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Let

$f = 4x^6 - 24x^5 - 2113x^4 + 8612x^3 + (-39114t - 12978)x^2 + (78228t + 8668)x + (58671t^2 + 18848602t - 2168)$

(I will explain its provenance if necessary, but there may be no need).

Is there a value of the parameter $t \in \mathbb{Z}$ for which $f$ has a rational root?

It is possible $t$ is very large, or even that there is no such $t$, e.g. if $f$ is always irreducible.

It seems $f$ is irreducible for all $t$ from $0$ up to $10^7$, I didn't search any higher. It would be especially interesting if there is a very large $t$ which works and is computable.

[Edit] Further to @Tito's comment, I will explain where this comes from. It was from an attempt to find polynomials with a given discriminant. Using my lattice idea from Polynomials with a given discriminant, I picked a random composite number $n=2173$, and constructed the polynomial

$f_0 = tx^3 + (m^2 - 2m + 1)x^2 + 2173x + 2173$

which has discriminant divisible by $n$ for some $m \in \mathbb{Z}$. It has discriminant equal to $n$ if $\text{Disc}(f_0)/n = 1$, i.e.

$-4m^6 + 24m^5 + 2113m^4 - 8612m^3 + (39114t + 12978)m^2 + (-78228t - 8668)m + (-58671t^2 - 18848602t + 2169) = 1$

So @Benjamin has shown there is no combination of $m,t \in \mathbb{Z}$ that solves this equation, in this case.

Oisin Robinson
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    By the rational root theorem, if $x=p/q$ is a root of $f$, then $q|4$ and $p|58671t^2+18848602t-2168$. So we replace $x$ by $X=4x$ to get a new monic polynomial with integer coefficients, say $g(X,t)$. Now, for a fixed integer $t$, all rational roots $X$ of $g$ will be integers. So instead of thinking of $g$ as a sextic in $X$, you can think of it as a quadratic in $t$ with integer coefficients, which is much easier to deal with. – Chris Lewis Jan 30 '25 at 15:57
  • I wonder then can we decide if there exists some suitable $t$ based on whether the discriminant of the quadratic is a square. Thanks @ChrisLewis – Oisin Robinson Jan 30 '25 at 16:02
  • @OisínRobinson Yes, the provenance of the equation should be interesting. Where is it from? – Tito Piezas III Jan 30 '25 at 22:09
  • @TitoPiezasIII I edited the question to show where I got the poly. I actually made a mistake in applying my lattice method, it slipped through because I didn't double check and it so happens that $f_0$ has discriminant divisible by $n$ anyway. – Oisin Robinson Jan 30 '25 at 22:53
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    @OisínRobinson Ok. Courtesy of Benjamin Wright's answer, your curve has a simplified form $4u^3-27 = 2173v^2,$ but $u$ is a quadratic expression in the variable $x$, hence making it hyperelliptic. – Tito Piezas III Jan 31 '25 at 03:34
  • Just to explain my mistake, to apply the lattice idea correctly I should have taken e.g. $f_0=(x^3-3m^2x+2m^3)+nx+tn$ (linear combination of polynomials corresponding to 3 columns of lattice). It boils down to the hyperelliptic curve $y^2 = 11664x^6 - 25345872x^4 + 18358859952x^2 - 4432644976428$ which has positive leading coefficient of $x$. But I appreciate the comments/answer regardless. – Oisin Robinson Jan 31 '25 at 11:30

2 Answers2

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The discriminant in your answer is an even-degree polynomial in $x$ with negative leading coefficient, so it is negative (and hence not a perfect square) for all sufficiently large $x$. Since we want $x$ to be rational with denominator dividing $4$, this reduces to a finite search.

Incidentally, this discriminant has a particularly nice form:

$$\Delta = -2^4\cdot 3^3\cdot 41\cdot 53\cdot\left(\left((x-1)^2 -\frac{41\cdot 53}{3}\right)^3 + \frac{1}{4}\right)$$ with real roots at $$ x= 1 \pm \sqrt{\frac{2173}{3} - \frac{\sqrt[3]{2}}{2}}.$$

This polynomial is negative outside of the interval $[-26, 28]$. None of the rational $x$ in this interval with denominator dividing $4$ make $\Delta$ a perfect square, hence there are no solutions to the original problem.

Benjamin Wright
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This seems to reduce to finding a rational point on a hyperelliptic curve.

Let

$a = 58671$

$b = -39114x^2 + 78228x + 18848602$

$c = 4x^6 - 24x^5 - 2113x^4 + 8612x^3 - 12978x^2 + 8668x - 2168$,

then

$g = at^2 + bt + c$

is a quadratic in $t$, and $g = f$. For $g$ to have a rational root, its discriminant $\Delta = b^2-4ac$ must be a square. So we are left with finding (or not) a rational point on the hyperelliptic curve

$y^2 = -938736x^6 + 5632416x^5 + 2025792288x^4 - 8140718592x^3 - 1465323088320x^2 + 2946942633600x + 355270306149316$

which at least clarifies the problem. Equivalently,

$y^2/4 = 4\times 2173\, (-3 x^2 + 6x + 2170)^3 - 27\times 2173$

where $n=2173=41\times53\,$ is the random number chosen.

Tito Piezas III
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Oisin Robinson
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