Split prime factors on typical cubic outputs

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Statement and scope

Put $F=x^3+2y^3+4z^3$, and define

$$\Omega(n)=\#\{p\parallel n:\ p\equiv1\pmod3,\ 2\text{ is a cube modulo }p\}.$$

For uniform $\mathbf x\in[1,N]^3\cap\mathbb Z^3$,

$$\frac{\Omega(F(\mathbf x))}{\log\log N} \longrightarrow\frac16 \quad\text{in probability}. \tag{1}$$

The proof below uses Chebotarev for the fixed splitting field of $T^3-2$, the Hasse bound for an elliptic curve, and elementary residue-box counting.

This is an arithmetic statistic on inputs, not a bound for representation multiplicity. It arose in an attempted secant-orbit descent: deleting outputs with more than a fixed number of these prime factors cannot retain positive input mass.

First and second moments

Let $\mathcal P$ denote the displayed prime set. These are exactly the completely split primes of the degree-six splitting field of $T^3-2$. Chebotarev gives

$$\sum_{\substack{p\le Z\\p\in\mathcal P}}\frac1p =\frac16\log\log Z+O(1).$$

At each such prime, scaling the coordinates identifies the projective curve $F=0$ with the Fermat cubic. The Hasse bound gives

$$\beta_p:=p^{-3}\#\{\mathbf u\bmod p:F(\mathbf u)=0\} =p^{-1}+O(p^{-3/2}).$$

For $d=p$ or $pq$ with distinct primes in $\mathcal P$, CRT and counting residue classes in the box imply

$$\Pr(d\mid F(\mathbf x)) =\prod_{p\mid d}\beta_p+ O\!\left(N^{-1}+dN^{-2}+d^2N^{-3}\right). \tag{2}$$

For the error, there are $O(d^2)$ root classes modulo $d$, and each class has $(N/d)^3+O((N/d)^2+N/d+1)$ representatives.

Take $Z=N^{1/8}$ and put

$$T=\sum_{\substack{p\le Z\\p\in\mathcal P}} \mathbf1_{p\mid F(\mathbf x)}.$$

The errors in (2), summed over single primes or pairs, tend to zero. Since $\sum_p(\beta_p-p^{-1})$ and $\sum_p\beta_p^2$ converge, it follows that

$$\mathbb ET=\frac16\log\log N+O(1),\qquad \operatorname{Var}T=\frac16\log\log N+O(1). \tag{3}$$

Exact valuations and the large primes

The only singular zero of $F$ modulo these primes is the origin. Every nonsingular zero has $p^2$ lifts to a zero modulo $p^2$; the origin has $p^3$ such lifts. Consequently

$$\#\{\mathbf u\bmod p^2:F(\mathbf u)=0\}\ll p^4.$$

Using the same box estimate with modulus $p^2$ gives

$$\mathbb E \sum_{\substack{p\le Z\\p\in\mathcal P}} \mathbf1_{p^2\mid F(\mathbf x)} \ll \sum_p p^{-2} \;+\;\frac ZN+\frac{Z^3}{N^2}+\frac{Z^5}{N^3} =O(1). \tag{4}$$

Also $F(\mathbf x)\le7N^3$, so the number of distinct prime divisors greater than $Z$ is bounded by $\log(7N^3)/\log Z=O(1)$. Subtract (4) from the small-prime count and use (3), Chebyshev's inequality, and Markov's inequality. This proves (1).

Exact limitation for the orbit approach

For every fixed $K$, restricting to $\Omega(F(\mathbf x))\le K$ retains $o(N^3)$ inputs. More generally, convergence (1) persists after conditioning on any sequence of subsets of the root box with a fixed positive lower proportion. This includes the established positive-mass cube-free restriction; it would also include any positive-mass squarefree restriction.

Thus an orbit argument cannot simply discard all outputs of unbounded split-prime complexity. This statement does not assert that each prime yields a small integral partner, that different local classes lift simultaneously to rational points, or that any such lifts have height $O(n^{1/3})$. No global generator count or integer multiplicity bound was proved. The unresolved step is height-sensitive control of the integral representations, not the normal-order calculation above.