# When do roots of three quadratic polynomials multiply to 1?

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.

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 $$c_3=0$$.

• 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! Jul 14 '20 at 18:33
• I am curious: what does $f\big(C(g)\big)$ mean? Jul 14 '20 at 18:35
• 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. Jul 14 '20 at 18:36
• $C(g)$ 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. Jul 14 '20 at 18:39

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.

• 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 $\tilde{q}\left(r_3^{\pm}\right)=0$. Jul 13 '20 at 18:08
• @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? Jul 13 '20 at 18:35
• If the first two equations are $$x^2+ax+b=0$$ and $$y^2+cy+d=0\,,$$ then the quartic is given by $$q(t):=t^4-act^3+(a^2d-2bd+bc^2)t^2+abcdt+b^2d^2\,.$$ Jul 13 '20 at 18:40
• I messed up the sign: $$q(t):=t^4-act^3+(a^2d-2bd+bc^2)t^2{\color{red}-}abcdt+b^2d^2\,.$$ Jul 13 '20 at 18:47