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Let $c_0, \cdots, c_{n-1}$ be elements of $\mathbb{F_{2^k}}$, find the sum $\sum\limits_{0\leq i < j <n}c_ic_j$.

It is true that if $k=1$ and $d$ be the number of non-zero elements, then $$\sum\limits_{0\leq i < j <n}c_ic_j=\bigg(\begin{matrix}d\\2\end{matrix}\bigg) \mod 2$$

What happens if $k>1$.

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  • $\begingroup$ I don't get it. Pretty much anything might happen. With $d=2$ you can get any non-zero element $x$ of the field $\Bbb{F}_{2^k}$ as the result by selecting $c_0=1,c_1=x$, and $c_i=0$ for all $i\ge2$. Could you make this clearer? Where does this come from? $\endgroup$ Mar 24, 2015 at 12:27
  • $\begingroup$ It is the second coefficient of $(x-c_1)\cdots (x-c_n)$. $\endgroup$
    – vudu vucu
    Mar 24, 2015 at 12:33
  • $\begingroup$ Yes. What do you want to know about it? $\endgroup$ Mar 24, 2015 at 12:38
  • $\begingroup$ As you said, we can find any elements. I think we can just count this numbers, "$|sum=x|$" but can not find a class of elements such $sum=x$. But maybe we can.. I am not sure about it.. $\endgroup$
    – vudu vucu
    Mar 24, 2015 at 12:46

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