Trying multiplicative composition through cubic matrices

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The idea and its outcome

This attempt uses multiplication, rather than additive collision estimates or progression first moments. Realize a positive diagonal cubic as a matrix determinant, multiply two representatives, and try to return the product to the diagonal family by an input-dependent conjugation. Unlike a fixed number-field norm, a matrix determinant can restrict to an absolutely irreducible diagonal cubic, so the existing constant-field norm obstruction does not exclude this starting point.

The attempt fails for a precise arithmetic reason. Ordinary products have a trace obstruction; modifying the product removes that obstruction but changes its determinant. Even allowing arbitrary pointwise rational scalars to repair the determinant yields only $O_\varepsilon(T^{17/18+\varepsilon})$ candidate product values up to $T$. These statements are proved below. They concern this particular composition mechanism, not arbitrary rational composition laws or the density of the original cubic values. No target case is resolved.

1. A determinant model

Fix positive integers $a,b$, put $c=(a,b,a^2b)$, and write

$$F(\mathbf x)=\sum_{i=1}^3c_ix_i^3.$$

The choice $a=2,b=1$ gives $x^3+2y^3+4z^3$ after permuting variables. Let $\zeta$ be a primitive cube root of unity and work over $L=\mathbb Q(\zeta,\alpha,\beta)$, with $\alpha^3=a$, $\beta^3=b$. Define

$$U=\alpha\operatorname{diag}(1,\zeta^2,\zeta),\qquad V=\beta\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}.$$

Then $U^3=aI$, $V^3=bI$, and $VU=\zeta UV$. The nine matrices $U^iV^j$, $0\le i,j\le2$, form a basis of $M_3(L)$, with multiplication

$$(U^iV^j)(U^kV^l) =\zeta^{jk}a^{\lfloor(i+k)/3\rfloor}b^{\lfloor(j+l)/3\rfloor} U^{(i+k)\bmod3}V^{(j+l)\bmod3}. \tag{1}$$

All these basis matrices except $I$ have trace zero.

Set $e_1=U$, $e_2=V$, $e_3=U^2V$, and $X(\mathbf x)=\sum_i x_i e_i$. The relation $e_je_i=\zeta e_ie_j$ for $i<j$, together with $1+\zeta+\zeta^2=0$, gives

$$X(\mathbf x)^3=F(\mathbf x)I,\quad \operatorname{tr}X(\mathbf x)=\operatorname{tr}(X(\mathbf x)^2)=0, \quad\det X(\mathbf x)=F(\mathbf x). \tag{2}$$

The last equality also follows from Newton's identity $\det X=\operatorname{tr}(X^3)/3$ when the first two traces vanish. Call a matrix pure when those traces vanish. Purity is necessary for a nonzero scalar multiple of a matrix to be conjugate into $\mathcal V=\operatorname{span}_L(e_1,e_2,e_3)$. Purity alone does not give a conjugate with positive rational coordinates.

2. Ordinary products and sandwiches

Write $X=X(\mathbf x)$ and $Y=X(\mathbf y)$. Multiplication using (1) gives

$$\operatorname{tr}(XY)=0, \qquad \operatorname{tr}((XY)^2) =6ab(1+\zeta)(x_1y_2-x_2y_1)(x_2y_3-x_3y_2). \tag{3}$$

For a direct check, the six coefficients of $XY$, at basis elements $(U^2,UV,V,V^2,U^2V^2,UV^2)$, are respectively

$$x_1y_1,\quad x_1y_2+\zeta x_2y_1,\quad a(x_1y_3+\zeta x_3y_1),\quad x_2y_2,\quad x_3y_2+\zeta^2x_2y_3,\quad a\zeta^2x_3y_3.$$

In the square, only the pairs $(UV,U^2V^2)$ and $(V,V^2)$ have scalar products. Their two orders each contribute $\zeta^2ab$ and $b$, respectively, proving (3).

Thus an arbitrary input-dependent conjugation and nonzero scalar can return $XY$ to $\mathcal V$ only if

$$x_1y_2=x_2y_1\quad\text{or}\quad x_2y_3=x_3y_2. \tag{4}$$

For positive integer inputs, the distinct products $F(\mathbf x)F(\mathbf y) \le T$ satisfying (4) number

$$O_{a,b}(T^{2/3}(\log(2T))^2). \tag{5}$$

Indeed, in boxes $x_i\le M$, $y_i\le N$, write proportional coordinate pairs as $s(h,k)$ and $t(h,k)$ with $\gcd(h,k)=1$. There are at most $2j$ choices with $\max(h,k)=j$, so these pairs number $O(MN\log(2\min(M,N)))$. The two free coordinates contribute $MN$. In dyadic maximum-coordinate shells, positivity forces $MN\ll_{a,b}T^{1/3}$. Summing the resulting $M^2N^2\log(2\min(M,N))$ bound proves (5).

Sandwiches fare worse. Polarization of (2) gives

$$\operatorname{tr}(X^2Y)=3\sum_i c_ix_i^2y_i>0 \tag{6}$$

on positive inputs. Since $XYX$ is similar to $X^2Y$, it cannot return to $\mathcal V$ by conjugation and a nonzero scalar. More generally $X^kYX^l$ is similar to $X^{k+l}Y$. Reduction of $k+l$ modulo three leaves either an existing cube dilation of $F(\mathbf y)$, the product case (4), or the impossible case (6). Conjugation $XYX^{-1}$ preserves the old determinant rather than creating a new one.

3. A modified product really does remove the trace obstruction

The natural repair is

$$D=\frac{XY-\zeta YX}{1-\zeta}. \tag{7}$$

This matrix is pure for every pair of inputs. To see this without a large expansion, the coefficient of $t$ in $(X+tY)^3=F(\mathbf x+t\mathbf y)I$ is scalar. Multiply it by $Y$ and take traces, using $\operatorname{tr}Y=0$:

$$\operatorname{tr}((XY)^2)=-2\operatorname{tr}(X^2Y^2).$$

It follows that

$$\operatorname{tr}((XY-\zeta YX)^2) =(1+\zeta^2)\operatorname{tr}((XY)^2) -2\zeta\operatorname{tr}(X^2Y^2)=0.$$

Cayley–Hamilton now gives $D^3=(\det D)I$.

The determinant, however, is the following sextic rather than $F(\mathbf x)F(\mathbf y)$:

$$H(\mathbf x,\mathbf y)= \sum_i c_i^2x_i^3y_i^3+ \sum_{i<j}c_ic_j\bigl(3x_i^2x_jy_iy_j^2-x_i^3y_j^3\bigr), \qquad D^3=HI. \tag{8}$$

This follows by cubing $D=\sum_i x_iy_i e_i^2+(1+\zeta)\sum_{i<j}x_iy_j e_ie_j$ using (1). The preserved symbolic certificate checks all nine coefficients of (2), (3), (6), and (8) exactly; it is not a test on finitely many numerical inputs.

There is no rational-function scalar over $\mathbb Q(\zeta)(\mathbf x,\mathbf y)$ which repairs (8) to the desired product. The polynomial $F(\mathbf y)$ is absolutely irreducible (its projective plane cubic is smooth), and does not divide $H$: the coefficient of $x_3^3$ in $H$ is $c_3^2y_3^3$. Thus $F(\mathbf x)F(\mathbf y)/H$ has valuation one at the prime divisor $F(\mathbf y)$, and cannot be a cube.

This also exhausts the combinations $A\,XY+B\,YX$ on generic inputs. Their trace-square equals $\operatorname{tr}((XY)^2)(A^2-AB+B^2)$, so purity outside (4) requires $B/A=-\zeta$ or $-\zeta^2$. The second normalized choice is $D(\mathbf y,\mathbf x)$ and has determinant $H(\mathbf y,\mathbf x)$.

4. Even pointwise scalar repair gives only sparse product values

The failure of a rational-function identity does not by itself exclude scalars chosen separately at each integer input. The following does.

Theorem. For either $J(\mathbf x,\mathbf y)=H(\mathbf x,\mathbf y)$ or $H(\mathbf y,\mathbf x)$, the number of positive integer input pairs with

$$F(\mathbf x)F(\mathbf y)\le T,\qquad J\ne0,\qquad F(\mathbf x)F(\mathbf y)/J\in\mathbb Q^{\times3}$$

is $O_{a,b,\varepsilon}(T^{17/18+\varepsilon})$. The same bound therefore holds for their distinct product values.

Proof. If the displayed quotient is a rational cube, then

$$z^3=F(\mathbf x)F(\mathbf y)J(\mathbf x,\mathbf y)^2 \tag{9}$$

for an integer $z$. A rational number whose cube is an integer is itself an integer. Conversely (9) gives the required quotient when $J\ne0$. Allowing $J=0$ in the counting argument only enlarges the set.

Suppose first $J=H(\mathbf x,\mathbf y)$. Fix positive $x_1,x_2,\mathbf y$ and set $t=x_3$. Then

$$G(t)=F(\mathbf x)=c_3t^3+C_0,\qquad h(t)=H(\mathbf x,\mathbf y)=At^3+Bt+C,$$

where

$$C_0>0,\quad A=c_3^2y_3^3>0,\quad B=3c_3y_3^2(c_1x_1^2y_1+c_2x_2^2y_2)>0.$$

The polynomial $G$ has three simple complex roots. If $F(\mathbf y)G h^2$ were a cube in $\mathbb C(t)$, every root of $G$ would have to occur in $h$ with multiplicity $1\pmod3$. Since $\deg h=3$, this forces $h$ proportional to $G$, contradicting $B>0$. Consequently $z^3-F(\mathbf y)G(t)h(t)^2$ is absolutely irreducible of total degree nine. For $J=H(\mathbf y,\mathbf x)$, vary $t=x_1$ instead: its cubic and linear coefficients are $c_1^2y_1^3>0$ and $3c_1y_1^2(c_2y_2x_2^2+c_3y_3x_3^2)>0$, giving the same proof.

The coefficient-uniform Bombieri–Pila estimate in Browning–Verzobio, Lemma 2.2 states that an absolutely irreducible degree-$e$ integral plane curve has $O_{e,\epsilon}(Q^{1/e+\epsilon})$ integer points in $[-Q,Q]^2$. Its full statement, including uniformity, was checked.

In a root box $x_i\le M$, $y_i\le N$ with $M\ge N$, put $S=MN$. Equation (9) gives $|z|\ll_{a,b}S^3$. Applying that estimate to each fixed slice gives $O_{a,b,\epsilon}(S^{1/3+\epsilon})$ possibilities. There are $O(M^2N^3)$ choices of the frozen coordinates, hence

$$\#\text{pairs in the box}\ll_{a,b,\epsilon} M^2N^3S^{1/3+\epsilon}=S^{10/3+\epsilon}/M. \tag{10}$$

If $N>M$, interchange the tuples and the two orientations of $H$.

Use dyadic maximum-coordinate shells. At fixed dyadic product $S$, the sum of $1/M$ over $M\ge\sqrt S$ is $O(S^{-1/2})$. Thus (10) sums to $O(S^{17/6+\epsilon})$. Positivity imposes $S\ll_{a,b}T^{1/3}$, and summing over $S$ proves the theorem after renaming $\epsilon$. $\square$

Scalars in $\mathbb Q(\zeta)$ do not change this conclusion: if their cube is rational, they are rational. Otherwise an irreducible cubic polynomial over $\mathbb Q$ would have a root in a quadratic field.

What remains outside the obstruction

These bounds allow the conjugating matrix and the repairing scalar to depend arbitrarily on the input; they are not just another fixed-map bound. The second bound also treats unbalanced root heights, so it is stronger than a zero-density statement in balanced input boxes.

They do not exclude a different nonlinear operation, extra auxiliary variables, or rational inputs with uncontrolled denominators. Ordinary matrix multiplication retains the determinant on a nine-dimensional space, but returning to the three-dimensional diagonal family is the failed step. The modified product returns to the pure-matrix variety, not to a positive integral diagonal representation of the product. No positive-density theorem for $F$ or additional Erdős 940 case follows.