Fixed rational composition loses balanced input mass

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Statement and scope

Fix

$$F(\mathbf x)=a_1x_1^3+a_2x_2^3+a_3x_3^3, \qquad a_i\in\mathbb Q_{>0}.$$

The following theorem is proved. It strengthens the linear-in-one-input obstruction, but does not decide the density of the values of $F$.

Theorem. Suppose that a fixed rational map $C\in\mathbb Q(\mathbf x,\mathbf y)^3$ satisfies

$$F(C(\mathbf x,\mathbf y))=F(\mathbf x)F(\mathbf y). \tag{1}$$

Then $C$ is not polynomial of total degree at most two. Furthermore,

$$\#\{(\mathbf x,\mathbf y)\in[1,N]^6\cap\mathbb Z^6: C\text{ is defined and }C(\mathbf x,\mathbf y)\in\mathbb N^3\} \ll_{F,C}N^{17/3}. \tag{2}$$

Thus no fixed rational composition retains a positive fraction of all input pairs in balanced root boxes. In boxes with both root heights comparable to $N$, its product values also number $O(N^{17/3})$, compared with the ambient target scale $N^6$. The theorem is not a count of all product values with unrestricted, possibly very unbalanced factors. It also does not control maps whose complexity varies with $N$.

Excluding the pure quadratic terms

First suppose $P$ is a homogeneous quadratic polynomial map satisfying (1). Write

$$P=A(\mathbf x)+B(\mathbf x,\mathbf y)+D(\mathbf y),$$

where $B$ is bilinear. Setting $\mathbf y=0$ gives $F(A)=0$.

Suppose $A\ne0$. The plane cubic $E:F=0$ is smooth, since its three partial derivatives cannot vanish simultaneously at a projective point. It is therefore geometrically integral of genus one. The rational map $[A]:\mathbb P^2\dashrightarrow E$ must be constant: if two points in its domain had different images, restriction to their joining line would give a nonconstant rational map $\mathbb P^1\dashrightarrow E$. It extends to a morphism, since $E$ is proper and smooth, contradicting Riemann--Hurwitz. The constant image is a rational point $q$, as can also be seen by evaluating at a rational point where $A\ne0$. Hence

$$A=qH$$

for a nonzero rational homogeneous quadratic $H$.

Make a rational linear change of output coordinates taking $q$ to $(1,0,0)$ and its tangent line to $W=0$. The transformed cubic is

$$\begin{aligned} G(U,V,W)={}&\alpha U^2W+\beta UV^2+\gamma UVW+\delta UW^2\\ &+\epsilon V^3+\eta V^2W+\theta VW^2+\kappa W^3, \end{aligned} \tag{3}$$

with $\alpha\ne0$. The transformed coordinates of $P$ are

$$U=H+L+u,\qquad V=M+v,\qquad W=T+w,$$

where $L,M,T$ are bilinear and $u,v,w$ depend quadratically only on $\mathbf y$.

Compare bidegrees in $G(P)=F(\mathbf x)F(\mathbf y)$. The $(5,1)$ part gives $\alpha H^2T=0$, so $T=0$. The $(4,2)$ part gives

$$\alpha Hw+\beta M^2=0. \tag{4}$$

If $\beta=0$, then $w=0$, and $G(P)=\epsilon(M+v)^3$. This cannot be a nonzero scalar multiple of $F(\mathbf x)$ over $\mathbb Q(\mathbf y)$: $F$ is absolutely irreducible and not a cube. If $\beta\ne0$ but $w=0$, then $M=0$, and $G(P)=\beta Uv^2+\epsilon v^3$ has degree at most two in $\mathbf x$, again impossible.

In the remaining case $\beta w\ne0$, work over $K=\mathbb Q(\mathbf y)$. Equation (4) makes $H$ a scalar multiple of $M^2$. Thus all coordinates of $P$ belong to $K[L,M]$. Choose a nonzero $h\in K^3$ annihilated by the two linear forms $L,M$. Directional differentiation of (1) gives

$$0=F(\mathbf y)D_hF(\mathbf x) =3F(\mathbf y)\sum_i a_i h_i x_i^2.$$

This forces every $h_i=0$, a contradiction.

Consequently $A=0$. Interchanging the input blocks proves $D=0$, so the proposed quadratic composition must be bilinear.

Excluding the bilinear map and the lower-degree terms

Write $P=M(\mathbf y)\mathbf x$ over $K=\mathbb Q(\mathbf y)$, and put $m=F(\mathbf y)$. The matrix $M$ is invertible: a kernel vector would again give a nonzero direction $h$ with $D_hF=0$. The Hessian determinant is

$$\det\operatorname{Hess}F =216a_1a_2a_3x_1x_2x_3.$$

Applying the Hessian transformation formula to $F(M\mathbf x)=mF(\mathbf x)$ shows that the product of the row forms of $M$ is a nonzero scalar multiple of $x_1x_2x_3$. Unique factorization makes $M$ monomial. Coefficient comparison therefore gives

$$F(\mathbf y)=c\,g(\mathbf y)^3, \qquad c\in\mathbb Q^\times,\quad g\in K^\times.$$

This is impossible by the valuation at the irreducible divisor $F=0$: it is one on the left and divisible by three on the right. Absolute irreducibility follows from smoothness; distinct projective components of positive degree in the plane would intersect and give a singular point.

For a polynomial $C$ of total degree at most two, degree at most one is impossible by degrees. In degree two, comparison of the total degree-six parts gives

$$F(C_2)=F(\mathbf x)F(\mathbf y),$$

where $C_2$ is the homogeneous quadratic part. The preceding argument excludes this. Lower-degree terms therefore provide no exception.

Counting positive integral images

If some coordinate of $C$ is nonpolynomial, the resultant divisibility lemma in six variables gives $O_{C,\varepsilon}(N^{5+\varepsilon})$ integral input points. Choose a fixed $\varepsilon<2/3$.

Otherwise $C$ is polynomial of degree $D\ge3$. Choose a coordinate $P$ of degree $D$. At a counted point, positivity and (1) imply

$$|P(\mathbf x,\mathbf y)|\ll_F N^2. \tag{5}$$

Let $P_D$ be its highest homogeneous part. Choose a primitive integer vector $v$ with $P_D(v)\ne0$, and complete it to a unimodular integral basis. The transformed input box is contained in a fixed dilate of $[-N,N]^6$. Along each line in direction $v$, the polynomial $p(t)$ has degree $D$ and leading coefficient $c=P_D(v)\ne0$, independent of the other five coordinates.

Uniformly over the lower coefficients, a degree-$D$ polynomial with leading coefficient $c$ satisfies

$$\#\{t\in\mathbb Z:|p(t)|\le T\}\ll_{D,c}1+T^{1/D}. \tag{6}$$

Indeed, factoring over $\mathbb C$, every such $t$ lies within $(T/|c|)^{1/D}$ of at least one of the $D$ roots. The corresponding real intervals contain $O_D(1+(T/|c|)^{1/D})$ integers.

There are $O_C(N^5)$ choices for the other coordinates. Equations (5)--(6) give

$$O_{F,C}(N^{5+2/D})\le O_{F,C}(N^{17/3}),$$

which proves (2).

Remaining arithmetic question

The theorem concerns a prescribed rational rule, not the existence of other integral representations of the product. It does not bound ordinary integer fibers of $F$, and supplies neither density zero nor positive density for those values. The bounded-fiber criterion remains unproved.