Cube-free outputs and cubic multiplicity

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Status

The elementary statements below are proved and author-recorded. They concern $F(x,y,z)=x^3+2y^3+4z^3$. A positive proportion of input triples have cube-free output; this is not a positive-density theorem for distinct outputs. The remaining issue is still multiplicity.

A cube-free sieve at the critical scale

Theorem. Put

$$\rho_p=p^{-9}\#\{\mathbf u\bmod p^3:F(\mathbf u)\equiv0\pmod{p^3}\}.$$

Then

$$\#\{\mathbf x\in[1,N]^3\cap\mathbb Z^3:F(\mathbf x)\text{ is cube-free}\} =\delta_F N^3+o(N^3),\qquad \delta_F=\prod_p(1-\rho_p)>0. \tag{1}$$

Every triple counted in (1) is primitive.

For $p\ge5$, first count the triples with all coordinates divisible by $p$: there are at most $(N/p)^3$. Otherwise choose a unit coordinate. Its coefficient and the derivative of its cube term are units modulo $p$. For fixed other coordinates there are at most three possible unit residues modulo $p^3$, each lifting uniquely from a root modulo $p$. Union over the three possible choices of coordinate gives

$$B_p(N):=\#\{\mathbf x\in[1,N]^3:p^3\mid F(\mathbf x)\} \le 10N^3p^{-3}+9N^2. \tag{2}$$

Since $0<F(\mathbf x)\le7N^3$, only primes $p\le7^{1/3}N$ occur. The elementary bound $\pi(t)\ll t/\log t$ consequently gives, for $P\ge3$,

$$\#\{\mathbf x:\exists p>P,\ p^3\mid F(\mathbf x)\} \ll N^3P^{-2}+\frac{N^3}{\log N}. \tag{3}$$

The same unit-coordinate argument in a complete residue box gives $\rho_p\le10p^{-3}$ for $p\ge5$. Every factor $1-\rho_p$ is positive, because $F(1,0,0)=1$. Thus the product in (1) converges to a positive number.

For fixed $P$, CRT and elementary box counting give

$$\#\{\mathbf x\in[1,N]^3:p^3\nmid F(\mathbf x)\ (p\le P)\} =N^3\prod_{p\le P}(1-\rho_p)+O_P(N^2).$$

Use (3), let $N\to\infty$, and then let $P\to\infty$. This proves (1). If a prime divided all coordinates, its cube would divide the output, proving primitivity. At $p=2$, more explicitly, $8\mid F(\mathbf x)$ if and only if all three coordinates are even: the valuations of the nonzero terms are distinct modulo three. Hence $\rho_2=1/8$.

Bounded average logarithm of the cube part

For a positive integer $n$, define

$$D(n)=\prod_p p^{\lfloor v_p(n)/3\rfloor}.$$

Thus $D(n)^3$ is the largest perfect cube dividing $n$.

Theorem. Uniformly for $N\ge2$,

$$\sum_{\mathbf x\in[1,N]^3}\log D(F(\mathbf x))\ll N^3. \tag{4}$$

First restrict to primitive triples. At every prime, at least one coordinate is a unit. For $p\ge5$, the corresponding one-variable congruence modulo $p^k$ has at most three unit roots. This remains true at $p=3$, since cubing on $(\mathbb Z/3^k\mathbb Z)^\times$ has kernel of order at most three. The coefficients of $F$ are units there. At $p=2$, a primitive triple has $v_2(F(\mathbf x))\le2$, by the distinct-valuation argument above, so it contributes no cube factor. Consequently for every odd prime and $k\ge1$,

$$\#\{\mathbf x\in[1,N]^3:\gcd(\mathbf x)=1,\ p^k\mid F(\mathbf x)\} \ll N^3p^{-k}+N^2. \tag{5}$$

The absolute constant in (5) does not depend on $p,k,N$.

Write $C=7^{1/3}$ and $J_p=\lfloor\log(7N^3)/(3\log p)\rfloor$. Expanding $\log D(n)=\sum_p\sum_{j\ge1}(\log p)\mathbf1_{p^{3j}\mid n}$, equation (5) bounds the primitive contribution to (4) by a constant times

$$N^3\sum_p\frac{\log p}{p^3-1} +N^2\sum_{p\le CN}J_p\log p.$$

The first series converges. The second prime sum is at most $\tfrac13\log(7N^3)\pi(CN)\ll N$. The primitive contribution is therefore $O(N^3)$.

For an arbitrary triple write $\mathbf x=g\mathbf y$, with $\mathbf y$ primitive. Homogeneity gives $D(F(\mathbf x))=gD(F(\mathbf y))$. Sum the primitive estimate at $N/g$, together with the term $\log g$, and use $\sum_g g^{-3}(1+\log g)<\infty$. This proves (4). In particular,

$$\#\{\mathbf x\in[1,N]^3:D(F(\mathbf x))>H\} \ll\frac{N^3}{\log H}\qquad(H>1). \tag{6}$$

This controls the output's cube part, not the logarithm of its representation multiplicity.

Primitive collisions at cube-free outputs remain

The split identity already proved in the energy note is

$$\begin{aligned} F(27m+92s,212m+480s,171m+515s)\\ =F(27m+12s,212m+642s,171m+391s). \end{aligned} \tag{7}$$

Its common value is

$$G(m,s)=39076783m^3+310349304m^2s +837996084ms^2+768326188s^3. \tag{8}$$

There are $\gg T^2$ pairs $1\le m,s\le T$ for which (8) is cube-free. Here is a sieve proof, including the local conditions.

The binary cubic $G$ has no rational projective zero: the rank-two linear map on either side of (7) would turn such a zero into a nonzero rational zero of $F$, contrary to its distinct $2$-adic valuations. Thus $G$ is irreducible over $\mathbb Q$, and its discriminant is nonzero. Also

$$\gcd(G(1,0),G(0,1))=\gcd(39076783,768326188)=1.$$

For each prime, at least one of these two values is a unit, so $G$ has no fixed prime-cube divisor.

Outside finitely many primes dividing its discriminant, every nonzero zero of $G$ modulo $p$ has a nonzero partial derivative. Split off the pairs with $p\mid m,s$; these contribute at most $(T/p)^2$. For the other pairs, choose a coordinate with nonzero derivative and fix the other coordinate. There are at most three simple roots modulo $p^3$ for either choice. Hence

$$\#\{1\le m,s\le T:p^3\mid G(m,s)\} \le T^2p^{-2}+6T(Tp^{-3}+1). \tag{9}$$

Only primes $p\ll_G T$ can occur. Once $P$ contains the finitely many bad primes, (9) gives the tail bound

$$\#\{(m,s):\exists p>P,\ p^3\mid G(m,s)\} \ll_G T^2P^{-1}+\frac{T^2}{\log T}.$$

In complete residue boxes the local excluded proportions are $O(p^{-2})$. Their product of complements is positive by local nonobstruction. Finite CRT, followed by this tail estimate, proves the asserted positive input density of cube-free $G$-values.

For these parameters the two triples in (7) are positive, primitive and distinct. They lie in $[1,854T]^3$. The map from parameters to collision pairs is injective: their difference is $(80s,-162s,124s)$, which determines $s$, and then the first coordinate determines $m$. Thus this fixed split plane supplies $\gg N^2$ primitive collisions with cube-free output in a box of side $N$.

Consequence for the density route

The restriction to cube-free outputs retains $\gg N^3$ input triples, and $8F(\mathbf x)\in S_3$. A bound $O(N^3)$ for their restricted collision count would therefore give positive lower density of $S_3$. No such bound, nor the weaker bounded-fiber positive-mass criterion, has been established. Equations (4) and (7) show why merely deleting common dilations does not complete the argument: primitive split collisions survive, and the critical average over large directions is still uncontrolled.

The primitive-fiber analysis now separates the bounded entropy of common dilations from primitive multiplicity. It also shows that every curve defined over $\mathbb Q$ on the special cubic's output surface has degree divisible by three, excluding rational lines and conics even without cube-freeness. No uniform estimate for its higher-degree curves or primitive fibers was obtained.