# How to Find Solutions to a Multivariate Polynomial System

I have a system of polynomials, where the first one is a multivariate linear polynomial, but the rest are univariate quadratic polynomials. How would I solve such a system (finding one or all solutions, or showing there are no solutions)? For example,

$$17x+16y-5z-67=0 \\ x^2+3x-5=0 \\ 4y^2-7y-4=0 \\ z^2-6z-3=0$$

## 4 Answers

The system has no solution. This can be seen by computing a Groebner basis, for example. But also a direct approach is possible. We can compute $$x,y,z$$ from the second, third and last equation (two solutions each) and then substitute it into the first one. Even if we would replace the first equation by $$17x+16y-5z-a=0$$ where $$a$$ is an integer, there is no solution.

• How would you compute a Groebner basis? Also, since polynomials have at most two solutions, how would you know which solutions work? Must you just substitute each one and check? Aug 4, 2020 at 14:23
• Yes, indeed, we would need to check every combination. For Groebner bases do a google search. The Buchberger algorithm gives $\{1\}$ as Groebner base, so no solutions. Aug 4, 2020 at 14:24
• I looked at Wikipedia, which said "a the system is inconsistent if this Gröbner basis is reduced to 1", but I don't understand how to reduce it? Does this also mean that if it isn't reduced to 1, it must have solutions? Aug 4, 2020 at 14:29
• Where are you getting those $i$'s? The roots of the quadratics are all real. Aug 4, 2020 at 14:30
• @RobertIsrael This was a typo, sorry. I had $x^2+3x+5=0$. Aug 4, 2020 at 14:34

Well, one way is to just throw it into Wolfram. But, the procedure in this case is quite simple. Start by finding all solutions of the bottom three single variable equation. Then see if any combinations of solutions from these equations produce a solution to the first equation. In this case no such combinations exist.

• Since polynomials have at most two solutions, how would you know which solutions work? Must you just substitute each one and check? Aug 4, 2020 at 14:23
• @DUO Basically it's just guess and check. In general most nonlinear systems don't have straightforward, systematic solution methods. Aug 4, 2020 at 15:10

The discriminants of the three quadratics are $$29$$, $$113$$ and $$48 = 4^2 \cdot 3$$ respectively. A linear combination of $$x$$, $$y$$, $$z$$ and $$1$$ over the rationals with some nonzero coefficients is a linear combination of $$\sqrt{29}$$, $$\sqrt{113}$$, $$\sqrt{3}$$ and $$1$$ over the rationals with some nonzero coefficients. But in fact, since $$29$$, $$113$$ and $$3$$ are distinct primes, $$\sqrt{29}$$, $$\sqrt{113}$$, $$\sqrt{3}$$ and $$1$$ are linearly independent over the rationals: see e.g. this.

We have $$0=4(x^2+3x-5)+(4y^2-7y-4)+(z^2-6z-3)-(17x+6y-5z-67)=$$ $$=4x^2-5x+4y^2-13y+z^2-z+40=$$ $$=\left(2x-\frac{5}{4}\right)^2+\left(2y-\frac{13}{4}\right)^2+\left(z-\frac{1}{2}\right)^2+\frac{221}{8}>0,$$ which says that our system has no real solutions.

• This is interesting. What is it called? Is it some form of the determinant? Aug 4, 2020 at 15:09
• @DUO No. Since the number $67$ is a big enough, I just tried to get $0$ as sum of squares plus something positive and it turned out. Aug 4, 2020 at 15:11
• You just made me think of something. How about we combined the squares of all the polynomials and tried to find solutions to that? Would that work? Aug 4, 2020 at 15:15
• @DUO I think in the general we also can try to make something like this. But it would be not so easy. Aug 4, 2020 at 15:28