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Fix $0<\varepsilon<1$ and $\delta>0$. For every sufficiently large prime power $q$ and every

$$I\subseteq[q^\varepsilon,2q^\varepsilon]\cap\mathbb Z,\qquad |I|\ge\delta q^\varepsilon,\qquad (i,q)=1\ (i\in I),$$

each residue modulo $q$ is $s(B)$ for a subset $B\subseteq I$ with $|B|\le q^{\varepsilon/2}$. The threshold depends only on $\varepsilon,\delta$, so the statement is uniform in $I$ and the residue.

Source: published PDF, Theorem 2, pp. 7–8. The printed statement allows every $\varepsilon>0$. This page certifies the complete range $0<\varepsilon<1$ used by the paper's absorption theorem; it does not certify the unused larger range. The proof is complete relative to the exact external CFP and divisor inputs, with the restricted modular approximation and proper-dilate reduction explicitly repaired.

Bears on. Problem 297.

Proof

Use the progression, sets, and parameters from gap_symmetrization. If its dimension $k$ is at least 2, its volume estimate first gives $A<q$. Apply the corrected Lemma 4 to its integer steps; they need not individually be units. Then claim_1 gives

$$A\ge\left(\frac{\rho}{16k}\right)^k q^{1-k\kappa(q)}, \qquad \rho=\delta/4,\quad \kappa(q)\longrightarrow0.$$

On the other hand, $A\le C_k q s^{1-k}\le C_k' q^{1-(k-1)\varepsilon/2}$. For every fixed $2\le k\le d$ these inequalities contradict each other for large $q$, since $k\kappa(q)<(k-1)\varepsilon/4$ eventually. Thus the actual dimension is 1.

Now $P_*=[-a_1,a_1]d_1$ contains a member of $J$ coprime to $q$. It follows that $d_1$ is coprime to $q$. Choose $T$ to be its inverse modulo $q$ and take $d_1'=1$. The bound required by Claim 1 is valid because $2(q/a_1)(a_1/q)=2$. This proves the same volume lower bound for $k=1$ without any assumption $a_1\le q$.

The proper progression contained in $\Sigma(B_0)$ has at least $(\lambda/4)a_1$ points in an arithmetic progression with step $d_1$. Since $\lambda\asymp q^{\varepsilon/2}$ and $a_1\ge(\rho/16)q^{1-\kappa(q)}$, this number exceeds $q$ for large $q$. Any $q$ consecutive terms of a progression with step coprime to $q$ give every residue. Every such term is a subset sum of $B_0$, and $|B_0|\le s\le q^{\varepsilon/2}$. Each element of $B_0$ is the inverse of a distinct element of $I$. Lifting a subset of $B_0$ to those original denominators gives the required subset $B\subseteq I$.