Question about $\mathbb{Q}_2^{\ast}/(\mathbb{Q}_2^{\ast})^2$

Question

If we take the set $\{\pm 1, \pm 2, \pm 3, \pm 6\}$ as representatives for $\mathbb{Q}_2^{\ast}/(\mathbb{Q}_2^{\ast})^2$, then for instance $15$ is in the class of $-1$ because $-15 \in (\mathbb{Q}_2^{\ast})^2$ using Hensel's lemma: we have e.g. $$ |1^2+15|_2=1/16<1/4=|2\cdot 1|_2^2 $$ But now I'm not sure about, for instance, which class is $-10$ in? I couldn't find any solution using a Mathematica script that I could lift using Hensel's lemma, so I guess there must be something here that I'm not understanding. Any help would be appreciated.

Answer 1

For simple questions like this I really prefer an approach via series to the use of Hensel. The crucial fact now concerns the binomial expansion: $$ (1+4t)^{1/2}\in\mathbb Z[[t]]\,. $$ From this it follows directly that a $2$-adic unit congruent to $1$ modulo $8$ is a square. I know you know this, but the series makes the fact very transparent, I think. From this it follows that any unit $u\equiv3\pmod8$ is congruent to $3$ mod squares, with a similar result for $-1$ and $-3$. Equally, any prime element $\equiv2\pmod{16}$ is congruent to $2$ mod squares, etc. Such rules cover $\mathbb Z_2^*$ and $2\mathbb Z_2^*$, and everything in $\mathbb Q_2^*$ is congruent mod squares to something in $\mathbb Z_2^*\cup2\mathbb Z_2^*$.