Attack log — two smooth summands for every integer

2026-08-17 — literature (fetched before any search)

Fetched, in order:

• Green, 100 Open Problems (Dec 2025 update), Problem 59, p.29 of compute/refs/green-100-open-problems.pdf: “Is every $n\le N$ the sum of two integers, all of whose prime factors are at most $N^\varepsilon$?” Comments: asked by Erdős (1981 collection [110]): if $f(n)$ is the least integer not of the form $a+b$ with $P^+(ab)\le n$, then for every $k$ and $n>n_0(k)$ one should have $f(n)>n^k$. “This conjecture does not look hard but I could not get anywhere with it.” Record still Balog’s exponent $4/(9\sqrt{e})\approx 0.2695$ from 30 years ago [16]. Wooley: if a prime $p\equiv 3\pmod 4$ is a sum of two $p^\varepsilon$-smooth numbers then there is a quadratic non-residue $\le p^\varepsilon$. Solving the problem for every $\varepsilon>0$ is therefore at least as hard as Vinogradov’s least quadratic non-residue on that prime class. Best unconditional exponent there is $1/(4\sqrt{e})\approx 0.1516$. Known under GRH.
• Erdős Problem #334 (accessed 2026-08-17; last edited 03 April 2026). Status: OPEN, “cannot be resolved with a finite computation”. Zero claimed proofs. Two comments (my99n, 03 Mar 2026; Woett, 19 Nov 2025). Cites [Er76e, p.272], [ErGr80, p.70], [Er82d, p.55]. Related OEIS A062241, A045535.
• Balog, On additive representation of integers, Acta Math. Hungar. 54 (1989), 297–301, DOI 10.1007/BF01952060. Primary scan saved as compute/refs/balog-1989.pdf. Theorem: every $N>N_0(\varepsilon)$ is $a+b$ with $P^+(ab)\le N^{4/(9\sqrt{e})+\varepsilon}$. Opening paragraph: Erdős called $N^{1/3}$ “the first non-trivial question because the statement with $N^{1/2}$ in place of $N^{1/3}$ is almost trivially true.” The method “cannot be enough to prove even $N^\varepsilon$” and is not self-contained: it invokes Fouvry’s large-modulus distribution theorem. A weaker exponent is available from the large sieve alone.
• Hildebrand–Tenenbaum, Integers without large prime factors, J. Théor. Nombres Bordeaux 5 (1993), 411–484. Additive paragraph (p.469 of the scan): Balog–Sárközy 1984b give $P^+(n_i)\le N^{2/5}$; Balog 1989 improves to $0.2695\ldots$; it is not known whether $P^+(n_i)\le N^\varepsilon$ holds for any fixed $\varepsilon>0$. Ternary: Balog–Sárközy 1984a give $P^+(n_1n_2n_3)\le\exp(c\sqrt{\log N\log\log N})$.
• Sárközy, On sums with small prime factors, Acta Math. Hungar. 67 (1995), 333–345 (DOI 10.1007/BF01874496): ternary bound $\exp\bigl((\sqrt{3/2}+\varepsilon)\sqrt{\log N\log\log N}\bigr)$. Not a binary improvement.
• Balog–Sárközy, On sums of sequences of integers, I, Acta Arith. 44 (1984), 73–86 (compute/refs/balog-sarkozy-1984-acta-arith.pdf): this is the dense-sumset cousin (a sum $a+b$ from two dense sequences is smooth), not the two-smooth-summands theorem.
• OEIS A062241 (Schroeppel / Wilson / Johnson): smallest $n\ge 2$ that is not a sum of two $\mathrm{prime}(k)$-smooth positive integers. Terms through $a(29)=2570169839$, with $a(30)$ uncomputed as of Donovan Johnson, Aug 2010.
• OEIS A045535: least negative pseudosquare modulo the first $n$ odd primes (Lehmer–Lehmer–Shanks; b-file through $n=50$ by Dobbelaere, Feb 2021). A062241($k$) tracks A045535($k-1$) on the computed range; Touch Sungkawichai (Mar 2026) states $a(n)\le A045535(n-1)$.
• Dimitrov–Vigneri–Attias, arXiv:1912.11546: cryptographic “anti-Goldbach” conjecture $F(n)=O((\log n)^{2+\varepsilon})$. Heuristic / algorithmic; no theorem.
• Citation graph check (Balog 1989 DOI, Green #59, Bloom #334, Hildebrand–Tenenbaum, Baker 2009 survey, Győry–Hajdu–Sárközy arXiv:2006.15307, Blomer–Grimmelt–Li–Rydin Myerson arXiv:2111.01601): no later all-integers binary exponent. Scoped miss, not a proof that none exists.

Published record I must beat, after the fetch:

Here $P^+(1)=1$ and

Green’s “every $n\le N$ is a sum of two $N^\varepsilon$-smooth integers, for all large $N$” is the uniform statement $F(n)\le n^\varepsilon$ for large $n$. Isolated values of $G(y)$ are residue unless they force an infinite covering or a new exponent.

Do not claim a table of $G(y)$ as a dent. Do not claim Balog’s exponent with an implicit $N_0$.

2026-08-17 — tonight’s handle

Three live handles, in order of what would count as a dent:

1. An explicit $\varepsilon<1/2$ with a checkable covering template that writes every $n\ge N_0$ ( $N_0$ explicit) as a sum of two $n^\varepsilon$-smooth positive integers, with the range $n<N_0$ certified. The square-plus-remainder template is $\varepsilon=1/2$ and is not a dent.
2. A certified finite residue cover that implies an infinite statement: a modulus $M$ and, for every class $r\pmod M$, a uniform construction (not a lookup table of small $n$) producing two $n^\varepsilon$-smooth summands.
3. A documented obstruction to any such cover: a verifier that a searched family of templates fails, together with the quadratic / Fouvry reasons why the search cannot break Balog or the Vinogradov wall.

Balog’s own reduction (1989, p.298): it is enough to treat $N^\alpha$-rough $N$ (least prime factor $>N^\alpha$). If $N=N'N''$ with $P^+(N')\le N^\alpha<p(N'')$ and $N''=a+b$ is a good splitting, then $N=N'a+N'b$ is too. Hard inputs are primes and products of large primes.

Weight he actually uses, for the record (not a template I can certify without Fouvry):

The exponent $4/(9\sqrt{e})$ is the point where the $z$-sum $1-\log((4/9)/\alpha)$ stays positive after the Rankin / Mertens factor $e^{-1/2}$ hidden in Fouvry’s range.

2026-08-17 — trivial covering, recorded so I do not claim it

Let $m=\lfloor\sqrt n\rfloor$ and $r=n-m^2$, so $0\le r\le 2m$. If $r\ge 1$ then $n=m^2+r$ with $P^+(m^2)=P^+(m)\le m\le\sqrt n$ and $P^+(r)\le r\le 2m<2\sqrt n+1$. If $r=0$ then $n=(m-1)^2+(2m-1)$ and $2m-1<2\sqrt n+1$. Hence $F(n)<2\sqrt n+1$ for every integer $n\ge 2$. This is Balog’s “almost trivial” $N^{1/2}$ statement, fully explicit. Isolated improvements of the constant $2$ on a finite range are residue.

2026-08-17 — negative-pseudosquare obstruction (to be certified)

A number $m\equiv 7\pmod 8$ for which $(-m/q)=1$ for every odd prime $q\le y$ cannot be a sum of two positive $y$-smooth integers. Proof is elementary Jacobi / quadratic reciprocity and will be checked, not trusted from the July working report on #334. Consequence: $G(y)\le A045535(\pi(y)-1)$, and for primes $p\equiv 3\pmod 4$ one has $n(p)\le F(p)$. A uniform bound $F(n)\le n^{0.1}$ would beat Burgess on that prime class.

Next: write verifiers and a residue-cover search. The search product is a failure with a checker, unless a template actually lifts.

2026-08-17 — trivial covering certified, not claimed

compute/trivial_cover.py: $F(n)<2\sqrt n+1$ on $[2,200000]$ with zero failures; worst ratio $F_{\mathrm{split}}/\sqrt n=1.9977$ at $n=197136$. Large spot checks at $10^{12}$, $10^{18}+3$, $99991^2$, and $131486759$ all pass. This is the square-plus-remainder template. It is not a dent.

2026-08-17 — obstruction lemma, certified

compute/obstruction.py:

• A045535 prefix through $n=22$: every listed $m$ is $7\pmod 8$ and $(-m/q)=1$ for the first $n$ odd primes. Leastness checked through $m\le 10^7$.
• For $M=131486759$, explicit square roots of $-M$ modulo every odd prime $q\le 79$. Two-pointer: $M\notin S_{79}+S_{79}$. Displayed splitting $649+131486110$ gives $F(M)=83$.
• Smooth counts through $M$: $\lvert S_{73}\cap[1,M]\rvert=727473$, $\lvert S_{79}\cap[1,M]\rvert=808372$. Later matched the C bitset generator exactly.

The reciprocity argument (even summand $2^k A$, odd summand $B$, Jacobi symbols of $-M$ on $A$ and $B$, then quadratic reciprocity forces a sign $-1$ when $M\equiv 7\pmod 8$) was rechecked on every small A045535 value by a direct search for a forbidden splitting: none exists.

2026-08-17 — residue-cover search, failed

compute/covering_search.py on $[2,50000]$ and refined on $[2,80000]$:

• Short-interval template of length $n^\alpha$: $\alpha=1/2$ has only the tiny holes $\{2,3,7,8,23,24\}$ (of which $2,3$ are $n^\alpha<2$ artefacts). $\alpha=0.4$ still has 110 holes, last at $3496$. $\alpha=1/3$ has thousands.
• Window $2\sqrt n+1$: zero holes, as the trivial covering predicts.
• Finite shift lists on residue classes modulo $8,24,120$: at $\alpha=0.4$, modulo $8$, seven classes lift on $[2,20000]$ and class $5$ dies at $n=18181$. At $\alpha=1/3$ every class dies. A fixed list of offsets cannot track a growing smoothness bound.
• Square-adjustment $n=(m-k)^2+(n-(m-k)^2)$ with $k\le 256$ and $\alpha=0.4$: 309 holes remaining, last still at $79922$. The template does not close.
• Even reduction $F(n)\le\max(2,F(\mathrm{odd\ part}(n)))$ is an identity and does not remove the odd obstruction.

No finite residue cover lifted. Search residue, as required.

2026-08-17 — certified $G(y)$ through $79$

C word-bitset (g_of_y.c) independently reproduces A062241:

Every integer in $[2,G(y)-1]$ is a sum of two $y$-smooth positives; $G(y)$ itself is not. This is a finite table.

2026-08-17 — exact exception prefixes, still residue

Integer tests $F^q>n^p$, C sweep plus Python recomputation of every listed $F$:

• $F(n)\le n^{1/2}$ on $[2,10^6]$ except $\{3,7,23\}$.
• $F(n)\le n^{2/5}$ on $[2,10^6]$ except sixteen values, last $479=G(11)$.
• $F(n)\le n^{1/3}$ on $[2,3\cdot 10^5]$ except seventy-six values, last $18191=G(23)$.

The $n^{0.27}$ sweep (just above Balog) still has exceptions at the right endpoint of $[2,2\cdot 10^5]$. A last-exception on a prefix is not $N_0(\varepsilon)$ for Balog's theorem.

The published record is not beaten. Documented residue.

2026-08-27 — literature fetched again before q1

Opened, in order, before any q1 search:

• Green, 100 Open Problems, Problem 59, from https://people.maths.ox.ac.uk/greenbj/papers/open-problems.pdf (Dec 2025 text). Still the Balog exponent $4/(9\sqrt{e})\approx 0.2695$, still the Wooley / Vinogradov wall at $1/(4\sqrt{e})\approx 0.1516$, still a pointer to Erdős #334. No update since the 17 August fetch.
• Erdős Problem #334 live page is behind Cloudflare here. Wayback https://web.archive.org/web/20260824194602/https://www.erdosproblems.com/334 (captured 24 August 2026): OPEN, “cannot be resolved with a finite computation”, last edited 03 April 2026, zero claimed proofs, two comments. Statement and Balog exponent unchanged.
• Wayback forum thread, same date: my99n (03 Mar 2026) on A062241 versus A045535; Woett (19 Nov 2025) on the ternary Sárközy bound, the $n^{0.1}$ Vinogradov obstruction, and Erdős’s $\exp(c\sqrt{\log n\log\log n})$ binary conjecture.
• OEIS A062241 and A045535 via scripts/oeis_lookup.py. Terms match the certified $G(y)$ table. A062241 still ends at $a(29)=2570169839$ in the JSON prefix we received.
• Hildebrand–Tenenbaum, J. Théor. Nombres Bordeaux 5 (1993), PDF from https://jtnb.centre-mersenne.org/item/10.5802/jtnb.101.pdf, journal pp. 468–469: Balog–Sárközy 1984b give $N^{2/5}$; Balog 1989 improves to $0.2695$; $N^\varepsilon$ unknown for any fixed $\varepsilon>0$. Ternary: $\exp(c\sqrt{\log N\log\log N})$.
• arXiv:2006.15307 (Győry–Hajdu–Sárközy): decomposability of the set of smooth numbers, not an all-integers binary exponent.
• arXiv:2111.01601v5 (Blomer–Grimmelt–Li–Rydin Myerson): prime plus two almost-prime squares; not two smooth summands.
• arXiv:1912.11546v2 (Dimitrov–Vigneri–Attias): RSA / anti-Goldbach heuristic, not a theorem.
• Ki–Maier–Sankaranarayanan, Acta Arith. 175.4 (2016): cites Balog $0.2695$ as the all-integers binary record; their own theorems are square-plus-smooth and prime-plus-smooth almost-all results. They also record Balog–Sárközy’s exponential-sum remark that $2/5$ can be lowered to $0.392$, still above Balog 1989.
• Balog’s publication list https://www.renyi.hu/~balog/publist.pdf : item [24] is the 1989 paper; items [9]–[10] are the 1984 Studia I/II pair.

Published record unchanged. Still no later all-integers binary exponent. Scoped miss, not a proof that none exists.

2026-08-27 — parent replay

Started python3 compute/verify_all.py before any q1 search. Trivial covering and the obstruction lemma replayed immediately. The Python $G(y)$ bitset matched the published table through $y=53$. The $y=59$ and $y=61$ rows ($G=6077111$) are too large for a timely pure-Python bitset; the C program independently reproduced both:

cc -O3 -std=c11 -o g_of_y_c g_of_y.c
./g_of_y_c 61 6077111 200000
./g_of_y_c 59 6077111 200000

Both print first_uncovered=6077111, limit_is_sum=0, G_equals_limit=1, with smooth counts $114680$ and $102546$. Parent covering_search.py also replayed: no residue-class cover at any $\alpha<1/2$.

2026-08-27 — closed-form templates, all fail below $1/2$

New code in compute/q1/. Integer tests $F_{\mathrm{template}}^q>n^p$ on $[2,20000]$:

Holes persist to the right endpoint at every exponent below $1/2$. The square formula has zero failures of the true trivial bound $F_{\mathrm{split}}<2\sqrt n+1$ on this range, as it must.

Floor-divisor is the closed-form short-interval rewrite: $u=\lfloor n^{p/q}\rfloor$, $a=n-(n\bmod u)$. The remainder is automatically $n^{p/q}$-smooth by size. The first summand fails exactly when $\lfloor n/u\rfloor$ is $n^{p/q}$-rough. At $\varepsilon=1/2$ this is the trivial covering again. Below $1/2$ it does not lift.

2026-08-27 — infinite failure families

compute/q1/certs/infinite_family.json:

• Floor-divisor, 129 primes $11\le P\le 541$ and $10000\le P\le 10300$, at $2/5$, $1/3$, and $27/100$: a $u$ near $P^{\varepsilon/(1-\varepsilon)}$ makes $n=Pu+1$ fail the template, 129/129 hits, zero misses. The inequality $P>n^\varepsilon$ for this shape is exactly $\varepsilon<1/2$. Large sample at $2/5$: $P=10289$, $n=4455138$.
• Largest power of two, $k=4,\ldots,16$: $n=2^k+q$ with $q$ the next prime after $2^{k-1}$ fails even $n^{1/2}$ (so fails every smaller exponent). Example $k=16$: $n=98307=65536+32771$.

These kill those two formulas as infinite coverings. They are not a lower bound on $F$.

2026-08-27 — polynomial size obstruction

compute/q1/certs/poly_obstruction.json. For each of $x^2$, $x^3$, $x(x+1)$, $x(x+1)/2$, $x^2+x+1$, $x(x+1)(x+2)/6$, the gap $P(k+1)-P(k)$ at $k\in\{20,50,100,200,400\}$ is at least $k^{d-1}/2$. Remainder-by-size therefore has exponent $1-1/d\ge 1/2$. Cubes are worse than squares. Smoothness of a large remainder is the square-adjustment search already run on 17 August, which left holes.

2026-08-27 — two-factor search, same prefixes as $F$

Balog–Sárközy shape: $a=uv$ with $u\le n^{1/5}$ and both $v$ and $n-a$ equal to $n^{p/q}$-smooth. On $[2,20000]$ this matches $F$ exactly:

• $2/5$: sixteen holes, last $479$, equal to the known $F$ exceptions.
• $1/3$: seventy-six holes, last $18191$, equal to the known $F$ exceptions.

The $n^{1/5}$ factor restriction does not lose a splitting on this range. It also does not produce $N_0$. A last exception on a prefix is not Balog–Sárközy made effective.

No dent. Published record not beaten.