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let $F$ be a field of $p^b$ elements, $p$ prime and $b \in \mathbb{N}$. Suppose I have $(a_1, a_2, \ldots, a_s) \in F^s$ and an equation $$ 0 = a_1 x_1 + \dots + a_s x_s. $$ I was wondering if anybody could help me figure out what the number of solution $(x_1, x_2, \dots, x_s) \in F^s$ to this equation would be? (or estimate if exact number not available) Thanks!

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The set of solutions is in fact a subspace of $F^s$ (as is easy to check or known from linear algebra). As a vector space of dimension $n$ has $(p^b)^n$ elements it suffices to compute the dimension of the space of solutions. There are essentially two cases:

  1. $a_i=0$ for all $i$. In this case all elements of $F^s$ are solutions, i.e. the space of solutions is $s$-dimensional and hence has $p^{bs}$ elements.
  2. There is an $a_i\neq 0$. Then the formula can be rewritten as $$x_i=a_i^{-1}(\sum_{j\neq i}a_jx_j)$$ Thus the space of solutions is $(s-1)$-dimensional and hence has $p^{b(s-1)}$ elements.
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Hint:

The set of solutions to $\,a_1x_2+\ldots+a_sx_s=0\;$ is a subspace of dimension $\,s-1\,$ in $\,\Bbb F^s\,$ (assuming at least one$\,a_i\neq0\,$ ) ...

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  • $\begingroup$ Thank you very much! This would have been fine with me too, but there was a solution below.. $\endgroup$ – Tom Mosher Jun 25 '13 at 23:05
  • $\begingroup$ I know: sometimes there are hints for the OP to do some of the job, sometimes the whole solution is given. You're welcome.:) $\endgroup$ – DonAntonio Jun 26 '13 at 5:36

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