I'm trying to figure out the number of solutions to a particular system of equations. The variables here are $A_1,A_2,A_3,B_1,B_2,B_3,C_1,C_2,C_3$, all in $\mathbb{F}_2$, constraint to

(a) $B_1^2 = A_1 C_1$,

(b) $A_1 C_2 + A_2 C_1 = 0$, and

(c) $A_1 C_3 + A_3 C_1 = 0$.

This problem actually relates to something I'm doing in algebraic geometry right now--I just suck at counting, so forgive me if they I have been approaching this isn't optimal. Here's what I've got so far:

$B_2, B_3$ are free, so that is 4 options.

I split the problem into two main cases, case 1 being when $B_1 = 1$ and when $B_1 = 0$.

Case $B_1 = 1$: Then $A_1 = C_1 = 1$ by (a). By (b), we must then have $A_2 + C_2 = 0$, and since we are in $\mathbb{F}_2$, this means that $A_2 = C_2$, so we have 2 options here. Likewise we must have $A_3 = C_3$ by (c), giving us 2 more options, independent from what we did in (b). So for this case, we have

(4 [from $B_2, B_3$]) $\times$ (2 [from $A_2, C_2$]) $\times$ (2 [from $A_3, C_3$]) = 16.

So it is just the case when $B_1^2 = 0$ that I am having difficulty with, as this implies either $A_1 = 0$ or $C_1 = 0$ (or both), splitting the problem into many sub-cases whhich might intersect.

  • $\begingroup$ Of course you might as well replace $B_1^2$ with $B_1$ throughout, in $\mathbb F_2$. $\endgroup$ May 15, 2013 at 23:44

1 Answer 1


I don't think the $B_1=0$ subcases are that bad. If $A_1=C_1=0$, then $A_2,A_3,C_2,C_3$ can be whatever they want. If $A_1=0$ but $C_1=1$, then $A_2=A_3=0$ but $C_2,C_3$ can be whatever. The case $C_1=0$ but $A_1=1$ is similar.

(If you really just care about this one specific problem, you can check all $2^9=512$ options by brute force!)

  • $\begingroup$ I wrote a Python program to do this by brute force and found 128 solutions. Hope this is correct. $\endgroup$
    – Chris
    May 15, 2013 at 23:51

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