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I am trying to solve this problem:

Let $\alpha_1x_1 + \alpha_2 x_2 + \alpha_3x_3 + \alpha_4x_4 + \alpha_5 = 0$ and $x_1x_2+x_3x_4=0$ be equations over field of size $2$. Show that we can't choose $\alpha_1,\ldots,\alpha_5$ so that the first equation has solution and second equation holds for each solution of first.

I need hint to prove this.

Ashot
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    Well, $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$, and $\alpha_5 = 1$ is such a choice. You probably want a condition to guarantee that the first has solutions. – Daniel Fischer Mar 16 '14 at 11:54
  • @DanielFischer Sure, corrected. – Ashot Mar 16 '14 at 11:56
  • How did this question arise, please? – Gerry Myerson Mar 16 '14 at 11:58
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    You don't have many cases to consider, and the main issue is organising them to cover everything. – Mark Bennet Mar 16 '14 at 12:01
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    @GerryMyerson I am finding minimal number of cosets to cover solutions of this equation $x_1 x_2 + x_3 x_4 + x_5 x_6$ – Ashot Mar 16 '14 at 12:10
  • If the first equation has solutions, either all $\alpha_i$ are $0$, in which case it's easy to prove, or one of the $\alpha_i,, i \leqslant 4$ is $1$. Up to a renumbering, we can assume $\alpha_1 = 1$. Then what can you say about the set of solutions of the first equation? – Daniel Fischer Mar 16 '14 at 12:20
  • @DanielFischer it will be subspace of dimension $3$. Will have $8$ elements, but why for some elements second equation doesn't hold? – Ashot Mar 16 '14 at 12:27
  • How can you parametrise the eight elements? – Daniel Fischer Mar 16 '14 at 12:28
  • This is probaly related to general properties of bent functions. At least I strongly suspect that $x_1x_2+x_3x_4$ being bent should imply the claim. Bent functions are (w.r.t. Hamming distance) as far away from all linear functions as can be, and it does loook like that should come to the fore here. I'm too tired to check everything now. – Jyrki Lahtonen Mar 16 '14 at 20:07

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