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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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    @CamMcLeman: sorry yes of course2012-01-11

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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\}. $

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    Thanks, Cam, I do like to educate the OP, but that takes time, and often they want just enough to do the homework, but quickly. I see, both answers there were mine, after the OP said the first answer wasn't good enough. Sigh.2012-01-11