Many representations with large ratios in all three traces

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Scope

For every fixed $K\ge2$, there are arbitrarily large $N$ and $K$ distinct positive representations

$$x_j^3+y_j^3+z_j^3=N^3,\qquad (x_j,y_j,z_j)\in[N/3,N]^3,$$

such that, for each of the three chosen traces $x+y,x+z,y+z$, every reduced ratio between two representations has height $\asymp_K N$. Constants may depend on the fixed construction.

This rules out forcing a small ratio in at least one of the three traces merely from a large fixed number of interior representations. No primitivity assertion is made. The constructed family is sparse and has cube targets; it does not rule out a typical-fiber estimate after deleting negligible input families.

A conic on the equal-sums surface

In the plane $x+y=3(w-z)$, put $d=x-y$. The equation $x^3+y^3+z^3=w^3$ contains the residual conic

$$(23z-29w)^2+23(3d)^2=312w^2. \tag{1}$$

Indeed, substituting $x=(3(w-z)+d)/2$ and $y=(3(w-z)-d)/2$ reduces the cubic equation to

$$(w-z)\bigl(23w^2-58wz+23z^2+9d^2\bigr)=0.$$

The quadratic factor is equivalent to (1).

Let $\rho=\sqrt{-23}$ and $\alpha_0=-105-3\rho$, whose norm is $11232=312\cdot6^2$. Choose $L$ with $2^{L-1}\ge K$, and choose $K$ distinct sign vectors $\epsilon\in\{\pm1\}^L$ whose last coordinate is $+1$. Fix even integers $c_1,\ldots,c_L$, to be selected below, and take positive even integers $M$. Define

$$G(M)=\prod_{i=1}^L((M+c_i)^2+23),\qquad w=6G(M),$$ $$t_\epsilon+e_\epsilon\rho =\alpha_0\prod_{i=1}^L(M+c_i+\epsilon_i\rho)^2,$$

and

$$z_\epsilon=\frac{t_\epsilon+29w}{23},\qquad d_\epsilon=e_\epsilon/3,\qquad x_\epsilon=\frac{3(w-z_\epsilon)+d_\epsilon}{2},\qquad y_\epsilon=\frac{3(w-z_\epsilon)-d_\epsilon}{2}.$$

These coordinates are integral. The coefficient $e_\epsilon$ is divisible by three. Modulo $23$, one has $t_\epsilon\equiv-105G$ and $29w=174G$, so the formula for $z_\epsilon$ is integral. Since $M+c_i$ is even, $t_\epsilon,e_\epsilon,z_\epsilon,d_\epsilon$ are odd and $w$ is even, which verifies the divisions by two. Taking norms gives (1), hence

$$x_\epsilon^3+y_\epsilon^3+z_\epsilon^3=w^3.$$

As $M\to\infty$ with the $c_i$ fixed,

$$(x_\epsilon,y_\epsilon,z_\epsilon)/w \longrightarrow(2/3,5/6,1/2).$$

Thus all the points eventually lie in the claimed interior box.

Choosing constants so that the trace gcds are bounded

Before specializing $X_i=M+c_i$, put

$$U_\epsilon=\prod_i(X_i+\epsilon_i\rho),\qquad V_\epsilon=\prod_i(X_i-\epsilon_i\rho).$$

Each of the three trace formulas is a binary quadratic expression

$$A U_\epsilon^2+B U_\epsilon V_\epsilon+C V_\epsilon^2, \qquad A C\ne0.$$

This follows directly from its nonzero real-linear dependence on $t_\epsilon,e_\epsilon$, together with $w=6U_\epsilon V_\epsilon$. Over an algebraic closure it therefore factors into expressions $U_\epsilon-\lambda V_\epsilon$, with $\lambda\ne0$.

On the open set $X_i\ne\pm\rho$, use the coordinates $u_i=(X_i+\rho)/(X_i-\rho)$. Each such factor becomes, up to a unit, a Laurent binomial

$$\prod_i u_i^{\epsilon_i}-\lambda.$$

It is irreducible because the exponent vector is primitive. Binomials from two selected sign vectors are not associates: their exponent vectors are neither equal nor opposite. Also no trace polynomial has a factor $X_i-\rho$ or $X_i+\rho$, because $A C\ne0$. Consequently two trace polynomials for the same chosen trace and different sign vectors are coprime.

Their common-zero locus in affine $L$-space therefore has codimension at least two. A generic line parallel to $(1,\ldots,1)$ misses it: the exceptional translations lie in the image of this locus times an affine line, of dimension at most $L-1$. There are only finitely many pairs and traces. Thus fixed even integral $c_i$ can be chosen outside all exceptional algebraic sets. For that choice the corresponding one-variable trace polynomials in $M$ are pairwise coprime.

After clearing their fixed denominators, Bezout identities for these polynomials show that the gcds of their integer values are bounded independently of $M$. Each trace has size $\asymp w$, so all the reduced trace-ratio heights are $\asymp w$. In particular the represented triples are distinct for all sufficiently large $M$. Set $N=w$ to finish the proof.

What remains possible

The fixed-ratio collision bound does not acquire a deterministic rich-fiber consequence by simply using all three traces. This construction shows why. But it produces only $O_K(T^{1/(2L)})$ inputs of root height at most $T$, since $w\asymp M^{2L}$. It is therefore a zero-mass input family. It neither establishes unbounded typical multiplicity nor prevents a statistical estimate which first excludes sparse exceptional families.