Walkthrough — a long gap in a dilate modulo a prime

Discovery notes, not a paper. Empty sections would mean not done.

• Problem: problems/long-gap-dilate
• Quest: Green 100 #32, improve Shakan’s universal 2
• Model: SuperGrok CLI grok-4.6
• Date: 2026-08-17
• Argument status: no certified $C>2$. Leftover is a verified SAT table through $p=71$, a degree-wall (homogeneous and rising-factorial), a exact $G(73,9)=24$, SAT upper bounds at $p=79,83,89$, and failed lifts
• Problem status: open

0. What was actually missing

The missing degree of freedom was not a SAT solver and not a larger small-prime table. Shakan’s published argument already gives

for every $A\subset\mathbb F_p$ with $|A|>1$. The 2 is the sum of two degree bounds on a Rédei slice $\chi^m=t^pg+h$. At the scale $m\sim Cp/|A|$ with $C>2$ that slice has no middle coefficients forced to vanish. The interval that defines a gap is discarded when the product $\prod_{j=1}^m(t+j+da)$ is replaced by its top homogeneous part $(t+da)^m$. Whatever beats 2 has to keep that interval, or leave Rédei entirely.

Isolated $G(p,\sqrt p)$ for $p\le 71$ cannot be that argument.

1. False starts (named obstacles)

The list citation. The proposed-50 source arXiv:2205.14038, “On higher moments of Fourier coefficients”, is a quant-ph paper. The result Green cites is Shakan, arXiv:2004.14828, SIDMA 34 (2020). Fetched first.

Tweaking the Wronskian. Cancelled leading terms give $\deg W\le 2k-2$ instead of $2k-1$. Discrete derivatives of rising factorials do the same comparison. Both rearrange $O(1)$. Not a leading constant.

Assuming $0\in A$. Legal by translation. Then $\deg_d w=1+(n-1)m$. At $C=2$ this is still larger than $p$, so the $d^p-d$ half of Alon does not drop.

Dirichlet clustering. Any fixed-size $n$-set has a dilate of diameter $O(p^{1-1/(n-1)})$, so $G(p,n)/(p/n)\to n$. This is Green’s Bohr-set remark. At $n\sim\sqrt p$ the box principle needs more than $p$ boxes. No improvement.

Energy / Freiman. $|A-A|\le 3n$ makes some dilate an interval and the gap is $p-O(n)$, enormous. Random / Sidon / small multiplicative subgroups have $\max g\sim(p/n)\log n$, also enormous. The sets that could keep the leading constant near 2 are exactly the ones the polynomial method already treats. Rank-2 Bohr sets collapse to diameter $p^{3/4}$.

Cauchy–Schwarz on occupancies. $\sum N$ and $\sum N^2$ over pairs $(d,t)$ recover only the pigeonhole $C\ge 1$. Pairwise inclusion-exclusion with a matching of $A$ is the same.

Singer as a $C\to 2$ family. For $q=3,5$ the Singer set in $\mathbb Z/(q^2+q+1)$ matches the SAT value of $G$. At $q=17$, $p=307$, $n=18$ it has gap 90 against Shakan 32 (ratio $5.28$). Not a near-extremal family.

Calling the SAT table a new bound. For $17\le p\le 71$ one has $G(p,\mathrm{round}\sqrt p)/\sqrt p\in[2.18,3.09]$. House rule: isolated small-$p$ tables are not a universal $C>2$. Extra above Shakan is order $n$ on this range; that could be a hidden $+n$ (which would be leading 3 at $n\sim\sqrt p$) or slack that dies.

2. The useful failure

The homogeneous Rédei slice is a wall, not a knob. Once $m\ge 2p/n$, $\chi^m=t^pg+h$ is an identity that a general degree- $n$ polynomial can satisfy for dimension reasons. There is nothing left to contradict. The calculation that looks like it might give $2+\varepsilon$ is the same calculation with no hypotheses left.

That failure is useful because it names the discarded information: $B$ is an interval, $w(d,t)=d\prod_a(t+da+1)_m$ is a product of rising factorials, and the lower Stirling terms are exactly what the top-degree slice throws away. A Python expansion of the full $w$ vanishes on $\mathbb F_p^2$ at the same $m$ the gap definition requires. The extra terms do not automatically vanish earlier.

3. The click

Two translations, then a stop.

First: $\max_d g(dA)$ is the length of the longest arithmetic progression in the complement. Green #32 is the hitting-set statement “every $\sqrt p$-set misses some $100\sqrt p$-AP”. Shakan is $H(p,T)>2p/(T+2)$. A new bound is $H(p,C\sqrt p)>\sqrt p$ for some fixed $C>2$.

Second: the only published proof of the 2 uses a homogeneous slice that is information-theoretically unable to see $C>2$. Beating 2 is a different argument, or a use of the rising factorials that this session did not find.

There is no third click that produces a $C>2$ proof.

4. The argument, in the order it was found

1. Fetch Green #32 and Shakan (the real arXiv id). Record the published constant 2.
2. Re-derive the Wronskian. Confirm that $\alpha=2$ in $\deg g+\deg h\le\alpha k$ is forced by the slice, and that $C>2$ makes the middle-coefficient list empty.
3. Write the Dirichlet bound and check it against exact $G(p,3)$ and $G(p,4)$: the ratios climb toward 3 and 4, as they should, and this does not speak to $n\sim\sqrt p$.
4. Encode “exists an $n$-set hitting every $T$-AP” in SAT, binary- search $T$, get exact $G(p,\mathrm{round}\sqrt p)$ through $p=71$. Independently recompute every witness. Enumerate all affine-normalised $n$-sets through $p=41$; they match SAT.
5. Hunt a $C=2+o(1)$ family (grids, squares, subgroups, Singer, jitter, greedy hitting sets, local search). None approaches 2 at $n\sim\sqrt p$. Local-search upper bounds live at ratio $3$–$3.5$.
6. Stop before calling the finite table a universal constant.

5. Computer search

• compute/certs/sat_G.jsonl — exact $G$, witnesses, Glucose logs of the binary search.
• compute/certs/sat_G_verified.json — independent recomputation of every witness gap.
• compute/certs/enum_diagonal.json — brute-force $G$ through $p=41$.
• compute/certs/G_n3.json, G_n4.jsonl, G_n5.jsonl — fixed-$n$ exact values.
• compute/certs/local_upper.jsonl, constructions.jsonl, singer.json, greedy_hit.json — upper bounds / families.
• figures/sat_ratios.png — $G/(p/n)$ and $G/\sqrt p$ against Shakan’s 2.
• compute/verify.py — Shakan on every proper nonempty subset for $p\le 13$, and on the listed constructions.

Replay:

cd problems/long-gap-dilate
sh compute/run_all.sh

6. What is proved vs still open

Proved, published, not improved. Shakan: every $A$ of size $n>1$ has a dilate missing $2p/n-2$ consecutive residues. For $n=\lfloor\sqrt p\rfloor$ this is a gap of $2\sqrt p-2$. The leading constant 2 is the record.

Proved, elementary, already in Green’s comments. For each fixed $n$, Dirichlet gives $G(p,n)=p-o(p)$, so the Shakan constant $2$ is not sharp at bounded size. This is the regime $|A|\le c\log p$.

Certified computation, not a bound. For every prime $17\le p\le 73$ and $n=\mathrm{round}\sqrt p$, $G(p,n)\ge 2.1\sqrt p$, with an independently checked witness of size $n$ and (for $p\le 41$) an exhaustive proof that nothing smaller-gap exists. $G(73,9)=24$ is SAT plus Cadical unsat, not enum. This is a finite list.

Still open. Green #32: is there a universal $C=100$, or even a universal $C=2.01$, such that every $A$ of size $\sim\sqrt p$ has a dilate missing $C\sqrt p$? No such $C>2$ is proved here. No construction shows that $2$ is sharp either.

7. 2026-08-27 continuation

The unused handle was named: keep the interval in $w(d,t)=d\prod_a(t+da+1)_m$. Expanding it does not help. On the stored witnesses the Alon degrees of the rising $w$ equal the worst-case $k=nm-p+1$, and at the actual $G$ one already has $k\ge p$. The extra Stirling slice is the derivative of the top slice. There is nothing left in that polynomial to contradict at $C>2$.

SAT past $p=71$ is cheap on the SAT side. Cadical finished the UNSAT at $T=24$ for $p=73$ in 165s, so $G(73,9)=24$ is exact. $G(79,9)\le 26$, $G(83,9)\le 27$, $G(89,9)\le 30$ are still only witnesses. One extra exact row does not lift to a universal $C>2$. Constructions through $p=199$ drift up to ratio $4$–$5$, not down to $2$. The new ratios stay near $3$, same as $p=71$.

Replay: sh problems/long-gap-dilate/compute/run_all.sh and cd problems/long-gap-dilate/compute/q1 && ./run_all.sh.