Research log — a smaller constant in the two-squares gap

Status (accessed 2026-08-17)

• Green, 100 Open Problems, Problem 66 (Dec 2025): “Is there always a sum of two squares between $X-\tfrac1{10}X^{1/4}$ and $X$?” Still open. Comments point to Littlewood (Problem 2), Montgomery, and Erdős 1957 Problem 15. Green–Lindqvist give two almost-equal squares in the same $O(X^{1/4})$ scale. No later coefficient is recorded.
• The proposed-50 pointer “Erdős #143” is a numbering error. Bloom #143 is primitive-set sparsity. The matching page is Erdős Problem #222 (accessed 2026-08-17; cites [Er57], [Er61]). OPEN, “cannot be resolved with a finite computation”. Zero comments, zero claimed proofs. Differences listed at OEIS A256435.
• Erdős, Some unsolved problems, Michigan Math. J. 4 (1957), Problem 15: Bambah–Chowla give $a_{k+1}-a_k=O(a_k^{1/4})$; “it has not yet been shown that the $O$ can be replaced by $o$”.

Do not claim the Landau–Ramanujan density.

Published explicit upper bounds on $G(n)$

Write $G(n)=\min\{s-n:s\ge n,\ s=a^2+b^2\}$ (non-negative squares, $0$ allowed). Consecutive differences of two-squares are the same sequence, so Green’s backward interval is the same constant.

Log-scale lower bounds (Richards $1/4$; Dietmann–Elsholtz–Kalmynin– Konyagin–Maynard $0.868\cdots$) do not move the $X^{1/4}$ coefficient.

What we certified

Theorem. $G(n)<2\sqrt{2}\,n^{1/4}-3$ for every integer $n$ with $2\le n\le 1.024\cdot 10^{15}$, except $n\in\{3,6,21,91\}$.

• $n=1$ makes the right-hand side negative ($2\sqrt{2}-3<0$); excluded.
• The four exceptions are exact, with $G(3)=1,\;G(6)=2,\;G(21)=4,\;G(91)=6$. Each satisfies Jameson’s $a=2$ and fails $a=3$ by the integer test $(G+3)^4\ge 64n>(G+2)^4$.
• Independently, an exhaustive two-square table shows that these (plus $n=1$) are the only $a=3$ failures on $[1,5\cdot 10^6]$, and that Jameson’s $a=2$ has no failure on that range.

This is a computer-assisted improvement of the published additive constant $-2$ (Jameson) to $-3$ (Shiu’s next open increment), on an explicit range. It is not a proof for all large $n$, and it is not Green’s $1/10$.

Independently verified stored witnesses: every danger-zone top with $m\le 250$ in compute/a3_cert_m250.json. The $m\le 2000$ dump (~1e6 witnesses) is regenerable with certify_a3.py --m-max 2000 and is not in the tree. The same searcher reports zero unsaved tops for $m\le 8000$, which is the $1.024\cdot 10^{15}$ range.

Classification (integer)

Let $u=\lfloor\sqrt n\rfloor$, $\mathrm{rem}=n-u^2$, write $\mathrm{rem}=m^2+\ell$ with $0\le\ell\le 2m$. The Bambah–Chowla point is $(u,v)$ with $v=\lceil\sqrt{\mathrm{rem}}\rceil$. Set $h=u^2+v^2-n$ and $h_2=(u+1)^2-n$. Then $G(n)\le\min(h,h_2)$, and two-point $a=3$ fails if and only if $(\min(h,h_2)+3)^4\ge 64n$ (equivalently $\min(h,h_2)\ge\Phi-3$ with $\Phi=2\sqrt{2}\,n^{1/4}$).

Shiu’s ladder lemma: after the two zeros at the start of $I_u$, the values of $h$ form descending unit steps of length $2m$. The top of the $m$-th ladder is $n_0=u^2+m^2+1$ ($\ell=1$), with $h=2m$ and $h_2=k:=2u-m^2$. Rung $t\ge 1$ has $n=n_0+t$, leftovers $2m-t$ and $k-t$.

Integer bounds, for $n\ge 2$:

1. If $\ell=0$, then $n$ itself is a two-square and $G(n)=0$. For $n\ge 2$ one has $\Phi-3>0$, so $a=3$ holds.
2. If $k\le 2m-3$, then $\min\le 2m-3$ and $(\min+3)^4\le(2m)^4=16m^4<64n$, so two-point $a=3$ holds.
3. If $k\ge 3m+3$, then $\min=2m$ and $64n\ge 16(m^2+3m+3)^2+64m^2+64>(2m+3)^4$, so two-point $a=3$ holds.
4. If $t\ge 1$ and $k\in[2m-2,3m+2]$, write $\mu=\min(2m,k)$. Direct expansion gives $(\mu+2)^4<64(n_0+1)$ in each of the three subcases $k=2m-2$, $k=2m-1$, $k\ge 2m$. So no rung $t\ge 1$ is a two-point $a=3$ failure.

Thus every two-point $a=3$ failure with $n\ge 2$ is a ladder top $n=u^2+m^2+1$ with $2u=m^2+k$ and $2m-2\le k\le 3m+2$. On that set, $a=3$ is exactly the existence of some lattice point with leftover in $\{0,1,\dots,W\}$, $W=\max\{w:(w+3)^4<64n\}$.

Also $m\le\lfloor\sqrt{2u}\rfloor\le\lfloor\sqrt{2}\,n^{1/4}\rfloor$. So $m\le 8000$ covers every integer $n\le(8000/\sqrt{2})^4=1.024\cdot 10^{15}$.

The four exceptions

Each row has $(G+3)^4\ge 64n>(G+2)^4$.

Green’s $1/10$

Not proved, and not disproved for all sufficiently large $X$.

A gap census through $N=2\cdot 10^7$ (generate every two-square, no density claim):

• running-max of $(s_{k+1}-s_k)/s_{k+1}^{1/4}$ is $15/1508^{1/4}\approx 2.407$, attained at the empty ladder $38^2+7^2=1493$ to $38^2+8^2=1508$;
• that is $0.851$ of the Bambah–Chowla coefficient $2\sqrt{2}$;
• $1.526\cdot 10^6$ consecutive pairs still exceed $1/10$, including at $X=2\cdot 10^7$ itself (gap $7>0.1\cdot(2\cdot 10^7)^{1/4}\approx 6.69$).

Richards-type gaps of size $\asymp\log X$ live at $X=\exp(\Theta(k))$ and are far below $X^{1/4}$ at that scale; they do not produce infinite-family counterexamples to $1/10$. Isolated gap tables are not a bound. We do not claim that the running-max ratio tends to $0$.

Replay

See compute/README.md. The integer test $(G+3)^4<64n$ never uses floating $\Phi$ in the certificate checker.

Sources fetched

• Green, 100 Open Problems, Problem 66, Dec 2025 PDF.
• Erdős, Michigan Math. J. 4 (1957), Problem 15 (scan).
• Bloom, Erdős Problem #222, 2026-08-17.
• Bambah–Chowla, Proc. Nat. Inst. Sci. India 13 (1947), via Shiu’s arithmetic transcription.
• Shiu, Math. Gazette 97 (2013); Integers 19 (2019) #A48.
• Jameson, Math. Gazette 103 (2019), as cited by Shiu.
• Uchiyama, J. Fac. Sci. Hokkaido Univ. 18 (1964/5).
• Richards, Adv. Math. 1982; Dietmann–Elsholtz arXiv:1810.03203v2 and the IMRN 2023 constant quoted on Bloom #222.
• Guo–Ilyin, J. Number Theory 2024; Iyer, arXiv:2503.15789 (2025).