Restricting root variables by large prime factors

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The new mechanism tested

Instead of changing the cubic's coefficients, or bounding all of its fibers, restrict the inputs to integers with one large prime factor. This retains positive input mass and gives a bilinear description of the exponential sums. The hope was to use the prime factors to control alternative representations. The elementary estimates below are proved; the required restricted multiplicity bound is not. This attempt has not resolved any density or higher-exponent infinitude case.

A restriction that does retain positive input mass

Fix $0<\delta<1$ and $1/2<\theta<1$. Set

$$\mathcal A_\theta(N)= \{x\in(\delta N,N]\cap\mathbb Z: p\mid x\text{ for a prime }p>N^\theta\}.$$

There is at most one such prime, since $N^{2\theta}>N$. Consequently

$$|\mathcal A_\theta(N)| =\sum_{N^\theta<p\le N} \left(\left\lfloor N/p\right\rfloor -\left\lfloor\delta N/p\right\rfloor\right) =\bigl((1-\delta)\log(1/\theta)+o(1)\bigr)N. \tag{1}$$

The total floor error is $O(\pi(N))=o(N)$, and $\sum_{N^\theta<p\le N}1/p=\log(1/\theta)+o(1)$. Thus three independently restricted roots still supply $\gg N^3$ inputs to the fixed form $F=x^3+2y^3+4z^3$. This is not coefficient averaging with escaping arithmetic height. Multiplication of its values by $8$ puts them in $S_3$.

Writing $x=mp$ gives exactly

$$W_{\theta,N}(\alpha)= \sum_{m\le N^{1-\theta}} \ \sum_{\substack{p>N^\theta\\ \delta N<mp\le N}} e(\alpha m^3p^3).$$

For the resulting representation function $R_{\theta,N}$,

$$\sum_nR_{\theta,N}(n)^2 =\int_0^1 |W_{\theta,N}(\alpha)W_{\theta,N}(2\alpha) W_{\theta,N}(4\alpha)|^2\,d\alpha. \tag{2}$$

An $O(N^3)$ bound in (2), or a bounded logarithmic moment on these inputs, would give positive density by the existing multiplicity criteria. Neither was proved.

Where the elementary Type-II estimate loses the endpoint

On a rectangular block $m\asymp M$, $p\asymp P$, $MP\asymp N$, put

$$T(\alpha)=\sum_{m\asymp M}\sum_{p\asymp P}e(\alpha m^3p^3).$$

Cauchy–Schwarz in $m$, followed by a triangle inequality, gives

$$|T(\alpha)|^2 \ll M^2\#\{p\asymp P\} +M\sum_{\substack{p,q\asymp P\\p\ne q}} \left|\sum_{m\asymp M} e\bigl(\alpha(p^3-q^3)m^3\bigr)\right|. \tag{3}$$

The first term in this upper-bound method is $\asymp MN/\log P$. It is a factor $M$ above the square-root scale $N/\log P$ for the block. It persists even if the off-diagonal part of this particular bound were negligible. Applying such pointwise estimates and a fourth moment therefore leaves a polynomial cofactor loss when $M=N^\varepsilon$. This is a limitation of that estimate, not a lower bound on the actual sum or a refutation of a joint sixth-moment method.

Restricting all cofactors to $m\le M=N^{o(1)}$ sacrifices the positive input mass:

$$\#\{mp\in(\delta N,N]:m\le M,\ p\text{ prime}\} \ll\frac{N(1+\log M)}{\log(N/M)}=o(N). \tag{4}$$

For example, sum the elementary prime-counting upper bound $\pi(N/m)\ll(N/m)/\log(N/m)$ over $m\le M$.

The prime factors do not propagate to competing roots

An input prime $p\mid x$ imposes an equation on the competing value, not divisibility of a competing coordinate. This can be checked without an asymptotic heuristic. If $p>3$ and $p\equiv2\pmod3$, cubing permutes $\mathbb F_p$, so for every $n$,

$$\#\{(u,v,w)\bmod p:u^3+2v^3+4w^3=n\}=p^2.$$

At most $3p$ of these roots have a zero coordinate. Thus all-unit solutions exist for every target, and are nonsingular. They lift to every $p^k$, including when the original input has a coordinate divisible by $p$. If $p\nmid n$, all roots are nonsingular and the full number modulo $p^k$ is exactly $p^{2k}$.

The same elementary character estimate gives all-unit nonsingular roots at every sufficiently large good prime, not just in the displayed progression. CRT therefore does not make several input-prime conditions independent of the original output equation. A modulus exceeding the output range turns their conjunction back into exact equality; its very incomplete-box count has not become easier.

Target roughness does not force large root primes

There is also an exact higher-exponent caution. Fix $r\ge3$, put $A=\operatorname{rad}(r)$, and choose a prime $q\nmid r$. For any $y$ and any $K$, there is $k$ such that

$$U_k=A^{rk}+(r-1)q^{rk}$$

has no prime divisor at most $y$ and has at least $K$ distinct prime divisors. Nevertheless it is the sum of $r$ positive $r$th powers whose roots use only the fixed primes dividing $Aq$.

First choose $k$ divisible by every $p-1$ with $p\le y$, $p\nmid Aq$. Then $U_k\equiv r\not\equiv0\pmod p$ for these primes. At $p\mid A$ the second term is a unit, and at $p=q$ the first term is a unit, so $U_k$ is $y$-rough and is always coprime to $Aq$.

To add prime divisors while retaining all existing ones, let $e_p=v_p(U_k)$ for the finitely many $p\mid U_k$. Choose $L$ divisible by the earlier $p-1$ conditions and by the multiplicative orders of $A^r$ and $q^r$ modulo every $p^{e_p+1}$. Replacing $k$ by $k+Lt$ keeps all these valuations fixed and retains $y$-roughness. Since $U_{k+Lt}\to\infty$, some sufficiently large $t$ introduces a new prime divisor. Iterate.

For any fixed $\theta>0$, these roots have no prime factor greater than $N^\theta$ once $k$ is large, where $N=\max(A,q)^k$. This excludes a deterministic inference from arbitrarily large fixed-threshold roughness and many target prime factors to a large root-prime factor. It does not exclude a power-scale roughness condition, an average argument, or a condition requiring every root to have comparable size.

A power-scale test with comparable roots

The following stronger, but sparse, construction tests power-scale roughness. Fix $r\ge5$, $k\ge1$, and $0<\eta<1/2$. For all sufficiently large $Z$ there are

$$\gg_{r,k,\eta}\frac{Z^r}{\log Z}$$

tuples $Z<z_i\le2Z$ such that

$$P^-\left(\sum_{i=1}^r z_i^{rk}\right)\ge Z^\eta. \tag{5}$$

The target is represented by the $r$ positive $r$th powers of the roots $x_i=z_i^k$. These roots are comparable and all their prime factors are at most $2Z$.

Here is a proof with the sieve hypotheses explicit. Write $G(\mathbf z)=\sum_i z_i^{rk}$ and

$$\rho(d)=\#\{\mathbf z\bmod d:G(\mathbf z)=0\bmod d\}, \qquad g(d)=\rho(d)/d^r$$

for squarefree $d$. Both functions are multiplicative. At a prime $p\nmid rk$, expansion in additive characters and the monomial Gauss-sum bound give

$$\rho(p)=p^{r-1}+O_{r,k}(p^{r/2}),\qquad g(p)=p^{-1}+O_{r,k}(p^{-r/2}).$$

There are only finitely many other primes. Also $g(p)<1$ at every prime, since $G(1,0,\ldots,0)=1$. Since $r\ge5$, the product comparison $\sum_p p^{1-r/2}<\infty$ gives the uniform bound $\rho(d)\ll_{r,k}d^{r-1}$ for squarefree $d$.

Counting the box in residue classes now gives, uniformly for squarefree $d\le Z$,

$$\#\{\mathbf z\in(Z,2Z]^r:d\mid G(\mathbf z)\} =g(d)Z^r+O_{r,k}(Z^{r-1}). \tag{6}$$

For each residue vector the error is $O_r((Z/d)^{r-1})$, which explains the uniform remainder after summing over $\rho(d)$ roots.

Choose $2\eta<\gamma<1$ and use sieve level $D=Z^\gamma$ with sifting bound $z=Z^\eta$. The summed remainders in (6) are $O(Z^{r-1+\gamma})=o(Z^r/\log Z)$. The dimension-one lower linear sieve has positive lower factor when $\log D/\log z=\gamma/\eta>2$; the applicable statement, hypotheses, and positive lower factor were checked in Tao, The linear sieve, Theorem 2, equations (1)–(3), (11), (13). Its local product satisfies $\prod_{p<z}(1-g(p))\asymp_{r,k}1/\log z$. It therefore proves (5).

In terms of root height $N=(2Z)^k$, the targets in (5) have no prime factor below $\tfrac12N^{\eta/k}$, while every root prime is at most $N^{1/k}$. In particular, choosing $1/k<\theta$ produces represented power-rough targets with comparable roots and no prime anchor above $N^\theta$. The tuples form a sparse input set when $k>1$; (5) is not a positive-density assertion about distinct targets and does not exclude a statistical use of roughness.

Outcome

The restriction (1) avoids the immediate zero-density problem with prime roots alone. But neither prime divisibility nor the elementary bilinear estimate controls its fibers. A useful continuation would have to estimate the six bilinear sums jointly, before the Cauchy step in (3), or supply a different arithmetic mechanism. No such estimate is established here; another conditional moment bound is not a resolution.