Summable rational resonances for the special cubic

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Status

The results below are proved. They retain and estimate the rational resonances that obstruct the uniform contraction proposed in the carry-spectrum note. They do not bound the full collision energy or establish positive density. In particular, the minor-arc correlation with high-multiplicity outputs remains uncontrolled.

Put $e(t)=\exp(2\pi it)$ and

$$S_N(\alpha)=\sum_{1\le x\le N}e(\alpha x^3),\qquad W_N(\alpha)=S_N(\alpha)S_N(2\alpha)S_N(4\alpha).$$

For $F(x,y,z)=x^3+2y^3+4z^3$, Parseval identifies $\int_0^1|W_N(\alpha)|^2\,d\alpha$ with the box collision count. Define

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

with $A(1)=1$.

Theorem. The positive series $\mathfrak S=\sum_{q\ge1}A(q)$ converges, and

$$\sum_{q>Q}A(q)\sim c\,Q^{-2/3}\quad(Q\longrightarrow\infty),\qquad c>0. \tag{1}$$

For integers $N\ge2$ and $1\le Q\le N^{1/2}$, let

$$\mathfrak M(N,Q)= \bigcup_{\substack{1\le q\le Q\\(a,q)=1}} \left\{\alpha\in\mathbb R/\mathbb Z: \left|\alpha-\frac aq\right|\le\frac{Q}{qN^3}\right\}.$$

Put

$$I_j(t)=\int_0^1e(jtu^3)\,du,\qquad \mathfrak J=\int_{\mathbb R}|I_1(t)I_2(t)I_4(t)|^2\,dt.$$

Then $0<\mathfrak J<\infty$, and

$$\int_{\mathfrak M(N,Q)}|W_N(\alpha)|^2\,d\alpha =N^3\mathfrak J\mathfrak S+ O\left(N^3Q^{-2/3}+N^2Q^2\right). \tag{2}$$

For example, $Q=N^{3/8}$ gives an error $O(N^{11/4})$.

Exact prime-power weights

CRT shows that $A$ is multiplicative. The unit frequency is reparametrized by units when a complete sum splits over coprime moduli, compatibly for the three coefficients.

At $2$, the proved exact uniformity of $F$ modulo every $2^k$ gives

$$S(2^k,a)S(2^k,2a)S(2^k,4a)=0\quad(a\text{ odd}), \qquad A(2^k)=0\quad(k\ge1). \tag{3}$$

This uses uniformity of the three-variable form, not the false assertion that cubing permutes every residue modulo $2^k$.

For $p\ge5$ and $p\nmid b$,

$$S(p^2,b)=p,\qquad S(p^m,b)=p^2S(p^{m-3},b)\quad(m\ge3). \tag{4}$$

To prove this, write $x=y+p^{m-1}z$ and first sum over $z\bmod p$. The sum vanishes unless $p\mid y$. For $m\ge3$, substitute $y=pu$; the remaining summand has one redundant period. The case $m=2$ is immediate from the same calculation. Consequently

$$A(p^2)=(1-p^{-1})p^{-4},\quad A(p^3)=(1-p^{-1})p^{-3},\quad A(p^{k+3})=p^{-3}A(p^k)\quad(k\ge1). \tag{5}$$

The last recurrence must not be applied at $k=0$.

At modulus $p$, the cubic-character expansion and the ordinary Gauss-sum identity give $|S(p,b)|\le2\sqrt p$ for $p\equiv1\pmod3$. If $p\equiv2\pmod3$, cubing is a permutation and the sum vanishes. Thus

$$0\le A(p)\le64p^{-2}. \tag{6}$$

For the exceptional odd prime,

$$S(3,b)=0,\qquad S(9,b)=3\bigl(1+2\cos(2\pi b/9)\bigr) \quad(3\nmid b).$$

The numbers $2\cos(2\pi j/9)$, $j=1,2,4$, are the roots of $z^3-3z+1$. Their translates by one have product $-3$. Every unit modulo $9$ permutes the classes $\{\pm1,\pm2,\pm4\}$, so

$$S(9,a)S(9,2a)S(9,4a)=-81,\qquad A(9)=2/27. \tag{7}$$

Also

$$S(3^m,b)=9S(3^{m-3},b)\quad(m\ge3,\ 3\nmid b). \tag{8}$$

Indeed, write $x=y+3^{m-2}z$ with $z\bmod9$. Modulo $3^m$, $x^3\equiv y^3+3^{m-1}y^2z$. The $z$-sum is zero unless $3\mid y$, in which case it equals nine. Substitute $y=3u$. Thus

$$A(3)=0,\quad A(27)=2/81,\quad A(3^{k+3})=3^{-3}A(3^k)\quad(k\ge1).$$

In particular the exact local factor is

$$\sum_{k\ge0}A(3^k) =1+\frac{2/27+2/81}{1-1/27}=\frac{43}{39}. \tag{9}$$

The sharp resonance tail

For every odd prime the preceding formulas give the formal identity

$$\sum_{k\ge0}A(p^k)t^k =\frac{1+A(p)t+A(p^2)t^2-p^{-4}t^3}{1-p^{-3}t^3}. \tag{10}$$

Let $B$ be the nonnegative multiplicative function supported on odd cube-free integers, with $B(p)=A(p)$ and $B(p^2)=A(p^2)$. Dropping the negative numerator term coefficientwise gives

$$A(q)\le\sum_{dm^3=q}B(d)m^{-3}, \tag{11}$$

where the sum may include even $m$ without harming the upper bound. Equations (5)–(7) imply

$$\sum_dB(d)d^{2/3} =\prod_{p\text{ odd}}\left(1+A(p)p^{2/3} +A(p^2)p^{4/3}\right)<\infty. \tag{12}$$

Since $\sum_{m>T}m^{-3}\ll T^{-2}$ uniformly for $T>0$,

$$\sum_{q>Q}A(q) \le\sum_dB(d)\sum_{m>(Q/d)^{1/3}}m^{-3} \ll Q^{-2/3}\sum_dB(d)d^{2/3}. \tag{13}$$

For the matching lower bound, every odd $m$ satisfies exactly

$$A(m^3)=\frac{\varphi(m)}{m^4}. \tag{14}$$

The identity $\varphi(m)/m=\sum_{d\mid m}\mu(d)/d$ yields

$$\sum_{\substack{m\le X\\m\text{ odd}}}\frac{\varphi(m)}m =\frac4{\pi^2}X+O(\log(2X)).$$

Partial summation gives

$$\sum_{\substack{m>L\\m\text{ odd}}}\frac{\varphi(m)}{m^4} \sim\frac2{\pi^2}L^{-2}.$$

Retaining only $q=m^3>Q$ gives the matching lower bound in order of magnitude. The upper bound also implies

$$\sum_{q\le Q}qA(q)\ll Q^{1/3}; \tag{15}$$

integrate $\sum_{q>t}A(q)$ over $0\le t\le Q$.

For the exact asymptotic in (1), define a signed multiplicative function $C$, supported on odd integers, by the local values

$$C(p)=A(p),\quad C(p^2)=A(p^2),\quad C(p^3)=-p^{-4}, \quad C(p^k)=0\ (k\ge4).$$

Equation (10) gives the exact convolution

$$A(q)=\sum_{\substack{dm^3=q\\d,m\text{ odd}}}C(d)m^{-3}.$$

Moreover $\sum_d|C(d)|d^{2/3}<\infty$, by the same Euler-product comparison as (12), with the additional summable term $p^{-2}$. Since

$$\sum_{\substack{m>T\\m\text{ odd}}}m^{-3}\sim\tfrac14T^{-2}$$

and this tail is $O(T^{-2})$ for every $T>0$, absolute rearrangement and dominated convergence prove (1), with

$$c=\frac14\prod_{p\text{ odd}} \left(1+A(p)p^{2/3}+A(p^2)p^{4/3}-p^{-2}\right).$$

The deviations of these factors from one are absolutely summable. Each factor is at least $1-p^{-2}>0$, so the product is positive.

Major-arc approximation without an epsilon loss

On $\alpha=a/q+\beta$, residue-class summation and partial summation give

$$S_N(j\alpha)=\frac Nq S(q,ja)I_j(N^3\beta) +O\bigl(q(1+N^3|\beta|)\bigr)\qquad(j=1,2,4). \tag{16}$$

The implicit constants are absolute here. On the stated arcs the error is $O(Q)$. Every exact factor and every main factor is at most $N$ in absolute value, so telescoping the product yields

$$W_N(\alpha)=M_N(\alpha)+O(QN^2), \tag{17}$$

where, on the arc at $a/q$,

$$M_N(a/q+\beta)=N^3q^{-3}S(q,a)S(q,2a)S(q,4a) I_1(N^3\beta)I_2(N^3\beta)I_4(N^3\beta).$$

The arcs are disjoint for $N\ge2$: distinct reduced fractions are separated by at least $1/(qq')$, and the sum of their radii is at most $2Q^2/(qq'N^3)$. Their total measure is $O(Q^2/N^3)$. Consequently

$$\int_{\mathfrak M}|W_N-M_N|^2\ll NQ^4. \tag{18}$$

In particular, no factor $q^\varepsilon$ or $N^\varepsilon$ is hidden in the approximation error.

Substitution $v=u^3$, followed by splitting at $v=|t|^{-1}$ and integrating by parts on the remaining interval, proves $I_j(t)\ll(1+|t|)^{-1/3}$. Hence

$$\Phi(t):=|I_1(t)I_2(t)I_4(t)|^2\ll(1+|t|)^{-2}.$$

This proves finiteness of $\mathfrak J$, while continuity at zero proves positivity. Direct integration of the disjoint main terms gives

$$N^{-3}\int_{\mathfrak M}|M_N|^2 =\sum_{q\le Q}A(q)\int_{|t|\le Q/q}\Phi(t)\,dt.$$

The omitted denominators cost $O(Q^{-2/3})$ by (13), and the omitted integral tails cost $O(Q^{-1}\sum_{q\le Q}qA(q))=O(Q^{-2/3})$ by (15). Therefore

$$\int_{\mathfrak M}|M_N|^2 =N^3\mathfrak J\mathfrak S+O(N^3Q^{-2/3}).$$

Cauchy–Schwarz bounds the cross term in (17) by $O(N^2Q^2)$, using (18). The squared error $O(NQ^4)$ is no larger because $Q\le N^{1/2}$. This proves (2).

High-multiplicity outputs: the remaining minor-arc term

For a finite set $E$ of output integers put $P_E(\alpha)=\sum_{n\in E}e(-n\alpha)$. Equation (2), Cauchy–Schwarz, and $\int_0^1|P_E|^2=|E|$ imply

$$\left|\int_{\mathfrak M(N,Q)}W_N(\alpha)P_E(\alpha)\,d\alpha\right| \ll N^{3/2}|E|^{1/2}. \tag{19}$$

If $E=\{n:R_N(n)>K\}$, then $|E|\le N^3/K$ because $\sum_nR_N(n)=N^3$. Thus the major-arc contribution to the input mass over these fibers is $O(N^3K^{-1/2})$, uniformly in $N$.

The exact full input mass is $\sum_{n\in E}R_N(n)=\int_0^1W_NP_E$. The part over the complement of $\mathfrak M$ has not been bounded sufficiently. In particular, (19) is not a bound for this full nonnegative sum: either restricted Fourier integral can have either sign. A small-set correlation estimate for the minor arcs would supply a new route to tightness, but no such estimate is established here. Bounded major-arc energy alone is compatible with concentration on a density-zero set of outputs.

This compatibility is demonstrated by the congruence-preserving sampling construction. Its synthetic integer sequence preserves (2), (17), and (18), as well as all fixed congruences and the joint profinite limit, but has $O(N^2\log N)$ support and a heavy-fiber minor-arc correlation $(1+o(1))N^3$. It need not factor into cubic Weyl sums, so a proof using that additional structure is not excluded.

Truncated restriction estimates leave a moderate-amplitude range

Here the degree is denoted by $k\ge3$. Put

$$f_N(\alpha)=\sum_{x=1}^N e(\alpha x^k),\qquad \tau=\max\{2^{1-k},1/(k(k-1))\},\qquad 0<\zeta<\tau/2.$$

Henriot--Hughes, Theorem 1.1 and (1.5) give, for every fixed $\delta>0$,

$$\int_{|f_N|\ge N^{1-\zeta}}|f_N|^{2k+\delta} \ll_{k,\delta,\zeta}N^{k+\delta}. \tag{20}$$

The full statement was checked: the cutoff includes the coefficient $\ell^2$ norm, which is $\sqrt N$ here. In particular, it is not a square-root cutoff. No uniform endpoint limit in $\delta$ is asserted. The direct consequence at the critical moment is only

$$\int_{|f_N|\ge N^{1-\zeta}}|f_N|^{2k} \ll_{k,\delta,\zeta}N^{k+\zeta\delta}. \tag{21}$$

One can isolate the further estimate needed even if (21) were improved. Let $R_N$ count ordered sums of $k$ positive $k$th powers from $[1,N]$, let $E_K=\{n:R_N(n)>K\}$, and put $P_K(\alpha)=\sum_{n\in E_K}e(-n\alpha)$. Then

$$|E_K|\le N^k/K,\qquad \sum_{n\in E_K}R_N(n)=\int_0^1 f_N^kP_K.$$

If an available moment estimate is $\int_0^1|f_N|^{2s}\ll N^{s+\Delta_s}$, $1\le s\le k$, Cauchy--Schwarz and the cutoff give exactly

$$\left|\int_{|f_N|<N^{1-\zeta}}f_N^kP_K\right| \ll \frac{N^k}{\sqrt K} N^{\{(k-s)(1-2\zeta)+\Delta_s\}/2}. \tag{22}$$

Indeed, bound $|f_N|^{2(k-s)}$ by $N^{2(k-s)(1-\zeta)}$, and use $\|P_K\|_2=|E_K|^{1/2}$. Even a hypothetical $O(N^k)$ bound for the high-amplitude critical moment would control only its contribution by $O(N^k/\sqrt K)$. For $s<k$ and $\Delta_s\ge0$, the exponent in (22) is strictly positive. At $s=k$ it vanishes when $\Delta_k=0$, but that assumption already supplies the sufficient full critical-moment bound. The estimates used in this attempt leave a positive exponent and hence do not give a fixed-$K$ heavy-fiber estimate. Improving the endpoint only above the cutoff would not remove this loss. No new density conclusion follows from this application of truncated restriction; other uses of its weighted version are not ruled out.