$2$-adic valuation on $\mathbb Q (\sqrt{-15})$

Question

I was reading the book "The structure of groups of prime power order". In the book (page 226, Example 10.1.18) it is proved that

$-15$ has a square root is $\mathbb Z_2$ where $\mathbb Z_2$ denotes the ring of $2$-adic integers.

[I have attached a screen-shot of Example 10.1.18 at the end of my post]

So suppose $a=\sqrt{-15}$, then for any integers $x,y$ we have $x+ya\in \mathbb Z_2$ since $\mathbb Z \subset \mathbb Z_2$. So my question is

Given $x,y\in \mathbb Z$, is it possible to determine explicitly $\bar x, \bar y \in \mathbb Z$ and a maximum possible $l\ge 0$ such that

$$x+ya = 2^l(\bar x + \bar y a) \quad \text{ and } \quad \bar x + \bar y a \text{ is a unit in } \mathbb Z_2$$

Though not exactly related to the question but I think I should also say why I have this question; I am thinking of finding some kind of normal form of a matrix whose entries are from $\mathbb Z_2$ and has the form $x+ya$. So if I get a method for my above mentioned question then it will possibly help to write an algorithm for the normal form.

But I apologise for not showing much effort from my side for the question, I am actually not sure how to handle this. Any help will be greatly appreciated.

Thanks.

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Screen-shot of the example:

Answer 1

A way to do this is to write $\sqrt{-15}=\sqrt{1-16}$ from which you get the $2$-adic expansion of $a$, and thus of $x+ya$. To get an a priori bound on $l$, you can use the fact that $v_2(x+ya) \leq v_2((x+ya)(x-ya)) \leq v_2(x^2+15y^2)$ which you can compute since $x^2+15y^2 \in \mathbf{Z}$.