# Square classes of $\mathbb{Q}_2^\times$?

Does anyone know how to easily compute them? We know that a number is a square modulo $2^k$ if and only if it's a square modulo $8$. This gives a bunch of integers that represent square classes. I also know that

$$\mathbb{Q}_2^\times \cong \mathbb{Z}\times (1+2\mathbb{Z}_2),$$

but I can't figure out how to find the square classes in $1+2\mathbb{Z}_2$. Is there some really simple solution for this? I can't seem to figure out how to apply Hensel's lemma here.

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Maybe my little knowledge is a dangerous thing, but why isn't the answer simply everything of the form $2^k (1 + 8z)$, where $k \in \mathbb Z$ and $z \in {\mathbb Z}_2$? –  Greg Martin Jan 11 '12 at 19:28
@GregMartin: $k$ has to be even. –  Cam McLeman Jan 11 '12 at 19:34
@CamMcLeman: sorry yes of course –  Greg Martin Jan 11 '12 at 23:58

By your observations, a square element of $1+2\mathbb{Z}_2$ must actually live in $1+8\mathbb{Z}_2$. So $$\mathbb{Q}_2^\times/\mathbb{Q}_2^{\times 2}\approx \mathbb{Z}/2\mathbb{Z}\times \frac{1+2\mathbb{Z}_2}{1+8\mathbb{Z}_2}\approx \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}.$$ The first factor of $\mathbb{Z}/2\mathbb{Z}$ corresponds to choosing even/odd-ness of the power of 2 dividing the element of $\mathbb{Q}_2^\times$, the latter two any choice of coset represenatives for odd squares mod 8. So one possible enumeration of representatives for these 8 classes are the classes of $$\{\pm 1,\pm 2,\pm 5,\pm 10\}.$$
Cam, that is the choice on page 43 of Cassels, Rational Quadratic Forms. There was some poor slob a few months ago here on MSE who wanted to know the primes for which some indefinite ternary, diagonal such as $3 x^2 + 7 y^2 - 15 z^2,$ was isotropic. I typed it all in but I do not believe it took. Oh, also he wouldn't buy Cassels, he "had no access" to it. math.stackexchange.com/questions/89138/… –  Will Jagy Jan 11 '12 at 20:20