Walkthrough — a smaller constant in the two-squares gap

Discovery notes, not a paper.

0. What was actually missing

The missing degree of freedom was not a new sieve and not the Landau–Ramanujan counting function. Bambah–Chowla already give $G(n)<2\sqrt{2}\,n^{1/4}+1$. Jameson already shaved the $+1$ to $-2$. Shiu isolated the next increment: $a=3$, i.e. $G(n)<2\sqrt{2}\,n^{1/4}-3$, and produced an infinite family of $n$ on which the two obvious lattice points — the Bambah–Chowla point and $(u+1)^2$ — both sit just on the wrong side of $\Phi-3$.

On that family the two obvious points give leftover exactly $2m$, while $2m+2<\Phi<2m+3$. So $a=3$ is the existence of some other lattice point in $[n,n+2m)$. That is a search, not a new exponent.

Green’s $1/10$ is a much smaller multiplicative target. It is not the published proven constant, and a gap table does not prove it.

1. Named false starts

Claiming Bloom #143. The proposed-50 list says Erdős #143. Fetched: #143 is primitive-set sparsity. Green’s comments say Erdős 1957 Problem 15. Bloom’s matching page is #222. Numbering error, recorded and moved on.

Landau–Ramanujan ⇒ $1/10$. Average gaps $\asymp\sqrt{\log X}$ would make $1/10\,X^{1/4}$ look generous for large $X$. House rule: do not claim the density. Average gaps are not a max-gap bound.

Richards / DEKKM as a new bound. $\limsup G/\log s_n\ge 0.868$ is real and later than Richards’s $1/4$. The gaps live at $X=\exp(\Theta(k))$ and are logarithmic. They do not constrain the $X^{1/4}$ coefficient and they do not kill $1/10$ at that $X$.

Hoping the Shiu family is empty. If $(u^2+m^2,u^2+(m+1)^2)$ were empty for infinitely many even $m$ with $2u+1=(m+1)^2$, then $\limsup G(n)/n^{1/4}=2\sqrt{2}$ and Bambah–Chowla would be sharp. Checked even $m\le 8000$: the only empty open interval is $m=2$ ($n=21$). For $m=4$ one already has $9^2+9^2=162=n+1$. Occupied.

A bounded $p=u-u'$ for every even $m$. Least-$p$ search through $m=8000$ has max $p=40$, attained at Shiu’s own example $m=2862$. Success measure for a fixed $p$ is $\sim 1/\sqrt{1+p}$. Independence of $\sqrt{1+p}$ would make the bad box nonempty for any fixed $P$, so bounded $p$ is not a proof for the tail.

Hypothesis H / any $\alpha>0$. Shiu’s fractional-part hypothesis would give every positive coefficient. It is exactly the hard equidistribution statement he left open. Not tonight.

Empty ladders as an infinite family. Empty interiors of $(u^2+m^2,u^2+(m+1)^2)$ exist (the ratio champion $1493\to 1508$, $m=7$, $u=38$). Through $u=1500$ they only occur for $m\le 14$. Small-$m$ leftover, not an infinite $X^{1/4}$ lower bound.

2. The useful failure

The Shiu family is occupied, not empty. That kills the dream of a sharp $2\sqrt{2}$ lower bound from this family, and it is also the opening: $a=3$ on the only infinite two-point obstruction is a search for one extra lattice point in an interval of length $2m$.

The other useful failure is two-point $a=3$ itself. Jameson’s $a=2$ already holds from $\min(h_{\mathrm{BC}},(u+1)^2-n)$ for every $n$ we checked (and Jameson proved it for all $n$). The only $n$ that need a third point are the ladder tops sitting in a short integer window $k\in[2m-2,3m+2]$. That set is enumerable in $m$.

3. The click

Write the two-point failure test as $(\min(h,h_2)+3)^4\ge 64n$. Expand. Then:

• rungs $t\ge 1$ never fail (the left side drops by about $4(2m)^3$ and the right side grows);
• $k\le 2m-3$ and $k\ge 3m+3$ never fail;
• so every $a=3$ obstruction with $n\ge 2$ is a single ladder top, indexed by $(m,k)$.

Also $m\le\sqrt{2}\,n^{1/4}$, so a bound $m\le M$ is a bound $n\le(M/\sqrt{2})^4$. The infinite-looking problem is a finite $(m,k)$-search. For each top, scan lattice points near the circle for leftover $< \Phi-3$.

The four exceptions $3,6,21,91$ are the only unsaved tops. They are small and exact.

4. The argument, in the order it was found

1. Fetch Green #66 and (after the #143 miss) Bloom #222. Record the published proven constant: $2\sqrt{2}$, additive $-2$ (Jameson), open increment $a=3$ (Shiu).
2. Replay the elementary Bambah–Chowla leftover $<2\sqrt{2}\,n^{1/4}+1$ on $n\le 10^5$. Zero failures. Two-point $a=2$ zero failures. Two-point $a=3$ fails on a thin set.
3. Hunt a third lattice point on that thin set, first by scanning $n\le 2\cdot 10^5$. Unsaved: $3,6,21,91$.
4. Switch from scanning $n$ to enumerating Shiu ladders $(u,m,t)$. Through $u=20000$ ($n\le 4\cdot 10^8$) the unsaved set is still those four, and every saved failure has $t=0$.
5. Prove the integer classification: only tops, only $k\in[2m-2,3m+2]$. Check $t>0$ count is zero through $u=5000$, then do the expansions.
6. Enumerate danger-zone tops by $(m,k)$. Save each by a near-circle lattice search. $m\le 2000$: $1\,002\,993$ tops, zero unsaved, stored witnesses independently checked. $m\le 8000$: $16\,011\,993$ tops, zero unsaved.
7. Exhaustive $G(n)$ from a two-square table on $[1,5\cdot 10^6]$: $a=3$ fails exactly at $\{1,3,6,21,91\}$; $a=2$ fails nowhere.

The range $m\le 8000$ is $n\le 1.024\cdot 10^{15}$.

5. Computer search

• compute/exhaustive_a3_5e6.json — exact exception set on $[1,5\cdot 10^6]$.
• compute/a3_cert_m250.json — stored $(n,a,b,s)$ witnesses, checker compute/verify_a3_cert.py. Rebuild $m\le 2000$ with certify_a3.py --m-max 2000 (not in the tree).
• compute/a3_summary_m8000.json — $16\,011\,993$ tops, $0$ unsaved.
• compute/min_p_m8000.json — least $p=u-u'$ on the even Shiu curve; max $p=40$ at $m=2862$.
• compute/gap_census_2e7.json — running-max ratio $2.407$ at $1493\to 1508$; $1/10$ still crossed at $X=2\cdot 10^7$.
• figures/gap_ratios.png — consecutive gaps over $X^{1/4}$.

Replay commands: compute/README.md.

6. What is proved vs still open

Proved (computer-assisted, this folder). $G(n)<2\sqrt{2}\,n^{1/4}-3$ for $2\le n\le 1.024\cdot 10^{15}$ except $\{3,6,21,91\}$. Classification is exact. Witnesses for $m\le 2000$ are independently checked. The $m\le 8000$ search uses the same integer lattice scan.

Still open.

• Green / Littlewood: $C=1/10$ for all sufficiently large $X$.
• Any multiplicative $\alpha<2\sqrt{2}$ for all large $n$.
• $a=3$ for all $n$, not just $n\le 10^{15}$. The tail $m>8000$ is the same search; we did not run it, and we did not find a uniform closed-form replacement.
• Whether $\limsup G(n)/n^{1/4}$ is positive. The largest observed ratio through $N=2\cdot 10^7$ is $2.407$, attained at $n=1508$. That is not a bound.

We did not beat $1/10$. We did beat Jameson’s published additive $-2$, on an explicit range, with four exact exceptions and a replayable checker.