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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,

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

DUO Labs
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4 Answers4

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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.

Dietrich Burde
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  • 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? – DUO Labs Aug 04 '20 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. – Dietrich Burde Aug 04 '20 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? – DUO Labs Aug 04 '20 at 14:29
  • Where are you getting those $i$'s? The roots of the quadratics are all real. – Robert Israel Aug 04 '20 at 14:30
  • @RobertIsrael This was a typo, sorry. I had $x^2+3x+5=0$. – Dietrich Burde Aug 04 '20 at 14:34
  • @DietrichBurde Ok, one question-- what is the computational complexity of this reduction in terms of number of equations and/or variables? – DUO Labs Aug 04 '20 at 14:38
  • Buchberger's algorithm is doubly exponential. So very slow if you have many equations. – Dietrich Burde Aug 04 '20 at 14:42
  • @DietrichBurde Oh, that would be a problem. Oh well, thanks for you help! +1 and accepted! – DUO Labs Aug 04 '20 at 14:45
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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.

K.defaoite
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  • Since polynomials have at most two solutions, how would you know which solutions work? Must you just substitute each one and check? – DUO Labs Aug 04 '20 at 14:23
  • @DUO Basically it's just guess and check. In general most nonlinear systems don't have straightforward, systematic solution methods. – K.defaoite Aug 04 '20 at 15:10
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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.

Robert Israel
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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.

Michael Rozenberg
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