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Let $Q_s(x)\in\Bbb Z[x_1,x_2,\dots,x_s],\hat{Q}_{\hat s}(y)\in \Bbb Z[y_1,y_2,\dots,y_\hat s]$ be pair of homogeneous purely non-diagonal (every term of form $x_ix_j$ or $y_iy_j$) quadratic forms and having coefficients bounded by functions $f(s),\hat{f}(\hat s)$ such that $f(s)+\hat{f}(\hat s) \leq (s+\hat{s})B$ for a fixed $B\in\Bbb Z_{\geq 2}$ .

Fix a char $0$ ground field $\Bbb F=\Bbb C$ or $\Bbb R$.

Let $S_{r}(x-p)=0$ and $\hat{S}_{\hat{r}}(y-\hat{p})=0$ be spheres of radii $r$ and $\hat r$ centered at $p \in \Bbb F^s$ and at $\hat p \in \Bbb F^\hat{s}$ respectively.

Let $Z_Q$, $Z_S$ be zero sets in $\Bbb F^s$ of $Q_s(x)$, $S_{r}(x-p)$ respectively and $Z_{\hat{Q}}$,$Z_\hat S$ be zero sets in $\Bbb F^\hat{s}$ of $\hat{Q}_{\hat s}(y)$ and $\hat{S}_{\hat{r}}(y-\hat{p})$ respectively.

Let $G_1(x)$ be either sum of coordinates function and let $G_2(x)$ be sum of squares of coordinates function.

We call $[Q_s(x),\hat{Q}_\hat s(y)]_{(Z_S,Z_\hat{S},i)}$ a $(Z_S,Z_\hat S,i)$-pair if $x \in Z_S\cap Z_Q\iff y \in Z_\hat S\cap Z_\hat{Q}$ such that $G_i(x)+G_i(y)=s$ for either $i=1$ or $2$.

Given a homogeneous purely non-diagonal $Q_s(x)$, is there always a homogeneous purely non-diagonal $\hat{Q}_\hat s(y)$? If there is a pair, how does one find a homogeneous purely non-diagonal $\hat{Q}_\hat s(y)$ that is a pair? Does the minimum $\hat{s}$ required grow atmost as fast a $O(s^c)$ for some $c>0$?

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  • $\begingroup$ Some motivation and clearer notation (e.g., what is $\mathbb{F}$??) would improve this a lot. The specification of what is allowed to vary (e.g., $t$?) needs to be clarified a lot too. Try to present this question to someone in person to find out how imprecise its formulation is. $\endgroup$
    – Marguax
    Commented Oct 9, 2013 at 16:59
  • $\begingroup$ $\Bbb F$ is just a char $0$ ground field. Say $\Bbb C$ or $\Bbb R$. $\hat{s}$ is allowed to vary. $t$ is fixed to be be an integer less than $s+\hat{s}$. $\endgroup$
    – Turbo
    Commented Oct 9, 2013 at 17:07
  • $\begingroup$ I realized that there is a parametrization of $t$ with $s'$ which makes things more complicated. This base case now itself is interesting to me. $\endgroup$
    – Turbo
    Commented Oct 9, 2013 at 17:19
  • $\begingroup$ @Marguax Sorry I am not a mathematician. If you could tell me what is imprecise, I can fix it. $\endgroup$
    – Turbo
    Commented Oct 9, 2013 at 17:30
  • $\begingroup$ Since you have completely eliminated $t$, now it is much clearer. But the definitions of $|x|_i$ aren't non-negative when $\mathbb{F}=\mathbb{C}$, and even $|x|_1$ isn't when $\mathbb{F}=\mathbb{R}$ since you do not speak of absolute values there, so please be much clearer about exactly what is $\mathbb{F}$ (no notion of absolute value for a general char. 0 field, by the way). I also recommend again that you give some motivation for this question, as it looks quite artificial otherwise. $\endgroup$
    – Marguax
    Commented Oct 9, 2013 at 17:42

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