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IMC 2020 problem 6 was based on the following idea: if $P(x)=x^3-3x+1$ then $Q(x)=x^2-2$ has the property that it cyclically rotates the roots of $P$. That is, if we call them $x_1$, $x_2$, $x_3$ then $Q(x_1)=x_2$, $Q(x_2)=x_3$ and $Q(x_3)=x_1$. I want to find polynomials like this in general.

If $P(x)=(x-x_1)(x-x_2)(x-x_3)=x^3-ax^2+bx-c$ and $Q(x)=kx^2+lx+m$ and $Q(x_1)=x_2$, $Q(x_2)=x_3$ and $Q(x_3)=x_1$ then

$$\begin{bmatrix} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ x_3^2 & x_3 & 1 \end{bmatrix} \begin{bmatrix} k \\ l \\ m \end{bmatrix}= \begin{bmatrix} x_2 \\ x_3 \\ x_1 \end{bmatrix}.$$

The determinant of the Vandermonde matrix is the product of the pairwise differences which is the square root of the discriminant, so it's expressible as a function of $a,b,c$. $k$ is expressible as a symmetric polynomial too, so we can write $k=\frac{a^2-3b}{\sqrt{|\Delta|}}$, however $l$ and $m$ have expressions as a function of the roots which are not symmetric, so we can't write them as a function of the coefficients:

$$l=\frac{-x_1^2-x_2^3-x_3^3+x_1^2x_3+x_2^2x_1+x_3^2x_2}{\sqrt{|\Delta|}}$$

$$m=\frac{x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2-x_1^3x_2-x_2^3x_3-x_3^3x_1}{\sqrt{|\Delta|}}.$$

The problem with $l$ is the $x_1^2x_3+x_2^2x_1+x_3^2x_2$ bit, because it's missing the other half $x_1^2x_2+x_2^2x_3+x_3^2x_1$ which we would need to add to make it symmetric. $m$ is similar.

So my questions are:

  • When are the expressions for $l$ and $m$ "nice"? For example if the coefficients are rational when are $l,m$ rational?

  • Can we require something of the roots which makes $l,m$ expressible as a function of the coefficients? It seems like they're "nearly" symmetric, so it feels like it shouldn't take much to get there.

  • What happens in higher degrees?

  • So far I've been thinking about $Q$ having degree one less than $P$. Does this always allow for the "cleanest" solution? In particular the case where $P(x)=(x-x_1)(x-x_2)=x^2-ax+b$ is easy: $Q(x)=a-x$.

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  • $\begingroup$ See math.stackexchange.com/questions/1767252/… $\endgroup$
    – lhf
    Jul 18, 2021 at 17:26
  • $\begingroup$ @Blue They cannot be polynomials in the coefficients of $P$. In fact, they depend on the square root of the discriminant, accounting for two choice of cyclic permutations. $\endgroup$
    – WimC
    Jul 18, 2021 at 19:29
  • $\begingroup$ @WimC: Ah. I seem to have been thinking (out loud) too narrowly. $\endgroup$
    – Blue
    Jul 18, 2021 at 20:59
  • $\begingroup$ "$Q$ having degree one less than $P$" $\,-\,$ That is possible when all roots of $P$ are real and distinct, since you can always interpolate a unique polynomial of degree $n-1$ between $n$ real points. It is not necessarily possible when some of the roots are complex, for example $P(x)=x^3-1$ has the "swapper polynomials" $Q_1(x) = \omega x$ and $Q_2(x) = \omega^2 x$ where $\omega$ is a complex cube root of unity. In this case $Q(x)$ $=Q_1(x)Q_2(x)$ $=x^2$ also permutes the roots, but not cyclically since $Q(1)=1$. $\endgroup$
    – dxiv
    Jul 18, 2021 at 22:42

1 Answer 1

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Both $l$ and $m$ are rational when the discriminant $d$ of the cubic $P$ is a rational square. For example, let $$\lambda = x_1^2x_2 + x_2^2x_3 + x_3^2x_1$$ and $\overline \lambda$ its “symmetric conjugate” by swapping any two roots. Then $ (x-\lambda)(x-\overline \lambda)$ is a rational polynomial with discriminant $d$, a rational square, so both $\lambda$ and $\overline \lambda$ are rational.

Similarly for $$\mu = x_1^3x_2 + x_2^3x_3 + x_3^3x_1$$ the discriminant of $(x - \mu)(x - \overline \mu)$ is $ d \cdot (x_1 + x_2 + x_3)^2$, a rational square. Conclusion: if $d$ is a rational square then the coefficients of the permutation polynomials can indeed be expressed in the coefficients of $P$. The choice $\pm \sqrt d$ leads to two non~trivial permutations.

The condition that $d$ is a rational square is also required to ensure that the splitting field of $P$ has degree either $1$ or $3$ over $\mathbb Q$.

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