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Many sources states that a system of equations could be solved by polynomials with degrees as worst as the product of degrees of the original SOE, but I wonder if this claim ever has a non-trivial numerical example.

Take for example, the system of equation below: $$\begin{cases}2x^2-3y^2+z^2-2xy+yz+4zx-x+y+5z+3=0\\8x^2+y^2-7z^2+5xy+2yz-3zx+2x+y-3z+10=0\\x^2-2y^2+z^2-8xy+zx+5x-3y+8z+4=0\end{cases}$$

I tried to take the resultant of the first & second, second & third to eliminate $y$, then the resultant of the new quartics to eliminate $z$, which I obtained a 16 degree equation in $x$: $$-1181348112832836544x^{16}+2787121601073141696x^{15}+40158391608794922720x^{14}+167365623087559226016x^{13}+425757328708669625092x^{12}+770691543273712475900x^{11}+998726850263304276512x^{10}+932049443279171335256x^9+596362292653912157020x^8+188749123583761008004x^7+61344849084516069268x^6+103877159608205310252x^5+165228427446941144592x^4+153143321851661654352x^3+3605328434835159708x^2+12416051024495562700x+1687817147597036192=0$$ I know this factors to $$- 4 \left(161964076 x^{8} + 775805376 x^{7} + 1744839233 x^{6} + 2307601210 x^{5} + 2166944383 x^{4} + 1176234951 x^{3} + 110354632 x^{2} + 99386663 x + 12082888\right) \left(1823472436 x^{8} - 13036471860 x^{7} - 19186348433 x^{6} - 51973458301 x^{5} - 40189995271 x^{4} - 2648883622 x^{3} - 5304407431 x^{2} + 30351410271 x - 34921641821\right)=0$$ But how do I even factorize it in the first place by hand without knowing its factor? Or is there a nicer way to transform the system above into octics?

Thinh Dinh
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    My Gröbner basis gives immediately $z=f(x)$, $y=g(x)$ with polynomials $f,g\in \Bbb Q[x]$ of degree $7$, where $x$ has to satisfy the equation $h(x)=0$ with $$h(t)=161964076t^8 + 775805376t^7 + 1744839233t^6 + 2307601210t^5 + 2166944383t^4 + 1176234951t^3 + 110354632t^2 + 99386663t + 12082888$$ The resultant introduces other factors, yes. But they do not lead to a solution. – Dietrich Burde Apr 13 '25 at 11:26

2 Answers2

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One way systems of polynomial equations can sometimes be solved is to compute a Gröbner basis for the ideal generated by the polynomials whose vanishing defines the system. The Magma script below will do this for your system. If you copy and paste it into the online Magma calculator and run it, it returns a Gröbner basis of the form $$ \big\{x-p_1(z),y-p_2(z),p_3(z)\big\}\ , $$ where \begin{align} p_3(z)\stackrel{\text{def}}{=}&\,z^8 - \frac{270923793}{80982038}z^7 - \frac{299897515}{161964076}z^6 + \frac{723130957}{40491019}z^5\\ &-\frac{141531269}{40491019}z^4 - \frac{5114911831}{161964076}z^3+ \frac{452759481}{80982038}z^2\\ &+\frac{519821722}{40491019}z - \frac{487452147}{80982038}\,^\color{red}{\dagger} \end{align} and $\ p_1,p_2\ $ are polynomials of degree $\ 7\ .$ The polynomial $\ p_3\ $ and its derivative $\ p_3^{\,\prime}\ ,$ are relatively prime, so the equation $\ p_3(z)=0\ $ has eight distinct complex solutions, $\ z=z_i\ $ for $\ i=1,2,\dots,8\ ,$ and your original system has eight solutions: $$ (x,y,z)=\big(p_1\big(z_i\big),p_2\big(z_i\big),z_i\big) $$ for $\ i=1,2,\dots,8\ .$ According to Wolfram alpha, the equation $\ p_3(z)=0\ $ has two real roots and three pairs of complex conjugate roots, so your system has just two real solutions, $$ \big(p_1\big(r_1\big),p_2\big(r_1\big),r_1\big),\ \big(p_1\big(r_2\big),p_2\big(r_2\big),r_2\big)\ ,$$ where $\ r_1,r_2\ $ are the two real roots of the equation $\ p_3(z)=0\ .$

Q := RationalField();
P<x, y, z> := PolynomialRing(Q, 3);
I := ideal<P | 2*x^2-3*y^2+z^2-2*x*y+y*z+4*z*x-x+y+5*z+3,
           8*x^2+y^2-7*z^2+5*x*y+2*y*z-3*z*x+2*x+y-3*z+10,
           x^2-2*y^2+z^2-8*x*y+z*x+5*x-3*y+8*z+4>;
B := GroebnerBasis(I);
B;
P<z> := PolynomialRing(Q);
b:=  z^8 - 270923793/80982038*z^7 - 299897515/161964076*z^6 +
    723130957/40491019*z^5 - 141531269/40491019*z^4 -
    5114911831/161964076*z^3 + 452759481/80982038*z^2 + 519821722/40491019*z
    - 487452147/80982038;
d:=Derivative(b);
Gcd(b,d);
161964076*x^8 +775805376*x^7+1744839233*x^6+2307601210*x^5+2166944383*x^4+1176234951*x^3+110354632*x^2+99386663*x+12082888 in I;

$\color{red}{\dagger}$ By not reading Dietrich Burde's comment carefully enough, I was puzzled for a lot longer than I should have been by the fact that $\ p_3\ $ isn't a rational multiple of the polynomial $\ h\ $ given in that comment, and the first factor of your degree $16$ polynomial. Reading the comment more carefully would have told me that $\ h(\color{red}{x})\ $ should lie in the ideal $$\big\langle x-p_1(z),y-p_2(z),p_3(z)\ \big\rangle\ ,$$ and the last line of the above Magma script confirms that this is indeed the case.

lonza leggiera
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  • (+1) Nice answer, but can you explain how the code works? I have explored how Groebner basis works but don't understand how the rest works. – Thinh Dinh Apr 14 '25 at 04:46
  • Thank you. I'm not sure exactly what you're asking for clarification of. The reason for computing a Gröbner basis is that the set of complex numbers on which the polynomials in the Gröbner basis all vanish is the same as the set on which the original set of polynomials all vanish, so the set of solutions of your original system of polynomial equations is the same as the set of solutions of $$x-p_1(z)=0\y-p_2(z)=0\p_3(z)=0\ ,$$ and these can be solved by finding all solutions of the third and then substituting these solutions for $\ z\ $ into the first two. – lonza leggiera Apr 14 '25 at 05:29
  • Thank you for the comment, I'll dive into the code for some time, after that, I'll tick – Thinh Dinh Apr 14 '25 at 05:34
  • The last line of the code simply checks whether $\ h(x)\ $ lies in the ideal generated by the polynomials on the left side of your original system. The preceding three lines merely calculate the gcd of $\ p_3\ $ and $\ p_3^{,\prime}\ ,$ which turns out to be $\ 1\ ,$ thus telling you that $\ p_3(z)=0\ $ has $8$ distinct roots. – lonza leggiera Apr 14 '25 at 05:37
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Factoring a 16th-degree polynomial like the one you obtained by hand without knowing its factors seems impractical. You could try the Rational Root Theorem, testing factors of the constant term $(1687817147597036192)$ divided by factors of the leading coefficient $(−1181348112832836544)$, but the large coefficients and high degree make this tedious and unlikely to yield simple roots. As suggested in the comments by Dietrich Burde, the Gröbner basis minimizes the degree of the polynomial in $x$ to the “true” degree of the system’s solution set, which is 8 here, reflecting the maximum number of real solutions (per Bézout’s theorem for three quadratic). Unlike resultants, the Gröbner basis avoids extraneous factors, directly giving $h(x)$, which is one of your octic factors. The resultant’s 16th-degree polynomial includes $h(x)$ multiplied by an extraneous octic, explaining the degree doubling.

Manually, you could try linear combinations of the equations to eliminate terms and reduce variables, but this is trial-and-error and less reliable. Long story short, without computational tools or specific insights (e.g., symmetry), it’s not feasible to factor into octics manually.

Random Math Enthusiast
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