The complete congruence distribution of the special cubic

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Status

The statements below are proved. They show that even the joint congruence distribution has bounded second moment and bounded entropy deficit. They do not establish either density conclusion for the positive integer values of

$$F(x,y,z)=x^3+2y^3+4z^3.$$

In particular, they do not estimate the microscopic representation function in a growing integer box.

The subsequent sampling theorem makes this limitation precise: synthetic integer multiplicities preserve the actual box marginals modulo $N^2$ and every fixed modulus, converge to this same profinite density, and retain all the established major-arc estimates, but have support $O(N^2\log N)$ and unbounded microscopic entropy. They are not asserted to come from a polynomial.

A second moment uniform in every modulus

For a positive integer $q$, put

$$\rho_q(n)=q^{-2}\#\{\mathbf x\bmod q:F(\mathbf x)\equiv n\pmod q\}.$$

Use the complete sums and nonnegative resonance weights from the major-arc note:

$$S(q,b)=\sum_{x\bmod q}e(bx^3/q),\qquad A(q)=q^{-6}\sum_{\substack{b\bmod q\\(b,q)=1}} |S(q,b)S(q,2b)S(q,4b)|^2.$$

That note proves

$$\mathfrak S:=\sum_{d\ge1}A(d)<\infty, \qquad \sum_{d>m}A(d)\ll m^{-2/3}. \tag{1}$$

Proposition. For every $q\ge1$,

$$\frac1q\sum_{n\bmod q}\rho_q(n)^2 =\sum_{d\mid q}A(d)\le\mathfrak S, \tag{2}$$

and, with $0\log0=0$,

$$0\le\frac1q\sum_{n\bmod q}\rho_q(n)\log\rho_q(n) \le\log\mathfrak S. \tag{3}$$

Proof. The normalized Fourier coefficient of $\rho_q$ at $b$ is

$$\widehat\rho_q(b)=q^{-3}S(q,-b)S(q,-2b)S(q,-4b).$$

Parseval therefore expresses the left side of (2) as the sum of the squares of these coefficients. Group frequencies by their exact conductor $d=q/(b,q)$. For $b=(q/d)c$,

$$S(q,jb)=(q/d)S(d,jc)\qquad(j=1,2,4).$$

The conductor-$d$ group is exactly $A(d)$, proving (2).

Also $q^{-1}\sum_n\rho_q(n)=1$. Convexity of $t\log t$ gives the lower bound in (3). Apply concavity of the logarithm to the probability weights $\rho_q(n)/q$, restricted to their nonzero support, to get

$$\frac1q\sum_n\rho_q(n)\log\rho_q(n) \le\log\left(\frac1q\sum_n\rho_q(n)^2\right).$$

This proves the upper bound. $\square$

Quantitative convergence on the profinite integers

Let $\mu$ be normalized Haar measure on $\widehat{\mathbb Z}$, and view each $\rho_q$ as a function on that group. For $q\mid Q$, counting residue lifts gives

$$\mathbb E_\mu(\rho_Q\mid n\bmod q)=\rho_q. \tag{4}$$

Put $Q_m=\operatorname{lcm}(1,\ldots,m)$. Equations (2) and (4) make $\rho_{Q_m}$ an $L^2$-bounded martingale. It consequently converges in $L^2$ to a nonnegative function $\rho$, with $\int\rho\,d\mu=1$. In fact

$$\|\rho\|_2^2=\mathfrak S, \qquad \|\rho-\rho_{Q_m}\|_2^2 =\mathfrak S-\sum_{d\mid Q_m}A(d) \le\sum_{d>m}A(d)\ll m^{-2/3}. \tag{5}$$

Here the equality of norms follows because every fixed positive integer eventually divides $Q_m$; the identity for the difference is orthogonality of conditional expectation. The inequality uses the fact that every $d\le m$ already divides $Q_m$.

The pushforward of Haar measure on $\widehat{\mathbb Z}^3$ under $F$ is exactly $\rho\,d\mu$. Indeed its mass on each cylinder $n\equiv a\pmod q$ is $\rho_q(a)/q$, and (4) gives the same mass for $\rho\,d\mu$. These cylinder sets determine a measure. Moreover

$$0\le\int\rho\log\rho\,d\mu\le\log\mathfrak S. \tag{6}$$

For example, the upper bound follows directly from Jensen under the probability measure $\rho\,d\mu$, using $\int\rho^2=\mathfrak S$. All quantities are integrable because $\rho\in L^2$.

Exact descent and density at the prime three

Write $r_F(n)$ for the number of all positive integral representations of $n$, without a box restriction.

Proposition. One has

$$9\mid F(x,y,z)\quad\Longrightarrow\quad3\mid x,y,z, \tag{7}$$

and hence

$$r_F(27n)=r_F(n),\qquad r_F(n)=0\quad\text{if }v_3(n)\equiv2\pmod3. \tag{8}$$

The pushforward of normalized Haar measure on $\mathbb Z_3^3$ under $F$ has density, away from zero,

$$f_3(n)= \begin{cases} 1,&v_3(n)\equiv0\pmod3,\\ 4/3,&v_3(n)\equiv1\pmod3,\\ 0,&v_3(n)\equiv2\pmod3. \end{cases} \tag{9}$$

Proof. Every cube modulo nine is $0,1$, or $-1$. For $\epsilon_i\in\{-1,0,1\}$, the integer $\epsilon_1+2\epsilon_2+4\epsilon_3$ lies between $-7$ and $7$. If it is zero modulo nine it is zero as an integer. Since $4>2+1$ and $2>1$, this forces all three $\epsilon_i$ to vanish. This proves (7). Dividing all three coordinates by three proves the bijection in (8); iterating it proves the vanishing assertion.

The 27 choices of coordinate residues modulo three give the following numbers of output occurrences modulo nine, in order from residue zero through residue eight:

$$(1,3,3,4,3,3,4,3,3). \tag{10}$$

These counts follow by multiplying $(1+t+t^{-1})(1+t^2+t^{-2})(1+t^4+t^{-4})$ modulo $t^9-1$. The sole input class giving output zero modulo nine is the all-zero class. Every other class has a unit coordinate. For $s\in\{1,-1\}$,

$$\frac{(s+3u)^3-s^3}{9}=s^2u+3su^2+3u^3$$

is a bijection of $\mathbb Z_3$: it reduces to $u$ modulo three and its derivative is a unit everywhere, so every residue lifts uniquely. Varying this unit coordinate therefore distributes the output uniformly within its fixed residue class modulo nine, even after the other two coordinates are fixed.

Each input residue triple has measure $1/27$; each output class modulo nine has measure $1/9$. Thus (10) gives density one on unit outputs and $4/3$ on outputs of valuation one. All remaining inputs have all coordinates divisible by three. Scaling them down costs a factor $1/27$ in input measure and scales output by 27, also a factor $1/27$ in output Haar measure. Their normalized density is consequently unchanged. Iterating proves (9). There is no atom at zero, since (7) forces an exact zero in $\mathbb Z_3^3$ to be the zero vector. $\square$

In particular, the measures of the three valuation classes modulo three are respectively $9/13,3/13,1/13$. Consequently

$$\mu_3(F(\mathbb Z_3^3))=\frac{12}{13},\qquad \int f_3^2\,d\mu_3=\frac{43}{39},\qquad \int f_3\log f_3\,d\mu_3=\frac4{13}\log\frac43. \tag{11}$$

The second identity agrees with the independently computed local resonance factor in the major-arc note.

What remains unproved

For a growing integer box, the required estimate would instead be, for example,

$$\sum_n R_N(n)\log R_N(n)\ll N^3, \qquad R_N(n)=\#\{\mathbf x\in[1,N]^3:F(\mathbf x)=n\}. \tag{12}$$

Equations (2)--(6) average over complete residue boxes, not over $[1,N]^3$ at moduli of the size of the output range $N^3$. No comparison supplying (12) has been proved. Taking a limit over fixed congruences cannot by itself control concentration on a sparse set of ordinary integers. The corresponding minor-arc correlation term in the major-arc note remains uncontrolled.

Thus the exact descent at three and the bounded joint congruence entropy do not settle the density question or any higher-exponent infinitude case.