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I know that it is possible to find all roots of a cubic using the cubic formula. Now when I have a cubic $P(x)$, I create the simultaneous equations:

$$P(x)=y \quad\text{and}\quad P(y)=x$$

I want to find all the roots using just algebraic manipulation and cubic formula.

My previous work: I know at most $3$ roots lie on $x=y$, and therefore I will set $P(x)=x$ and that is a cubic, which I can solve.

But I don't know how to get the remaining $6$ roots within the constraints of the problem. Can someone help me?

dxiv
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 21 '23 at 00:07
  • I know how to find cubic roots, so I need to solve the equation system using only cubic roots and algebraic manipulation/ –  Mar 21 '23 at 02:06
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    @dxiv For a general quadratic the Galois group remains $C_2$ (starting from $x^2 + bx + c$ the new discriminant is $(b - 1)^2 - 4(c + 1)$). – ronno Mar 21 '23 at 03:42
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    @WEIMINGCHEN You did not explain the motivation behind the question. For all practical purposes, the answer would be to solve it numerically. Any algebraic solution, even if it turns out that one exists (which hasn't been established yet), would be very painfully complicated. – dxiv Mar 21 '23 at 05:16

2 Answers2

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[tl;dr] $\;$ The resulting sextic is solvable, but the calculations are extremely laborious, so for all practical purposes the answer would be to solve it numerically.

Let the general depressed cubic be $P(x) = x^3 + a x + b$ then the period-$2$ periodic points of $P$ other than the fixed points are the roots of the sextic:

$$ \frac{P\left(P(x)\right) - x}{P(x) - x} = x^6 + (2 a + 1) x^4 + 2 b x^3 + (a^2 + a + 1) x^2 + (2 a + 1) b x + a + b^2 + 1 $$

The Tschirnhaus transformation $\,z = x^3 + (a - 1) x + b\,$ reduces it to the cubic in $t = z^2\,$:

$$ t^3 + 6 a \,t^2 + 3 (3 a^2 - 4 a - 4) \,t + 4 a^3 - 12 a^2 + 27 b^2 + 16 $$

It follows that the original sextic is solvable.

It would be technically possible to solve the latter cubic, then for each of the three $t$ roots calculate the corresponding $z = \pm t^{1/2}$ and solve each of the six Tschirnhaus cubics. That would give $18$ potential $x$ roots for the original sextic, out of which the $6$ actual roots would need to be identified and retained. Or, once established that the sextic is in fact solvable, follow the reference quoted on wikipedia's page on sextics T. R. Hagedorn, General formulas for solving solvable sextic equations, J. Algebra 233 (2000), 704-757, to the tune of 54 loaded pages. Either way, the exact calculations are very heavy, so numerical root-finding would be preferred in practical applications.

[ EDIT ] $\;$ Finding the Tschirnhaus transformation was essentially trial and error. I thought I'd look for a bicubic (if that's what a cubic in $z^2$ can be called) and started with a depressed cubic substitution $z = x^3 + u x + v$. Canceling the $x^5$ term gave $v = b$ and canceling the $x^3$ term required $u \in \{a-1, a+1/2, a+2\}$. Out of the three choices, $u=a-1$ turned out to cancel the linear term as well.

[ EDIT #2 ] $\;$ Another Tschirnhaus transformation that settles the question of solvability even more directly is $z = x^3 + (a+1)x +b$, which reduces the sextic to (the square of) a cubic:

$$ z^3+(a+2)z-b $$

In light of the related question Irreducibility and Galois group of the 2 -periodic points of a (cubic) polynomial it may be worth noting that something similar holds true for the reduced quartic $P(x)=x^4+ax+b$. In that case $\frac{P\left(P(x)\right)-x}{P(x)-x}$ has degree $12$, and the substitution $z=P(x)+x$ reduces it to (the square of) a sextic.

dxiv
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  • [+1} Very very interesting. – Jean Marie Mar 21 '23 at 07:09
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    A particular case found using approach0 once more. Interesting elements also in the answers to this question but for the quadratic case. – Jean Marie Mar 21 '23 at 07:55
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    As for lower bounds, I think $x \mapsto P(x)$ is an order $2$ element of $G$. – ronno Mar 21 '23 at 10:26
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    More empirics: let's call the combined polynomial $Q$. Then $Q$ is irreducible iff $R(x) = x^3 + (a + 2)x + b$ is. Let $G = \operatorname{Gal}(Q)$ and $H = \operatorname{Gal}(R)$. Then $G$ is one of: $C_2 \times S_4$, $C_2 \times A_4$, $S_4$ and $D_6 = C_2 \times S_3$. I am using the group name database notation, so $D_6$ has $12$ elements. If $G = C_2 \times S_4$ then $H = S_3$. When $G$ is smaller I see no correlation, except if $a = -1$ then $G \cong D_6$ and $H = A_3$. – ronno Mar 21 '23 at 11:07
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    @JeanMarie Thank you, and thanks for the pointers. The general cubic case looks to be a lot more challenging, though. A constructive answer could be to find a clever (or lucky ;-)) Tschirnhaus transformation of degree $\le 4$ to bring it down to a cubic or a quadratic in powers of the auxiliary variable, or an otherwise reducible sextic. But that's neither trivial to search, nor guaranteed to find. – dxiv Mar 21 '23 at 23:23
  • @ronno Thank you, interesting findings. Don't think it helps here, but speaking of discriminants which you mentioned in a previous comment, the original cubic has $\Delta_P=-(4a^3+27b^2)$ while the associated sextic has $\Delta_Q=-\left(4(a+1)(a-2)^2+27 b^2\right) \left(4(a+2)^3+27b^2\right)^2$. FWIW an alternative way to write it is $Q(x) = P^2(x)+x P(x)+x^2+P'(0)+1$ though that doesn't look useful, either. – dxiv Mar 21 '23 at 23:35
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    @JeanMarie May be just luck, but I have that nagging feeling there is more reason to it why the transformation worked. – dxiv Mar 22 '23 at 04:33
  • Great update! Could you add how you found the Tschirnhaus transformation? – ronno Mar 22 '23 at 09:01
  • @ronno Thanks. Edited that in at the end of the answer. Wish I had some deep insight why that worked, but I can't claim that. – dxiv Mar 22 '23 at 15:47
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    I've asked a follow-up question here. Also thought I should note explicitly that your Tschirnhaus transformation is $P(x) - x$ though I still don't know why it works without just computation. – ronno Mar 22 '23 at 17:33
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Disclaimer : this is not an answer, therefore not to be upvoted (neither downvoted...) : just a (Geogebra) figure to clarify the question :

enter image description here

Here we have only 5 intersections. In some other cases there will be only $3$ for example...

Jean Marie
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