 y$), we may instead study 
$$  5 x^2 - a \cong 0 \pmod{9}  $$
where a value of $a$ forcing $x \cong 0 \pmod{9}$ is a necessary condition that $a/5$ is not a square in $\mathbb{Q}_3$.  Applying that constraint on $x$, we find $a \cong 0 \pmod{9}$ may create an obstruction to Hensel's lemma.1  Consequently, any other choice of $a$ produces a square in $\mathbb{Q}_3$.
Since $-1 \cong 8 \not \cong 0 \pmod{9}$ and $2 \not \cong 0 \pmod{9}$, $-1/5$ and $2/5$ are squares in $\mathbb{Q}_3$.  With the lemma you state and a very short argument about transitivity, you get the three way field isomorphism.
1: The condition $a \cong 0 \pmod{9}$ is necessary, but is not sufficient to obstruct a lift, as described at the cited Wikipedia article.  Just because a $p$-adic prefix of a number is zero does not mean that the number is zero.  As long as the number is not zero, a slightly more complicated procedure extends the prefix.  The argument above allows the existence of $a \cong 0 \pmod{9}$ which produce squares in $\mathbb{Q}_3$, but the particular choices of $a$ you are considering do not require examining this further.
