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First I define the level of field. The level of a field $\mathbb K$ is the least $n$ such that $−1$ is a sum of $n$ squares in field, and is denoted by $S(\mathbb K)$. I know that the level of $\mathbb Q_2$ (the field of $2$-adic numbers) is $4$. It is an easy calculation:

You can see that every $2$-adic integer which is congruent to $1 \bmod 8$ is a square. So $-7$ is a square and hence $S(\mathbb K) \leqslant 4$. On other hand any integral square in $\mathbb{Q}_{2}$ is congruent to $0$, $1$, or $4 \bmod 8$ so $S(\mathbb K) = 4$.

I am looking for the level of $p$-adic number field. Any suggestions? Thanks.

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I made some $\TeX$ edits and fixed (I hope) some minor English errors. Feel free to revert! –  Dylan Moreland May 25 '12 at 6:33
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1 Answer 1

up vote 4 down vote accepted

For odd $p$, $x^2 + y^2 + 1 \equiv 0 \mod{p}$ always has a solution because the sets $\{x^2 : x \in \mathbb{F}_p\}$ and $\{-y^2-1 : y \in \mathbb{F}_p\}$ both have cardinality $(p+1)/2$ and hence have nonempty intersection. By Hensel's Lemma (since $p$ is odd), $x^2 + y^2 + 1 = 0$ has a solution in $\mathbb{Q}_p$. Therefore, $S(\mathbb{Q}_p) \le 2$.

$S(\mathbb{Q}_p) = 1$ iff -1 is a square mod $p$ iff $p \equiv 1 \mod{4}$ (again by Hensel's Lemma).

So $S(\mathbb{Q}_p) = 2$ iff $p \equiv 3 \mod{4}$.

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