Uniform lattice counting and moving-modulus quintic means

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Status

The exact inequalities below are proved and author-recorded. They extend the fixed-progression quintic theorem to moving moduli and intervals. They remain statements about representation counts, not distinct represented values, and do not decide either unresolved question for $S_5$.

An exact lattice bound

Let $r\ge1$ be an integer, let $d_i>0$ be integers, and put

$$F_{\mathbf d}(\mathbf x)=\sum_{i=1}^r d_i x_i^r,\qquad V_{\mathbf d}=\Gamma(1+1/r)^r\prod_i d_i^{-1/r},\qquad \kappa_{\mathbf d}=\left(\sum_i d_i\right)^{1/r}.$$

For $M\ge1$ and an integer $a$, write

$$\rho_M(a;\mathbf d)=M^{1-r} \#\{\mathbf b\bmod M:F_{\mathbf d}(\mathbf b)\equiv a\pmod M\},$$

and let $R_{\mathbf d}(X;a,M)$ count positive integral $\mathbf x$ with $F_{\mathbf d}(\mathbf x)\le X$ in this target class. For every real $X\ge0$,

$$\frac{\rho_M V_{\mathbf d}}M \bigl(X^{1/r}-\kappa_{\mathbf d}M\bigr)_+^r \le R_{\mathbf d}(X;a,M) \le\frac{\rho_M V_{\mathbf d}}M \bigl(X^{1/r}+\kappa_{\mathbf d}M\bigr)^r. \tag{1}$$

Proof. Use the weighted norm $\|\mathbf z\|_{\mathbf d}=(\sum_i d_i|z_i|^r)^{1/r}$. The volume of its radius-$t$ ball in the positive orthant is $V_{\mathbf d}t^r$.

Fix one admissible residue vector. Attach the disjoint cubes $\mathbf q+[0,M)^r$ to its positive lattice points $\mathbf q$ of norm at most $t=X^{1/r}$. Every such cube lies in the positive ball of radius $t+\kappa_{\mathbf d}M$, proving the upper bound for this residue vector.

For the lower bound, take any positive real vector of norm at most $(t-\kappa_{\mathbf d}M)_+$ and round each coordinate upwards to the least positive integer in the prescribed residue class. The resulting lattice point has norm at most $t$. Each lattice point has preimage volume at most $M^r$. Thus the number of counted points is at least $V_{\mathbf d}(t-\kappa_{\mathbf d}M)_+^r/M^r$. Summing over the $\rho_M M^{r-1}$ admissible residue vectors proves (1).

No constants in (1) depend implicitly on $a$ or $M$.

Consequence for the finite quintic certificate family

Use the finite set $\mathcal E$ and canonical family $\mathcal C$ of the quintic first-moment proof:

$$\mathcal E=\{p\le1000:p\text{ prime},\ p\ne5,\ p\not\equiv1\pmod5\},$$

and each prime exponent in a member of $\mathcal C$ is $0,6,7,8$, or $9$. That proof establishes the finite coefficient inequality

$$\frac1{120}\sum_{\mathbf d\in\mathcal C^5} V_{\mathbf d}\rho_M(a;\mathbf d)>\Lambda:=\frac{429}{400} \tag{2}$$

for every $M,a$, not merely after taking a fixed-modulus limit. Its finite certificate is used as established here, without a new computation.

Put

$$\kappa=5^{1/5}\prod_{p\in\mathcal E}p^{9/5}.$$

Then $\kappa_{\mathbf d}\le\kappa$ for all tuples in this family. Canonical uniqueness and the upper bound $120$ on the number of orderings of a five-element multiset give

$$U_5(X;a,M)\ge \Lambda\frac XM \left(1-\frac{\kappa M}{X^{1/5}}\right)_+^5. \tag{3}$$

The inequality is strict whenever the bracket is positive. Hence uniformly for any moving moduli $M=o(X^{1/5})$ and residues $a$,

$$\liminf\frac M X U_5(X;a,M)\ge\frac{429}{400}>1. \tag{4}$$

In fact a below-one first moment remains impossible for all $M\le cX^{1/5}$ with any fixed

$$0<c<\kappa^{-1}\left(1-\left(\frac{400}{429}\right)^{1/5}\right).$$

The constant is extremely small but positive and independent of $X,a,M$.

Moving intervals

Let $0\le A<B$ and $H=B-A$. Subtract the upper bound (1) at $A$ from the lower bound at $B$, using the same finite coefficient family. This gives

$$U_5((A,B];a,M)\ge\frac\Lambda M \left[\bigl(B^{1/5}-\kappa M\bigr)_+^5 -\bigl(A^{1/5}+\kappa M\bigr)^5\right]_+. \tag{5}$$

Again it is strict if the displayed bracket is positive. Write $h=\kappa M$ and suppose $h\le B^{1/5}$. The mean-value theorem gives

$$B-(B^{1/5}-h)^5\le5hB^{4/5}, \qquad (A^{1/5}+h)^5-A\le80hB^{4/5}.$$

Thus the inner difference in (5) is at least $H-85\kappa M B^{4/5}$. In particular,

$$H\ge3400\kappa M B^{4/5} \quad\Longrightarrow\quad U_5((A,B];a,M)> \frac{16731}{16000}\frac HM>\frac HM. \tag{6}$$

The hypothesis implies $h\le B^{1/5}/3400$, so the preceding bound is applicable. More generally the liminf in (4), with $X$ replaced by the interval length $H$, is at least $429/400$ whenever $H/(MB^{4/5})\to\infty$.

These estimates exclude a below-one first-moment argument throughout these ranges. They say nothing comparable when $M$ is larger than the coordinate scale or the interval is shorter than the boundary error, and they do not rule out arguments exploiting repeated representations.

Aligning finitely many axis intercepts does not force a gap

A different proposed gap construction was to make all fractional parts $(n/d)^{1/5}$, for a finite collection of cores, approach integers from a prescribed side. The following exact statement shows why qualitative alignment alone cannot work.

Proposition. For any finite set $D$ of canonical fifth-power cores, there is a positive integer $q=b^5+4$ such that

$$\left(\left\{(qm^5/d)^{1/5}\right\}\right)_{d\in D},\qquad m\ge1, \tag{7}$$

is equidistributed in the $|D|$-dimensional unit torus. Nevertheless every target $qm^5$ is a sum of five positive fifth powers.

Proof. Let $P_D$ denote the primes supporting $D$ and let

$$K=\mathbb Q(p^{1/5}:p\in P_D).$$

The monomials $\prod_{p\in P_D}p^{j_p/5}$, $0\le j_p<5$, form a basis over $\mathbb Q$. To check the degree, adjoin $5^{1/5}$ first if needed, then the remaining prime radicals one at a time. Before adjoining $p^{1/5}$, the preceding field is unramified at the fresh prime $p$: the discriminants of the previous defining polynomials are supported on $5$ and their respective defining primes, and composita of unramified extensions remain unramified. Therefore $T^5-p$ is Eisenstein at any prime above $p$, and each adjunction has degree five. The first adjunction is ordinary Eisenstein over $\mathbb Q$.

For a canonical core the exponents are $0,6,7,8,9$, giving distinct residues modulo five. Thus distinct $d^{-1/5}$, $d\in D$, are nonzero rational multiples of distinct basis monomials, and are linearly independent over $\mathbb Q$.

Set $P=\prod_{p\in P_D\cup\{2,5\}}p$ and $b_0=2P$. Choose a prime $\ell\mid 8P^5+1$. Then $\ell\nmid5P$ and $\ell\mid b_0^5+4$. The derivative of $f(T)=T^5+4$ at $b_0$ is a unit modulo $\ell$. If $v_\ell(f(b_0))=1$, take $b=b_0$; otherwise take $b=b_0+\ell$. Expansion modulo $\ell^2$ gives $v_\ell(q)=1$ for $q=b^5+4$ in either case.

The field $K$ is unramified at $\ell$, so $q^{1/5}\notin K$, by valuation. Consequently

$$1,\quad (q/d)^{1/5}\quad(d\in D)$$

are linearly independent over $\mathbb Q$: a relation with nonzero constant term would put $q^{1/5}$ in $K$, whereas a relation with zero constant term contradicts the independence of the core radicals. The geometric-series formula for each nontrivial integer Fourier character, followed by Fourier approximation, proves equidistribution of (7).

Finally,

$$qm^5=(bm)^5+m^5+m^5+m^5+m^5.$$

All five summands are positive, completing the proof.

For example, for every fixed $0<\varepsilon<1$, a proportion $\varepsilon^{|D|}$ of the parameters $m$ make all the roots lie within $\varepsilon$ of integers from below. More generally, every nonempty open pattern occurs among represented targets. Thus no exclusion criterion can depend solely on membership of these finitely many fractional parts in a fixed nonempty open region. This statement does not cover shrinking tolerances depending on $n$, extra arithmetic conditions, or control of the full curved lattice shell.

Representation-weighted root phases

There is a stronger quantitative statement for fixed continuous root-phase filters. It is proved here independently of the particular represented ray above.

Lemma. Fix $r\ge2$, positive integral coefficients $a_1,\ldots,a_r$, and integers $M\ge1,b$. Put $F(\mathbf x)=\sum_i a_ix_i^r$. For every fixed real $\lambda\ne0$,

$$\sum_{\substack{\mathbf x\in\mathbb N^r\\F(\mathbf x)\le X\\ F(\mathbf x)\equiv b\pmod M}} e\bigl(\lambda F(\mathbf x)^{1/r}\bigr)=o(X), \tag{8}$$

where $e(t)=\exp(2\pi it)$.

Proof. Let $N=X^{1/r}$ and fix $\delta>0$. Tuples with some $x_i<\delta N$ contribute $O_{F,r}(\delta N^r)$. For the other tuples, fix $x_2,\ldots,x_r$, set $B=\sum_{i>1}a_ix_i^r$, and split $x_1$ into residue classes modulo $M$. The congruence in (8) either holds on an entire such class or nowhere on that class; the size restrictions give one interval. The phase $f(x_1)=\lambda(a_1x_1^r+B)^{1/r}$ has

$$f''(x_1)=\lambda(r-1)a_1B x_1^{r-2} (a_1x_1^r+B)^{1/r-2}.$$

On the retained interval, $B\asymp_{F,\delta}N^r$ and $x_1\asymp_{F,\delta}N$, so $|f''|\asymp_{F,\delta}|\lambda|/N$. The standard second-derivative estimate for exponential sums on an interval, $\sum e(f(n))\ll L\sqrt\eta+\eta^{-1/2}$ when $\eta\le |f''|\ll\eta$, applies after parameterizing the residue class. It gives $O_{F,\delta,M,\lambda}(N^{1/2})$ per fiber. Thus the left side of (8) is

$$O_{F,r}(\delta N^r)+O_{F,\delta,M,\lambda}(N^{r-1/2}).$$

Let $N\to\infty$ and then $\delta\to0$. This proves (8).

For clarity, this argument also gives the following dependence on a nonzero varying frequency:

$$\frac{|S_F(X;\lambda)|}{X} \ll_{F,r}\delta+C_{F,\delta,M} \left(\sqrt{\frac{|\lambda|}{N}} +\frac1{\sqrt{N|\lambda|}}\right). \tag{9}$$

It does not give a useful relative error for every shrinking phase set.

Continuous weights have the same limiting mean

Fix positive real numbers $\alpha_1,\ldots,\alpha_k$, and let $H$ be the closure in $\mathbb T^k$ of the continuous line $\{t(\alpha_1,\ldots,\alpha_k):t\in\mathbb R\}$. Write $\mu_H$ for Haar probability measure. For any continuous $\psi:\mathbb T^k\to\mathbb C$, put

$$w(n)=\psi(\alpha_1n^{1/r},\ldots,\alpha_kn^{1/r}),\qquad \bar w=\int_H\psi\,d\mu_H.$$

Fourier approximation and (8) give

$$\sum_{\substack{\mathbf x\in\mathbb N^r\\F(\mathbf x)\le X\\ F(\mathbf x)\equiv b\pmod M}}w(F(\mathbf x)) =\bar w\,R_F(X;b,M)+o(X), \tag{10}$$

where $R_F$ is the unweighted tuple count. Indeed, a Fourier character indexed by $\mathbf h\in\mathbb Z^k$ is constant one precisely when $\sum_j h_j\alpha_j=0$; every other character is covered by (8).

Ordinary target counting has the identical mean:

$$\sum_{\substack{n\le X\\n\equiv b\pmod M}}w(n) =\bar w\,\frac X M+o(X). \tag{11}$$

For a nonzero Fourier frequency $\lambda$, comparison with the integral gives an error $O_{M,\lambda}(1+X^{1/r})$ for the exponential sum in (11). Substitution $u=t^{1/r}$ and integration by parts bound the integral by $O_{M,\lambda}(1+X^{1-1/r})$. Both are $o(X)$. Fourier approximation proves (11), including the case of rational dependencies among the $\alpha_j$.

Take $r=5$, $\alpha_j=d_j^{-1/5}$ for any fixed finite list of cores, and $\psi\ge0$. Apply (10) to the finite canonical family used in the quintic certificate, and divide its ordered counts by $5!$. Every unordered representation has at most $5!$ orderings, so nonnegative weighting preserves this lower bound. If $U_5(n)$ is the number of unordered representations of $n$ by exactly five positive $5$-powerful integers, the certificate and (11) imply

$$\liminf_{X\to\infty}\frac M X \sum_{\substack{n\le X\\n\equiv b\pmod M}}U_5(n)w(n) \ge\frac{429}{400}\bar w. \tag{12}$$

The normalized weighted target count tends to $\bar w$. Thus adding a fixed nonnegative continuous root-phase filter with positive mean cannot turn that first-moment method into a deficit below one. The same conclusion permits any fixed nonnegative periodic target weight, by summing residue classes. This is a limitation of those fixed filters, not a statement about distinct values, shrinking-target filters, or infinitude of the complement.