The following problem bears some similarity to the Additive Basis Conjecture [[ALM91](http://www.sciencedirect.com/science/article/pii/009731659190045I),[JLPT92](http://www.sciencedirect.com/science/article/pii/009589569290016Q)] saying (in characteristic $3$) that there is an absolute constant $N$ such that for any positive integer $m$, any bases $B\_1,\dotsc,B\_N$ of ${\mathbb F}\_3^m$, and any $g\in{\mathbb F}\_3^m$, there exist subsets $S\_i\subseteq B\_i$ with $\sum\_{i=1}^N \sum\_{b\in S\_i} b=g$. 



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> Does there exist a positive integer $m$ such that, letting $q:=3^m$, there are bases $B\_1,\dotsc,B\_{100}$ of the field ${\mathbb F}\_q$ (over $\mathbb F\_3$) with the following property: for any choice of the subsets $S\_i\subseteq B\_i$, we have $\sum\_{i=1}^{100}\sum\_{b\in S\_i}b^2\ne 0$, unless all subsets $S\_i$ are empty?
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Notice that, assuming the Additive Basis Conjecture is true with $N=99$, the squares $b^2$ cannot be replaced with the first powers. (I can also prove this unconditionally.) On the other hand, it would be fine with me to have the squares replaced with any powers $b^k$ where the exponents $k$ come from the set $K:=\{3^j+1\colon j\ge 0\}$:



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> Does there exist a positive integer $m$ such that, letting $q:=3^m$, there are bases $B\_1,\dotsc,B\_{100}$ of the field ${\mathbb F}\_q$ (over $\mathbb F\_3$) *and exponents $k\_1,\dotsc,k\_{100}\in K$* with the following property: for any choice of the subsets $S\_i\subseteq B\_i$, we have $\sum\_{i=1}^{100}\sum\_{b\in S\_i} b^{k\_i}\ne 0$, unless all subsets $S\_i$ are empty?
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Ensuring that a sum of squares is non-zero (in a finite field)
For any $k\geqslant 2m+1$ elements $t\_1,\dots,t\_{k}$ in the additive group $\mathbb{F}\_3^m$ there exist coefficients $\varepsilon\_1,\dots,\varepsilon\_{k}\in \{0,1\}$, not all of them equal to 0, such that $\sum \varepsilon\_i t\_i=0$. This is a result of Olson (John E. Olson. A Combinatorial Problem on Finite Abelian Groups I, Journal of Number Theory, 1(1) (1969), 8-10) proved using group rings. Thus you can not find 100 such bases, and even three bases. 


