Real coefficient averaging and the rational-coefficient obstruction

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Status

The metric theorem below is proved and author-recorded. It concerns rounded real values, not sums of powerful integers. In particular, it supplies no rational or integral coefficient tuple and does not resolve Erdős 940. The finite-core reduction still requires a fixed integral diagonal form.

A log-free averaged estimate

Theorem. Let $r\ge3$ be an integer. For Lebesgue-almost every $\mathbf a\in[1,2]^r$, the set

$$\left\{\left\lfloor\sum_{i=1}^r a_i x_i^r\right\rfloor: x_1,\ldots,x_r\in\mathbb N\right\}$$

has positive upper natural density.

For $N\ge1$, put

$$Q_{\mathbf a}(N)=\#\left\{\mathbf x,\mathbf y\in[1,N]^r\cap\mathbb Z^r: \left|\sum_i a_i(x_i^r-y_i^r)\right|\le1\right\}.$$

We prove the uniform estimate

$$\int_{[1,2]^r}Q_{\mathbf a}(N)\,d\mathbf a\ll_r N^r. \tag{1}$$

First, for $T\ge1$,

$$B(N,T):=\#\{1\le u,v\le N:0<|u^r-v^r|\le T\} \ll_r T^{2/r}. \tag{2}$$

By symmetry suppose $u>v$. If $v\le T^{1/r}$, then $u^r\le v^r+T\le2T$, giving $O_r(T^{2/r})$ pairs. Otherwise

$$u^r-v^r\ge r v^{r-1}(u-v),$$

so summing the upper bound $T/(rv^{r-1})$ over $v>T^{1/r}$ proves (2). Here $r>2$ is essential.

For distinct vectors define $D=\max_i|x_i^r-y_i^r|$. Equation (2) implies

$$\#\{\mathbf x\ne\mathbf y:D\le T\} \le (N+C_rT^{2/r})^r-N^r. \tag{3}$$

For any fixed distinct pair, hold all but a coefficient corresponding to a maximal difference fixed. The allowed interval in that coefficient has length at most $2/D$. Thus its slab in $[1,2]^r$ has measure at most $2/D$.

Divide $1\le D\le N^r$ into dyadic intervals and expand (3). The off-diagonal contribution to (1) is at most a constant depending on $r$ times

$$\sum_{j=1}^r N^{r-j} \sum_{k=0}^{\lceil\log_2 N^r\rceil}2^{k(2j/r-1)}. \tag{4}$$

The $j$th term is $O_r(N^{r-j})$ when $2j<r$, $O_r(N^{r/2}\log(2N))$ when $2j=r$, and $O_r(N^j)$ when $2j>r$. All are $O_r(N^r)$. The diagonal contributes exactly $N^r$, proving (1).

Fatou's lemma gives

$$\int_{[1,2]^r}\liminf_{N\to\infty}\frac{Q_{\mathbf a}(N)}{N^r}\,d\mathbf a<\infty.$$

Consequently, for almost every $\mathbf a$, there are a finite constant $C_{\mathbf a}$ and arbitrarily large $N$ with $Q_{\mathbf a}(N)\le C_{\mathbf a}N^r$. Equal rounded values have unrounded difference less than one. Cauchy--Schwarz therefore gives at least $N^r/C_{\mathbf a}$ distinct rounded values from $[1,N]^r$. They are at most $2rN^r$, proving positive upper density.

Why this does not produce integral coefficients

The exceptional set in the theorem may contain every rational vector. Approximating coefficients separately at finite scales gives no fixed rational vector that works along an unbounded sequence.

There is also a precise obstruction to replacing the continuous measure in the proof by a discrete distribution. Suppose a measure has an atom of mass $p>0$ at a rational vector $\mathbf a$. Choose a nonzero integer vector $\mathbf v$ with $\mathbf a\cdot\mathbf v=0$. Then for every positive integer $k$,

$$\mu\{\mathbf b:|\mathbf b\cdot(k\mathbf v)|\le1\}\ge p.$$

Thus no uniform slab bound $C/\|k\mathbf v\|_\infty$ is available. This excludes that termwise replacement, not every arithmetic averaging argument: requiring $k\mathbf v$ to be an actual vector of power differences restores a difficult Diophantine condition.

More directly, if a countable weighted average over rational coefficient vectors satisfied

$$\sum_j p_j E_{\mathbf a_j}(N)\le C N^r, \qquad p_j>0,$$

along a common unbounded sequence, then positivity would already imply $E_{\mathbf a_j}(N)\le(C/p_j)N^r$ for every supported vector. Such an estimate contains the individual arithmetic energy theorem; the probabilistic extraction does not bypass it.

What an adaptive arithmetic average would need

There is a precise conditional extraction criterion for the weaker logarithmic moment. For a positive rational coefficient tuple $\mathbf a$, let

$$L_{\mathbf a}(N)=N^{-r}\sum_v R_{\mathbf a,N}(v) \log R_{\mathbf a,N}(v),$$

where the sum is over its exact rational output values, not rounded ones. Suppose probability measures $\mu_N$ on the countable set of rational tuples satisfy $\sum_{\mathbf a}\mu_N(\mathbf a)L_{\mathbf a}(N)\le C$. If one fixed finite set $K$ has $\mu_N(K)\ge\eta>0$ at arbitrarily large $N$, then some fixed $\mathbf a\in K$ has $L_{\mathbf a}(N)\le C/\eta$ along an unbounded subsequence. Select such a tuple in $K$ at each scale and use finite pigeonhole. The logarithmic multiplicity criterion gives positive upper density after clearing its fixed denominators.

A uniformly bounded moment of a proper arithmetic height, whose bounded sublevel sets are finite, would guarantee this discrete tightness by Markov's inequality. Ordinary weak compactness in a real or $p$-adic coefficient space does not guarantee it: measures can escape every finite arithmetic set. No adaptive average satisfying both the moment estimate and this tightness condition was constructed. This is a conditional criterion, not a new positive-density theorem.

A fixed sequence for which every rational tuple is exceptional

The following countermodel has been checked. It concerns a perturbation of powers, not actual powerful numbers. It shows that the difference estimate used in the metric proof, very accurate approximation to powers, and all eventual fixed congruences do not suffice for rational extraction.

Proposition. For every integer $r\ge3$ there is a strictly increasing sequence of positive integers $f(n)$ with the following properties:

1. $f(n)=n^r+O((\log n)^{1+o(1)})$.
2. For each fixed $m$, $f(n)\equiv n^r\pmod m$ for all sufficiently large $n$.
3. Uniformly for $N,T\ge1$,

$$\#\{u,v\le N:0<|f(u)-f(v)|\le T\}\ll_r T^{2/r}. \tag{5}$$

4. For every fixed positive integral coefficient tuple $\mathbf a$,

$$\#\{v\le X:v=\textstyle\sum_{i=1}^r a_i f(x_i) \text{ for some }x_1,\ldots,x_r\in\mathbb N\}=o(X), \tag{6}$$

where the left side counts distinct values.

Nevertheless, the rounded values $\lfloor\sum_i a_i f(x_i)\rfloor$ have positive upper density for almost every real $\mathbf a\in[1,2]^r$.

Construction. For integers $k\ge0$, put

$$m_k=\max\{1,\lfloor\log\log(k+e^e)\rfloor\},\quad M_k=\operatorname{lcm}(1,\ldots,m_k),\quad q_k=2^{\lfloor\log_2(k+1)\rfloor}M_k.$$

Both $M_k$ and $q_k$ divide their successors. Also $M_k\le m_k!$ implies $\log M_k=o(\log k)$, whence $q_k=k^{1+o(1)}$. On $2^k\le n<2^{k+1}$, define, apart from finitely many initial inputs,

$$f(n)=q_k\left\lfloor\frac{n^r}{q_k}\right\rfloor +(n^r\bmod M_k), \tag{7}$$

using the least nonnegative residue. Since $M_k\mid q_k$,

$$0\le n^r-f(n)<q_k. \tag{8}$$

This proves property 1 and the required congruence modulo $M_k$. Since $m_k\to\infty$, property 2 follows.

For sufficiently large $n$, even across block boundaries, the total possible change of the error in (8) is smaller than half of $(n+1)^r-n^r$. Start (7) beyond such a point and set $f(n)=n^r$ at the finitely many earlier inputs. The transition also obeys this inequality after choosing the starting point sufficiently large. We obtain positive integers and

$$f(n+1)-f(n)\ge\tfrac12((n+1)^r-n^r)$$

for every $n$. Hence for $u>v$, $f(u)-f(v)\ge(u^r-v^r)/2$; the already proved power-difference estimate (2), with $2T$ in place of $T$, proves (5).

Rational-coefficient value count. Fix $\mathbf a$ and an integer $J\ge1$, and write $2^K\le N<2^{K+1}$. If all $x_i>2^{K-J}$, their block indices lie between $K-J$ and $K$. By nested divisibility, each $f(x_i)$ is a multiple of $q_{K-J}$ plus an integer in $[0,M_K)$. Consequently $\sum_i a_i f(x_i)$ occupies at most $O_{\mathbf a}(M_K)$ residue classes modulo $q_{K-J}$. All these sums are $O_{\mathbf a}(N^r)$, so they contribute at most

$$O_{\mathbf a}\left(N^r\frac{M_K}{q_{K-J}}+M_K\right) \tag{9}$$

distinct values.

For fixed $J$, eventually $m_K-m_{K-J}\le1$. Therefore $M_K/M_{K-J}\le m_K$, while $2^{\lfloor\log_2(K-J+1)\rfloor}\gg K$. Thus $M_K/q_{K-J}\ll m_K/K\to0$, and (9) is $o_{\mathbf a,J}(N^r)$. The tuples with some $x_i\le2^{K-J}$ number at most $r2^{-J}N^r$. Let $N\to\infty$ and then $J\to\infty$ to obtain

$$\#\{\textstyle\sum_i a_i f(x_i):1\le x_i\le N\}=o(N^r).$$

Since $f(n)\asymp n^r$ and the coefficients are positive, values up to $X$ have all $x_i=O_{\mathbf a}(X^{1/r})$. This proves (6). Clearing fixed denominators gives the analogous assertion for rational coefficients. Rounding rational outputs cannot increase their number, so every rational tuple is exceptional for the metric conclusion.

Finally (5) and $f(n)\ll n^r$ are exactly the two sequence estimates used in the slab argument (3)--(4). Repeating that argument gives the same log-free averaged near-collision estimate, and Fatou and Cauchy--Schwarz give the last assertion. $\square$

This countermodel does not preserve the exact polynomial identity $f(n)=n^r$, nor does it claim all growing-modulus or major-arc estimates of the actual powers. It excludes extraction arguments based only on the properties explicitly listed above. No fixed rational coefficient tuple for the original problem was obtained.

Coordinate power rescaling does not remove height escape

For actual powers, positive upper density is invariant under replacing $a_i$ by $a_i b_i^r$ with fixed $b_i\in\mathbb Q_{>0}$, interpreting rational value sets after clearing a fixed common denominator. For integral $b_i$, writing $V_{\mathbf a}$ for the positive value set and $M=\operatorname{lcm}(b_1,\ldots,b_r)$ gives

$$V_{(a_i b_i^r)}\subseteq V_{\mathbf a},\qquad M^r V_{\mathbf a}\subseteq V_{(a_i b_i^r)}.$$

The second inclusion uses the roots $(M/b_i)x_i$. General rational $b_i$ follow by a common denominator and fixed output dilation. This is a qualitative invariance for a fixed rescaling. If the rescaling grows with the counting parameter, the resulting density constants can tend to zero. No tight arithmetic coefficient average was constructed.