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Say I have a trio of quadratic polynomials $p_1,p_2,p_3$. Under what conditions will I have $r_1r_2r_3 = 1$ where each $p_i(r_i) = 0$?

In other words, when does the following nonlinear system have a solution?

$$ p_1(r_1) = 0, \ p_2(r_2) = 0, \ p_3(r_3) = 0, \ r_1r_2r_3 = 1 $$


First of all, I know that I can write out the three polynomials, solve them, multiply their coefficients, and set that equal to $1$ for a condition, but this turns out to be quite messy and includes $8$ different cases, corresponding to the choice of root from each polynomial. I'm hoping to find something a bit more elegant, if it exists.

I can see that this is equivalent to asking when there exists $r_1,r_2$ such that $p_3(r_1^{-1}r_2^{-1}) = 0$, or similarly for other combinations. However, this doesn't really change much as far as I can tell. Again, I can solve, invert, multiply, and substitute to get a condition, this time cleaner and with only $4$ cases, but it's still far messier than I'm hoping for.

This problem came up while looking for the conditions under which a bivariate quadratic has a factorization into two bivariate linears. I believe there is an equivalence between these two sets of conditions, so if there is a known condition for that, it should also be sufficient here.

I've poked around with this for quite a while trying to figure it out, but I haven't gotten anywhere. A particular set of equations I was working with is

$$ 6x^2−8x−1=0 \\ y^2−y−6=0 \\ z^2+3z+1=0 \\ xyz = 1 $$

I know that this does not have any solutions, but I can only show it by direct computation of the roots. This one isn't too bad since the $y$ equation has integer roots, but that's obviously not the case in general.

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Let $\mathbb{K}$ be a field with algebraic closure $\overline{\mathbb{K}}$. For constants $a_i,b_i,c_i\in\mathbb{K}$ for $i\in\{1,2,3\}$ such that none of $a_1$, $a_2$, and $a_3$ is equal to $0$, there exist $x_1,x_2,x_3\in\overline{\mathbb{K}}$ such that $x_1x_2x_3=1$ and $$a_i\,x_i^2+b_i\,x_i+c_i=0$$ for every $i=1,2,3$ if and only if

  • for no $i\in\{1,2,3\}$, $b_i=c_i=0$, and

  • the following riesengroße equality holds $$\begin{align}&a_1^4 a_2^4 c_3^4 +4 a_1^3 a_2^3 a_3 c_1 c_2 c_3^3 +a_1^3 a_2^3 b_1 b_2 b_3 c_3^3 -2 a_1^3 a_2^3 b_3^2 c_1 c_2 c_3^2 -2 a_1^3 a_2^2 a_3 b_2^2 c_1 c_3^3 \\ &\phantom{a}+a_1^3 a_2^2 b_2^2 b_3^2 c_1 c_3^2 -2 a_1^2 a_2^3 a_3 b_1^2 c_2 c_3^3 +a_1^2 a_2^3 b_1^2 b_3^2 c_2 c_3^2 +6 a_1^2 a_2^2 a_3^2 c_1^2 c_2^2 c_3^2 +a_1^2 a_2^2 a_3 b_1^2 b_2^2 c_3^3 \\&\phantom{aa}-5 a_1^2 a_2^2 a_3 b_1 b_2 b_3 c_1 c_2 c_3^2 -4 a_1^2 a_2^2 a_3 b_3^2 c_1^2 c_2^2 c_3 +a_1^2 a_2^2 b_1 b_2 b_3^3 c_1 c_2 c_3 +a_1^2 a_2^2 b_3^4 c_1^2 c_2^2 \\&\phantom{aaa} -4 a_1^2 a_2 a_3^2 b_2^2 c_1^2 c_2 c_3^2 +a_1^2 a_2 a_3 b_1 b_2^3 b_3 c_1 c_3^2 +a_1^2 a_3^2 b_2^4 c_1^2 c_3^2 -4 a_1 a_2^2 a_3^2 b_1^2 c_1 c_2^2 c_3^2 \\&\phantom{aaaa} +a_1 a_2^2 a_3 b_1^3 b_2 b_3 c_2 c_3^2 +4 a_1 a_2 a_3^3 c_1^3 c_2^3 c_3 -5 a_1 a_2 a_3^2 b_1 b_2 b_3 c_1^2 c_2^2 c_3 -2 a_1 a_2 a_3^2 b_3^2 c_1^3 c_2^3 \\&\phantom{aaaaa} +a_1 a_2 a_3 b_1^2 b_2^2 b_3^2 c_1 c_2 c_3 +a_1 a_2 a_3 b_1 b_2 b_3^3 c_1^2 c_2^2 -2 a_1 a_3^3 b_2^2 c_1^3 c_2^2 c_3 +a_1 a_3^2 b_1 b_2^3 b_3 c_1^2 c_2 c_3 \\&\phantom{aaaaaa} +a_1 a_3^2 b_2^2 b_3^2 c_1^3 c_2^2 +a_2^2 a_3^2 b_1^4 c_2^2 c_3^2 -2 a_2 a_3^3 b_1^2 c_1^2 c_2^3 c_3 +a_2 a_3^2 b_1^3 b_2 b_3 c_1 c_2^2 c_3 \\&\phantom{aaaaaaa}+a_2 a_3^2 b_1^2 b_3^2 c_1^2 c_2^3 +a_3^4 c_1^4 c_2^4 +a_3^3 b_1^2 b_2^2 c_1^2 c_2^2 c_3+a_3^3 b_1 b_2 b_3 c_1^3 c_2^3 \phantom{a}=\phantom{a}0\,.\tag{*}\end{align}$$

The idea is to note that the polynomial $q(t)$ with roots of the form $r_1^\pm r_2^\pm$, where $r_i^{+}$ and $r_i^-$ for $i=1,2,3$ are defined as in Somos's answer is given by $$q(t):=a_1^2a_2^2\,t^4-a_1a_2b_1b_2\,t^3+(a_1b_2^2c_1+a_2b_1^2c_2-2a_1a_2c_1c_2)\,t^2-b_1b_2c_1c_2\,t+c_1^2c_2^2\,.$$ (A proof of this claim can be inferred from Example V of this question.)

The polynomial $\tilde{q}(t)$ with roots $\dfrac{1}{r_1^{\pm}r_2^{\pm}}$ is given by $$\tilde{q}(t):=t^4\,q\left(\frac{1}{t}\right)=a_1^2a_2^2-a_1a_2b_1b_2\,t+(a_1b_2^2c_1+a_2b_1^2c_2-2a_1a_2c_1c_2)\,t^2-b_1b_2c_1c_2\,t^3+c_1^2c_2^2\,t^4\,.$$ Hence, there exist such $x_1$, $x_2$, and $x_3$ if and only if $\tilde{q}(r_3^+)=0$ or $\tilde{q}(r_3^-)=0$. Therefore, this is equivalent to saying $$a_3^8\,\tilde{q}(r_3^+)\,\tilde{q}(r_3^-)=0\,.$$ If $c_3\neq 0$, then the requirement is $$\frac{a_3^8}{c_3^4}\,\tilde{q}(r_3^+)\,\tilde{q}(r_3^-)=0\,,$$ which is precisely (*). If $c_3=0$, then $b_3\neq 0$ must hold and we need to check whether $\tilde{q}\left(-\dfrac{b_3}{a_3}\right)=0$, and this is equivalent to $$\dfrac{a_3^8}{b_3^4}\,\tilde{q}\left(-\dfrac{b_3}{a_3}\right)=0\,.$$ The equation above is precisely (*) when $c_3=0$.

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  • $\begingroup$ Nice! I also found a paper which claims that two polynomials $f,g$ are relatively prime iff at least one of $\text{det} \ f(C(g))$ and $\text{det} \ g(C(f))$ vanishes. I haven't done the working out but I suspect that applying that with $\tilde{q}$ and $p_3$ will likely yield the same results. I'm somewhat disappointed that there's not a clean way to get to the result, but that's math sometimes. Thanks for your answer! $\endgroup$
    – Alex Jones
    Jul 14 '20 at 18:33
  • $\begingroup$ I am curious: what does $f\big(C(g)\big)$ mean? $\endgroup$ Jul 14 '20 at 18:35
  • $\begingroup$ By the way, if Somos wrote the whole expansion of the product at the end, I think the answer is most likely the same as mine. $\endgroup$ Jul 14 '20 at 18:36
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    $\begingroup$ $C(g)$ is the companion matrix for the polynomial $g$: en.wikipedia.org/wiki/Companion_matrix. The paper I saw was this one: ieeexplore.ieee.org/document/1099465. And yes, I do believe the expansions are the same. I think your answer includes a bit more helpful detail which is why I chose it as the accepted one, but both answers are great. $\endgroup$
    – Alex Jones
    Jul 14 '20 at 18:39
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Your question asks if you have three quadratic polynomials $$ p_1(x)\!:=\!a_1 x^2+b_1 x+c_1, \;\; p_2(x)\!:=\!a_2 x^2+b_2 x+c_2, \;\; p_3(x)\!:=\!a_3 x^3+b_3 x+c_3 $$ with three pairs of roots $$ p_1(r_1^+) = p_1(r_1^-) = 0, \quad p_2(r_2^+) = p_2(r_2^-) = 0, \quad p_3(r_3^+) = p_3(r_3^-) = 0 $$ where for $\,n=1,2,3,\,$ $$ r_n^{\,\pm} := \frac{-b_n\pm\sqrt{b_n^2-4a_n c_n}}{2a_n}, $$ then what is the condition that $\, r_1 r_2 r_3 = 1\,$ for some choice of the roots as given in terms of the coefficients of the three polynomials? The answer is given by a homogeneous degree $12$ polynomial expanded out with $34$ monomial terms $$ P := (a_1a_2a_3)^4 \prod_{i,j,k=\pm} (1 - r_1^{\,i}\,r_2^{\,j}\,r_3^{\,k}) = (a_1a_2a_3)^4 + \dots + (c_1c_2c_3)^4 $$ where the $\,\dots\,$ represents the other $32$ degree-$12$ monomial terms. I used a computer algebra system to get the expansion. As stated in the question

this turns out to be quite messy

and I don't think it can be simplified except for special cases, but I have been wrong before, so maybe there is hope.

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  • $\begingroup$ I think the process can be simplified somewhat. Assuming that $a_1$ and $a_2$ are nonzero, there exists a monic quartic $q(x)$ whose roots are of the form $r_1^{\pm}r_2^{\pm}$. The coefficients of $q(x)$ can be written easily in terms of $a_i$, $b_i$, and $c_i$ for $i\in\{1,2\}$. All that is left is to check whether $\tilde{q}(x):=x^4\,q\left(\dfrac{1}{x}\right)$ has a common root with $p_3(x)$. One can perform a Euclidean algorithm to find the greatest common divisor between $\tilde{q}(x)$ and $p_3(x)$, for example. Or simply check whether $\tilde{q}\left(r_3^{\pm}\right)=0$. $\endgroup$ Jul 13 '20 at 18:08
  • $\begingroup$ @Batominovski Can you elaborate on how to "easily" write the coefficients of $q$? Do you mean something other than expanding the product of the four roots? $\endgroup$
    – Alex Jones
    Jul 13 '20 at 18:35
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    $\begingroup$ If the first two equations are $$x^2+ax+b=0$$ and $$y^2+cy+d=0\,,$$ then the quartic is given by $$q(t):=t^4-act^3+(a^2d-2bd+bc^2)t^2+abcdt+b^2d^2\,.$$ $\endgroup$ Jul 13 '20 at 18:40
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    $\begingroup$ I messed up the sign: $$q(t):=t^4-act^3+(a^2d-2bd+bc^2)t^2{\color{red}-}abcdt+b^2d^2\,.$$ $\endgroup$ Jul 13 '20 at 18:47

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