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I'm trying to figure out a way to solve a system of equations where variables are multiplied together. Fairly similar to this question, however the example used seems simpler than what I want to do.

As an example, let's say I have 3 variables and 3 equations. All my equations are in the form:

$$c_1x + c_2y + c_3z + c_4xy + c_5xz + c_6yz + c_7xyz = 0$$

Constants can be either $0$ or a positive whole number. Variables are either $0$ or $1$.

I've been trying to figure out if/how to use substitution and/or elimination. I've also seen some solutions that replace the multiplied variables with a single new variable. I'm not entirely sure whether that would help or how to properly do it.

How should I go about solving this system?

Edit 1: I looked back at the problem I'm trying to solve and the constants can be negative.

Dan
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  • If your variables can only be zero or one you onla have èight possible solutions. It is trivial to check them through. – Kurt G. May 14 '22 at 16:10
  • You can solve such systems of equations over, say, real or complex numbers by using resultants or Groebner bases. In your case it is over $\Bbb F_2$, which is trivial as Kurt said. But may be you want to consider real numbers instead. Can you write down the three equations in $x,y,z$? Also, are your constants fixed? If not, they are further variables. – Dietrich Burde May 14 '22 at 16:12
  • Potentially there are a lot more variables. Enough that I don't think checking all combinations is going to work. – Dan May 14 '22 at 16:33
  • @DietrichBurde when you say write down the equations in x,y,z are you wanting the actual equations? I don't know the actual equations. I guess the best explanation is that a system of equations can be given to me and I need to solve it. Essentially I want to write a computer program to do this. Constants are fixed. – Dan May 14 '22 at 16:37
  • I think it's going to be over 30 variables. I assumed that would be too slow for my needs, but I probably should test it just to see. Maybe it won't be. I'll Google for those and see if they can help. Thanks. – Dan May 14 '22 at 17:01
  • Actually, you could first consider new variables $x,y,z$, and $z_1=xy$, $z_2=xz$, $z_3=yz$ and $z_4=xyz$, then your system consists of linear equations. Solve this first, then the rest is much easier. – Dietrich Burde May 14 '22 at 17:10
  • @DietrichBurde, that's actually the path I was thinking about trying, but I haven't found a good explanation on what to do after I introduce new variables. With introducing the new variables, the original equations become linear, but now I have more variables than equations correct? – Dan May 14 '22 at 17:22
  • Why do you have more variables than equations? I count less than $10$ variables, and you have $100000$ equations? It is the other way around, probably. You see, therefore it would be good if you give a meaningful explicit example, which you want to know (and why). You cannot solve arbitrary such systems. It really depends on your choices. – Dietrich Burde May 14 '22 at 17:23
  • I think where I'm getting confused is where/how the equations created for the new variables come in. So, in the original example, I start with 3 variables and 3 equations. After creating variables, I now have 3 linear equations in 7 variables. And then I have 4 new equations which are in 7 variables. At least that's my understanding. – Dan May 14 '22 at 18:50
  • I guess the question I really have is whether the 4 new equations are now part of my system of equations? I tried an example problem a few days ago like this and it didn't really figure out like I thought it would. – Dan May 14 '22 at 18:58
  • @Dan "Constants can be either $0$ or a positive whole number. Variables are either $0$ or $1$" $;$ A sum of non-negative terms can only be zero iff each term is zero. If $,c_1 \ne 0,$ then $,x=0,$, and similar for $,c_2, c_3,$. Otherwise if $,c_1 = c_2 = c_3 = 0,$ and $,c_4 \ne 0,$ then $,xy=0,$ so the problem breaks down into solving two systems for $,x = 0,$and $,y=0,$ respectively, and similarly for $,c_5, c_6,$. Otherwise $\cdots$ – dxiv May 15 '22 at 02:54
  • I may have been mistaken about he constants being only positive. I believe they can be negative. – Dan May 16 '22 at 03:03

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