I was playing around a bit with quadratic forms for a different question here, and among other things had to decide when the form $X^2+Y^2+Z^2$ is isotropic over a certain field $k$, i.e. whether there is $(0,0,0) \neq (x,y,z) \in k \times k \times k$ such that $x^2+y^2+z^2=0$. This is trivially the case if $k$ contains a square root of $-1$, giving a positive answer for $k$ being any $p$-adic field $\mathbb Q_p$ with $p \equiv 1$ mod $4$, and hence all extensions of those too. Playing around a bit with Hensel's Lemma (and, funnily, using that $-1$ is not a square) led me to believe the form is isotropic over all $\mathbb Q_p$ with $p \equiv 3$ mod $4$ as well. As far as $p$-adic fields go, this left the case of extensions of $\mathbb Q_2$.
Here, I think I have an argument that the form has no nontrivial zeroes i.e. is anisotropic over $\mathbb Q_2$ itself. So next I went to its seven distinct quadratic extensions. I found the form is isotropic by coming up with individual solutions over $\mathbb Q_2(\sqrt d)$ for $d=-1, -3, -5, -2, -6, -10$: i.e. all the quadratic extensions except $\mathbb Q_2(\sqrt 2)$. For this one, I believe it is still anisotropic. (Edit: Thanks to the Keith Conrad's answer, I no longer believe this.)
Question 1: Are my results so far correct? (Edit: As just said, the form is actually isotropic over each of the seven quadratic extensions of $\mathbb Q_2$.)
Question 2: Are there easier criteria to decide the question for further extensions of $\mathbb Q_2$? In particular, if the results are correct, for which quadratic extensions of $\mathbb Q_2(\sqrt 2)$ is the form isotropic? Also, is there any high-brow explanation why the field extension $\mathbb Q_2(\sqrt 2)$ is singled out here? (I mean, if the unramified $\mathbb Q_2(\sqrt{-3})$ had behaved differently from the others, that would not have surprised me, but this …?)
I am aware and know the absolute basics of the theory of Witt groups (or Witt rings), but not enough to be able to apply them here. But I'm happy to learn if somebody can tell me how to apply that machinery here. (I know just enough to guess that the residue characteristic of $2$ here will complicate things.)
Added: There is a theorem of Witt's which I am comfortable with, which says that all quadratic forms in $\ge 5$ variables are isotropic over any $p$-adic field. For a field extension $K \vert \mathbb Q_p$ of degree $n$, it is therefore tempting to express a quadratic form in $d$ variables (here, $d=3$) over $K$ as one in $nd$ variables over $\mathbb Q_p$, however the algebra of the field extension introduces some relations between the coefficients which effectively reduce the number of variables again. Still it seems as if at least in many cases, a "big" enough extension should give us enough variables to just rely on that theorem. I don't have the mindpower to think through this right now, but again would be happy to learn if that works.
Best Answer
When $K$ is a field not of characteristic $2$, being able to solve $x^2 + y^2 + z^2 = 0$ nontrivially in $K$ is equivalent to being able to solve $x^2 + y^2 = -1$ in $K$, so I will think of the problem from that point of view.
I have checked explicitly that this can be done when $K$ is a quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$, as you say you did, and it can also be done when $K = \mathbf Q_2(\sqrt{2})$.
I'll list all seven quadratic extensions of $\mathbf Q_2$ in the order you did: $$ \mathbf Q_2(\sqrt{-1}), \ \mathbf Q_2(\sqrt{-3}), \ \mathbf Q_2(\sqrt{-5}), \ \mathbf Q_2(\sqrt{-2}), \ \mathbf Q_2(\sqrt{-6}), \ \mathbf Q_2(\sqrt{-10}), \ \mathbf Q_2(\sqrt{2}). $$ In the four fields $\mathbf Q_2(\sqrt{-1})$, $\mathbf Q_2(\sqrt{-5})$, $\mathbf Q_2(\sqrt{-2})$, and $\mathbf Q_2(\sqrt{-10}),$ it is very easy to see a solution to $x^2 + y^2 = -1$: $$ 0^2 + \sqrt{-1}^2, \ 2^2 + \sqrt{-5}^2, \ 1^2 + \sqrt{-2}^2, \ 3^2 + \sqrt{-10}^2. $$
In $\mathbf Q_2(\sqrt{-3})$ we have a cube root of unity $\zeta_3 = (-1 + \sqrt{-3})/2$, for which $1 + \zeta_3 + \zeta_3^2 = 0$, so $$ -1 = \zeta_3 + \zeta_3^2 = \zeta_3^4 + \zeta_3^2. $$
In $\mathbf Q_2(\sqrt{-6})$, after playing around a bit I got $$ \left(\frac{2+\sqrt{-6}}{2}\right)^2 + \left(\frac{2-\sqrt{-6}}{2}\right)^2 = -1. $$
That leaves us with $\mathbf Q_2(\sqrt{2})$, whose ring of integers is $\mathbf Z_2[\sqrt{2}]$. I am not going to record here all the numerical work I did (by hand), but just report the final result: we can write $$ -1 = x^2 + (\sqrt{2} + 2)^2 $$ for some $x \in \mathbf Z_2[\sqrt{2}]$ where $x \equiv 1 \bmod \sqrt{2}^3$. This follows from Hensel's lemma on $\mathbf Q_2(\sqrt{2})$ for the polynomial $f(X) = X^2 + 1 + (\sqrt{2}+2)^2 = X^2 + 7 + 4\sqrt{2}$: $$ f(1) = 8 + 4\sqrt{2} \Longrightarrow |f(1)|_2 = \frac{1}{4\sqrt{2}} $$ and $$ f'(1) = 2 \Longrightarrow |f'(1)|_2^2 = \frac{1}{4}. $$ Since $|f(1)|_2 < |f'(1)|_2^2$, Hensel's lemma tells us there is a unique solution to $f(x) = 0$ where $x \in \mathbf Z_2[\sqrt{2}]$ and $|x - 1|_2 < |f'(1)|_2 = 1/2$, or equivalently $x \equiv 1 \bmod (\sqrt{2}^3)$. A more precise form of Hensel's lemma tells us that $|x - 1|_2 = |f(1)/f'(1)|_2 = (1/4\sqrt{2})/(1/2) = 1/2\sqrt{2} = 1/\sqrt{2}^3$, so $x \not\equiv 1 \bmod \sqrt{2}^4$.
We can compute $x$ using the binomial series for the square root: $$ x = (-7 - 4\sqrt{2})^{1/2} = (1 - 4\sqrt{2} - 8)^{1/2} $$ so $$ x = \sum_{n \geq 0} \binom{1/2}{n}(-4\sqrt{2}-8)^n = 1 + \sum_{n \geq 1} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n $$ Since ${\rm ord}_2\binom{1/2}{n} = -2n + s_2(n)$, where $s_2(n)$ is the sum of the base $2$ digits of $n$, and ${\rm ord}_2((4\sqrt{2}+8)^n) = 5n/2$, the $n$th term of the series has $2$-adic valuation $5n/2 - 2n + s_2(n) = n/2 + s_2(n)$, so its $\pi$-adic valuation ($\pi = \sqrt{2}$ over the $2$-adics) is twice that: $n + 2s_2(n)$. Since $n+2s_2(n) \geq 7$ for all $n \geq 1$ other than $n = 1$, $2$, and $4$, $$ x \equiv 1 + \sum_{n = 1, 2, 4} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n \bmod \pi^7\mathbf Z_2[\pi]. $$ Working this out modulo $\pi^7$, $$ x \equiv 1 + \pi^3 + \pi^5 + \pi^6 \bmod \pi^7\mathbf Z_2[\pi]. $$ In "normal" notation, with $\pi = \sqrt{2}$, this approximate root of $f(X)$ is $9+6\sqrt{2} \bmod 8\sqrt{2}\mathbf Z_2[\sqrt{2}]$. As a check, $$ (9 + 6\sqrt{2})^2 + (7 + 4\sqrt{2}) = 160 + 112\sqrt{2} = 16\sqrt{2}(5\sqrt{2} + 7), $$ which is a multiple of $\pi^7 = 8\sqrt{2}$.
In what way did your work suggest that there might not be a solution to $x^2 + y^2 = -1$ in $\mathbf Q_2(\sqrt{2})$?
There is a high-brow explanation for why we found a solution in the particular quadratic extension $\mathbf Q_2(\sqrt{2})$. Being able to solve $x^2 + y^2 = -1$ in a local field $F$ (not of characteristic $2$) is the same as saying the Hilbert symbol $(-1,-1)_F$ is $1$ rather than $-1$. Letting $F = \mathbf Q(\sqrt{2})_v$ be a completion of $\mathbf Q(\sqrt{2})$, Hilbert reciprocity says the number of $v$ such that $(-1,-1)_{\mathbf Q_2(\sqrt{2})_v}$ is $-1$ is even. When $v$ is a place of odd residue field characteristic, $\mathbf Q(\sqrt{2})_v$ is a local field that is an extension of $\mathbf Q_p$ for an odd prime $p$, so that Hilbert symbol is $1$ since we can already solve $x^2 + y^2 = -1$ in $\mathbf Q_p$ (and actually in $\mathbf Z_p$). That leaves us with just three remaining places $v$ on $\mathbf Q(\sqrt{2})$: the two real places (corresponding to the two different embeddings of $\mathbf Q(\sqrt{2})$ into $\mathbf R$) and the unique place over $2$, corresponding to the prime $(\sqrt{2})$ in $\mathbf Z[\sqrt{2}]$. We obviously can't solve $x^2 + y^2 = -1$ in $\mathbf R$, so the Hilbert symbol $(-1,-1)_{\mathbf Q(\sqrt{2})_v}$ is $-1$ when $v$ is a real place. That leaves us with just one more place, over $2$, and since the Hilbert symbol is $-1$ at an even number of places it must be $1$ at the place over $2$.
By similar reasoning, if $K$ is an arbitrary real quadratic field, we can describe the non-archimedean places $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:
(1) If $v$ lies over an odd prime then there is a solution in $K_v$ since there already is one in $\mathbf Q_p$ for all odd $p$, and $K_v$ contains some $\mathbf Q_p$.
(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is just one $2$-adic place on $K$. We can solve $x^2 + y^2 = -1$ in $K_v$ by Hilbert reciprocity because we can solve it in every other completion except the two real completions, and two is even but three is not.
(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.
If $K$ is an arbitrary imaginary quadratic field, we can also determine the non-archimedean $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:
(1) If $v$ lies over an odd prime then there is a solution in $K_v$ for the same reason as in the real quadratic case above.
(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is exactly one $2$-adic place on $K$. There is also exactly one archimedean place, with completion $\mathbf C$. We can solve $x^2 + y^2 = -1$ in $\mathbf C$, so Hilbert reciprocity implies we can solve $x^2 + y^2 = -1$ in $K_v$. In other words, we can solve $x^2 + y^2 = -1$ in every completion of $K$, so by Hasse-Minkowski we can solve $x^2 + y^2 = -1$ in $K$.
(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.
The second item of each list for real and imaginary quadratic fields explains why we found solutions in every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ without needing to work in the completion itself (i.e., a solution in $\mathbf Q(i)$, $\mathbf Q(\sqrt{-3})$, and so on): you wrote every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ as a $2$-adic completion of an imaginary quadratic field, and in the imaginary quadratic fields that you chose, you can check that $2$ does not split completely (sometimes $2$ ramifies, sometimes $2$ is inert, but $2$ never splits completely). In the case of $\mathbf Q_2(\sqrt{2})$, we can write it as $\mathbf Q_2(\sqrt{-14})$, a completion of $\mathbf Q(\sqrt{-14})$ at its unique $2$-adic place. Then results above say we have to be to solve $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{-14})$, and in fact a solution is $x = (2+3\sqrt{-14})/10$ and $y = (6-\sqrt{-14})/10$.
More generally, if $\mathbf Q_2(\sqrt{2}) = \mathbf Q_2(\sqrt{d})$ for some integer $d$ then ${\rm ord}_2(d) = 1$, so $\mathbf Q(\sqrt{d})$ ramifies at $2$ and thus there is a unique $2$-adic place on $\mathbf Q(\sqrt{d})$, so for $d < 0$ we'd be in the same situation as $\mathbf Q(\sqrt{-14})$: there will be a solution to $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{d})$.