How to recover the roots $a$, $b$, $c$, $d$ of a quartic from the expressions $ab+cd$, $ac+bd$ and $ad+bc$?

Question

I was looking at the method of solving a quartic $x^4+px^3+qx^2+rx+s=0$ by the following Lagrange's resolvent: $$(y-(ab+cd))(y-(ac+bd))(y-(ad+bc))$$ where $a$, $b$, $c$, $d$ are the roots. I had a look at the answers to this question. I understood how we can obtain the values of $ab+cd$, $ad+bc$ and $ac+bd$ from solving the above cubic, and how it works etc. However, I have trouble understanding how one obtains the roots $a$, $b$, $c$, $d$ from these quantities. This answer to the same question describes a method to do so, which I will describe in my own words here (without depressing the quartic):

Consider $(a+b)^2+(c+d)^2=a^2+b^2+c^2+d^2+2(ab+cd)$. This can be rewritten as $$(a+b)^2+(p+(a+b))^2=(p^2-2q)+2(ab+cd)$$ which, on expanding, gives you a quadratic in $a+b$. Hence, one can recover the values of $a+b$ and $c+d$ as the roots of this quadratic. Similarly, we can recover $a+c$, $b+d$, $a+d$ and $b+c$.

I understand that given these values (and $a+b+c+d=-p$), we can find $a$, $b$, $c$ and $d$. However, I am not sure how one makes the choice of which root is which. I tried to find some sort of consistency relations between the roots (so as to describe how to make these choices), but couldn't find any.

So, that is pretty much my question: how do we choose which one is $a+b$, and which one is $c+d$, etc.?

Answer 1

Let $$P+Q=R+S=T+U=-p$$ And consider $$a+b=P,c+d=Q,a+c=R,b+d=S,a+d=T,b+c=U$$ If we swap $a,b$ with $c,d$, we would arrange from $(P,Q,R,S,T,U)$ to $(P,Q,S,R,U,T)$, which is just arrange $2$ tuples $(R,S)$ and $(T,U)$. By the same method we can arrange $2$ in $3$ tuples $(P,Q),(R,S),(T,U)$, which covers $4$ cases.

Similarly, consider $$a'+b'=\color{red}Q,c'+d'=\color{red}P,a'+c'=R,b'+d'=S,a'+d'=T,b'+c'=U$$ By the same method, we could arrange $2$ in $3$ tuples $(Q,P),(R,S),(T,U)$, which covers $4$ cases.

So it suffices to pick randomly $(P,Q,R,S,T,U)$ and then swapping to get $(Q,P,R,S,T,U)$, which is $$a+b=P,a+c=R,a+d=T$$ Or $$a+b=\color{red}Q,a+c=R,a+d=T$$ By adding them up and subtracting $a+b+c+d$, we either get $$a=\frac{P+R+T+p}{2}$$ As a solution, or $$a=\frac{\color{red}Q+R+T+p}{2}$$ By substituting into the quartic, we would get $1$ of them, and then just solve for $b,c,d$ using the SOE above.

An example $$x^4-10x^3-8x^2+138x+135=0$$ $$ab+cd+ac+bd+ad+bc=-8,$$ $$(ab+cd)(ac+bd)+(ab+cd)(ad+bc)+(ac+bd)(ad+bc)\\=(abc+abd+acd+bcd)(a+b+c+d)-4abcd\\=-138\cdot 10-4\cdot 135=-1920,$$ $$(ab+cd)(ac+bd)(ad+bc)\\=(abc+abd+acd+bcd)^2+abcd\left((a+b+c+d)^2-4(ab+ac+ad+bc+bd+cd)\right)\\=(-138)^2+135\cdot\left(10^2-4\cdot(-8)\right)=36864$$ So $ab+cd,ac+bd,ad+bc$ are roots of the cubic $$y^3+8y^2-1920y-36864=0$$ $$\implies ab+cd=-32,ac+bd=-24,ad+bc=48$$ Consider $$(a+b)^2+(-10+a+b)^2=10^2-2\cdot(-8)+2\cdot(-32)=52$$ $$\implies a+b=6$$ Or $$a+b=4$$ Consider $$(a+c)^2+(-10+a+c)^2=10^2-2\cdot(-8)+2\cdot(-24)=68$$ $$\implies a+c=8\,\text{(picked randomly)}$$ Consider $$(a+d)^2+(-10+a+d)^2=10^2-2\cdot(-8)+2\cdot 48=212$$ $$\implies a+d=-4\,\text{(picked randomly)}$$ So we either get $$a=\frac{6+8-4-10}{2}=0$$ Or $$a=\frac{4+8-4-10}{2}=-1$$ By checking, we find $a=-1$, so solving for $b,c,d$: $$b=4-(-1)=5,c=8-(-1)=9,d=-4-(-1)=-3$$ Therefore $x\in\{-3;-1;5;9\}$.