1
$\begingroup$

Some context.

While working on a larger proof, I needed to show that a particular homogeneous system of polynomial equations had no rational solution except for the trivial one.

I have reduced this task to the following problem, and if the polynomial stated below has no rational root, it's over.

The question.

Let's consider the following polynomial of $\mathbb Q[X,Y]$:

$$P(X,Y)=Y^3(45X^2+18X-9)+Y^2(6X^2-15X+6)+Y(5X^4+15X^3-14X^2-3X-1)-X^5+3X^4+2X^3-X+1.$$

Does $P$ have a rational root?

Some remarks.

  • I know that there is no general algorithm or method to answer such a question, but since this is a particular case, maybe there is an answer here.

  • I think Faltings' theorem (though I don't understand it that well) shows that the rational roots of $P$ are in finite number.

  • We can assume $Y\ne 0$, since there are no rational root of the form $(X,0)$. Thanks to a comment of Dietrich Burde, we can also assume $X\ne 0$ and $X\ne Y$.

  • I thought of computing the roots of $P$ for $Y$ in terms of $X$, and finding conditions on $X$ (conditions of the form "$\sqrt{X^4-X^3+1}\in\mathbb Q$"), but it does not lead me anywhere.


If you are curious about it, this is what the curve defined by $P$ looks like:

enter image description here


Any enlightenment on this problem would be much appreciated.


A comment asked me what the original system was, so here it is:

$$\begin{cases} -x_2^2+x_1x_3-x_4^2-x_2x_5-x_4x_5=0 \\ x_1^2-x_2^2+x_1x_4-3x_3x_4-x_2x_5-3x_3x_5=0 \\ x_1x_2-x_2x_3+x_2x_4-x_4x_5-x_5^2 =0 \\ 3x_2x_3-x_2x_4-x_1x_5=0\\ 3x_3^2-x_1x_4-x_4^2-x_2x_5-x_5^2=0 \end{cases}$$

for $x_1,\ldots,x_5\in\mathbb Q$.

$\endgroup$
  • 1
    $\begingroup$ In general, it is very difficult or impossible to find out whether or not $P(X,Y)$ has rational roots. Can you give your "particular system of polynomial equations" of the original problem? This might be easier to access. $\endgroup$ – Dietrich Burde Feb 21 at 15:28
  • $\begingroup$ @DietrichBurde I know this is difficult in general, but I was hoping there would be some trick that would work here. Ok, I have posted the system. $\endgroup$ – E. Joseph Feb 21 at 16:05
1
$\begingroup$

The OP's system in the $x_i$,

$$\begin{cases} -x_2^2+x_1x_3-x_4^2-x_2x_5-x_4x_5=0 \\ x_1^2-x_2^2+x_1x_4-3x_3x_4-x_2x_5-3x_3x_5=0 \\ x_1x_2-x_2x_3+x_2x_4-x_4x_5-x_5^2 =0 \\ 3x_2x_3-x_2x_4-x_1x_5=0\\ 3x_3^2-x_1x_4-x_4^2-x_2x_5-x_5^2=0 \end{cases}$$

DOES have non-trivial integer solutions. Some small examples $x_1, x_2, x_3, x_4, x_5$ are,

$$18, 7, -19, 15, -28$$

$$133, 57, 46, -37, 75$$

$$380, 665, -97, 81, -651$$

and probably an infinite more.


Added details:

Given a system of $n$ equations in $n$ unknowns, we can generally resolve it into one equation in one unknown using resultants.

The trick is to find the simplest equation to "cleave" the system. In the OP's case, it is the 4th one. Solve for its $x_5$, and we find,

$$x_5 =\frac{3 x_2 x_3 - x_2 x_4}{x_1}$$

Substitute this into the 1st and solve for $x_3$,

$$x_3 = \frac{x_1 x_2^2 - x_2^2 x_4 + x_1 x_4^2 - x_2 x_4^2}{x_1^2 - 3 x_2^2 - 3 x_2 x_4}$$

Substitute this $x_5, x_3$ and we find the remaining 3 equations become the SAME cubic in 3 variables $x_4, x_1, x_2$. Why this is so is a peculiarity of this system.

It was then easy to use Mathematica to test small values such that the cubic in $x_4$ factors.

$\endgroup$
  • $\begingroup$ Thanks for the answer! How did you find those solutions? $\endgroup$ – E. Joseph Feb 23 at 13:03
  • $\begingroup$ Your system is peculiar. It is 5 equations in 5 unknowns, yet I was able to resolve it into a 3-variable cubic equation. I'll add more details to my answer in several minutes $\endgroup$ – Tito Piezas III Feb 23 at 13:06
  • $\begingroup$ @E.Joseph: Details have been added. How did this system arise? It is quite peculiar. $\endgroup$ – Tito Piezas III Feb 23 at 13:23
  • $\begingroup$ Thank you for adding the details! This system comes from Plücker relations for a two dimensional subspace of $\mathbb R^5$, when some of the Plücker coordinates are express at $\mathbb Q$-linear combinaison of the others. I am trying to show that by fixing $5$ relations, you can provide the Plücker relations from having a rational solution. $\endgroup$ – E. Joseph Feb 23 at 13:35
  • $\begingroup$ I was expecting that the final equation in one variable would have a huge degree. What I find strange is that the system was reducible to its "conditional equation" (a cubic) so quickly. There might be clever transformations to show this cubic may contain an elliptic curve in disguise hence has infinitely many rational solutions. $\endgroup$ – Tito Piezas III Feb 23 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.