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Quartic rationally derived polynomials

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Statement

Definitions and mathematical statement

A polynomial $p\in\mathbb Q[x]$ of degree four is rationally derived if each nonconstant derivative $p^{(j)}$, for $0\le j\le3$, splits as a product of linear factors in $\mathbb Q[x]$. Roots are counted with multiplicity. Its fourth derivative is a nonzero constant and imposes no condition; the identically zero later derivatives are excluded. The conjecture is \[ \nexists p\in\mathbb Q[x],\quad \deg p=4, \quad p\text{ has four distinct roots}, \quad p,p',p'',p^{(3)}\text{ split over }\mathbb Q. \] Equivalently, decide whether there are rational numbers $a_1<a_2<a_3<a_4$ such that the derivatives of $\prod_{i=1}^4(x-a_i)$ all have rational roots. Multiplying the polynomial by a nonzero rational number does not affect the question. An affine change $x\mapsto ux+v$, with $u,v\in\mathbb Q$ and $u\ne0$, also preserves rational derivedness and distinctness. Consequently the search can be normalized to roots $0,1,s,t$ with $1<s<t$ rational. The rationality of all four original roots alone is insufficient: the cubic and quadratic derivatives must split as well. A counterexample must meet every one of these conditions simultaneously.

Short English statement

Can a quartic with four distinct rational roots also have only rational roots in every nonconstant derivative?

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Math-1151-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1151

Quartic rationally derived polynomials

Definitions and mathematical statement

A polynomial $p\in\mathbb Q[x]$ of degree four is rationally derived if each nonconstant derivative $p^{(j)}$, for $0\le j\le3$, splits as a product of linear factors in $\mathbb Q[x]$. Roots are counted with multiplicity. Its fourth derivative is a nonzero constant and imposes no condition; the identically zero later derivatives are excluded. The conjecture is \[ \nexists p\in\mathbb Q[x],\quad \deg p=4, \quad p\text{ has four distinct roots}, \quad p,p',p'',p^{(3)}\text{ split over }\mathbb Q. \] Equivalently, decide whether there are rational numbers $a_1<a_2<a_3<a_4$ such that the derivatives of $\prod_{i=1}^4(x-a_i)$ all have rational roots. Multiplying the polynomial by a nonzero rational number does not affect the question. An affine change $x\mapsto ux+v$, with $u,v\in\mathbb Q$ and $u\ne0$, also preserves rational derivedness and distinctness. Consequently the search can be normalized to roots $0,1,s,t$ with $1<s<t$ rational. The rationality of all four original roots alone is insufficient: the cubic and quadratic derivatives must split as well. A counterexample must meet every one of these conditions simultaneously.

Short English statement

Can a quartic with four distinct rational roots also have only rational roots in every nonconstant derivative?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 2 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Source check and normalization.
Flynn [R1] explicitly separates the conjecture about four distinct quartic roots from the conjecture about a quintic having a triple root; his theorem proves the latter. It must not be reported as a proof of the former. The primary text also justifies rational affine changes of variable. We use those changes to isolate a necessary conic and exclude every symmetric root configuration. The literature search for this attempt did not locate a proof settling the quartic target; that search is not a substitute for one.

Scale the leading coefficient to one and translate by the rational mean of the four roots. The result has the form \[ p(x)=x^4-Ax^2+Bx+C,\qquad A,B,C\in\mathbb Q. \] If its four roots are distinct and real, Rolle's theorem gives three distinct real roots of $p'$ and two of $p''$. Since $p''(x)=12x^2-2A$, necessarily $A>0$. Requiring its roots to be rational means $A=6t^2$ for some nonzero $t\in\mathbb Q$.

The remaining derivative condition.
Write the rational roots of $p'/4$ as $r,s,-r-s$, using its zero quadratic coefficient. Expanding the product shows that splitting of the first two derivatives requires \[ r^2+rs+s^2=3t^2,\qquad B=4rs(r+s),\qquad A=6t^2. \] For a candidate with four distinct roots the three displayed derivative roots must also be distinct. Conversely, any rational triple satisfying these equations gives a polynomial whose positive-degree derivatives split, whatever $C$ is. This converse deliberately says nothing about the splitting of $p$ itself. The fourth and later derivatives add no nonconstant condition, and the third derivative is $24x$.
Excluding symmetric roots.
Suppose the original four rational roots are symmetric about some centre. Their mean is that centre, so it is rational; after the translation they are $u,-u,v,-v$, with $uv\ne0$ and $u^2\ne v^2$. Then \[ p(x)=(x^2-u^2)(x^2-v^2),\qquad A=u^2+v^2>0, \] and $B=0$. The nonzero roots of $p'$ have square $A/2$, whereas the roots of $p''$ have square $A/6$. If both derivatives split over $\mathbb Q$, choosing a nonzero root from each would give two rational numbers whose ratio has square three. That is impossible: writing a putative square root of three as a reduced fraction forces both its numerator and denominator divisible by three. Thus no quartic with four distinct roots symmetric about a centre is rationally derived. In particular this excludes every four-term rational arithmetic progression. The nonzero condition on the selected roots is ensured by $A>0$.
A checked obstruction to stopping at the conic.
The derivative equations have many rational points that do not solve the original question. Consider the explicit polynomial \[ p(x)=x^4-6x^2-\frac{1144}{343}x. \] Its first derivative satisfies \[ \frac{p'(x)}4=(x+11/7)(x+2/7)(x-13/7), \qquad p''(x)=12(x-1)(x+1),\qquad p^{(3)}(x)=24x. \] Nevertheless, putting $z=7x$ gives \[ 2401p(z/7)=z(z^3-294z-1144). \] The cubic reduces modulo five to $z^3+z+1$. Its values at $0,1,2,3,4$ are respectively $1,3,1,1,4$ modulo five, so it has no root there. A rational root of a monic integer polynomial is an integer and would reduce to such a root; consequently this cubic has no rational root. The quartic therefore does not split over $\mathbb Q$, even though every nonconstant positive-order derivative does. This failure is not merely the absence of real roots: the cubic has sign changes on $(-20,-10)$, $(-10,0)$ and $(0,20)$. It consequently has three distinct real roots, none zero, so the quartic has four distinct real roots. Certificate [R2] checks the factorization, residues and signs with exact rational arithmetic.
Final status and first remaining gap.
UNRESOLVED. All symmetric configurations are excluded, and the derivative conditions are reduced to explicit rational equations. The missing step is to impose four rational roots on the original quartic simultaneously with those equations in the asymmetric case. Solving the conic alone, or finding real roots numerically, does not supply that step; the displayed example verifies the precise failure.

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