How the two-smooth-summands search came together

0. What was actually missing

The missing object was not another table of the first $n$ that fails to be a sum of two $y$-smooth numbers. That table is OEIS A062241, it is finite for each fixed $y$, and Green and Bloom both warn that the uniform statement cannot be settled by a finite computation.

What would have counted is a template that lifts: a finite list of residue classes, or an explicit arithmetic construction depending on $n$ in a checkable way, that writes every large $n$ as $a+b$ with $P^+(ab)\le n^\varepsilon$ for some $\varepsilon<1/2$. Isolated prefixes — even a prefix on which $F(n)\le n^{2/5}$ has only sixteen exceptions — are an incomplete search unless the template continues.

The analytic gap is the same object Balog named in 1989: positivity of the binary smooth convolution

at $y=N^\varepsilon$, uniformly in $N$. Fouvry's large-modulus theorem gets him down to $4/(9\sqrt e)$. A covering search cannot replace that input.

1. Named false starts, with the specific obstruction

Square-plus-remainder as a new bound. Write $n=m^2+r$ with $0\le r\le 2m$, or $(m-1)^2+(2m-1)$ when $n$ is square. Then $F(n)<2\sqrt n+1$ for every $n\ge 2$. The verifier in trivial_cover.py confirms this on $[2,2\cdot 10^5]$ and on several $10^{12}$–$10^{18}$ spots. Balog's first page already calls the $N^{1/2}$ statement “almost trivially true.” The obstruction is not mathematical; it is bibliographic. This is the covering we were told not to claim.

A fixed list of shifts, one per residue class. For each $r\pmod M$ try the first forty $M$-smooth offsets $a$ and ask whether $n-a$ is $n^\alpha$-smooth. Modulo $8$ at $\alpha=0.4$ this almost works on $[2,20000]$: seven classes lift and class $5$ dies at $n=18181$. At $\alpha=1/3$ every class dies, the first deaths sitting between $937$ and $4282$. A fixed offset cannot follow a smoothness bound that grows with $n$. That is the obstruction.

Square-adjustment. Replace $m=\lfloor\sqrt n\rfloor$ by $m-k$ for $k\le 256$ and demand both $m-k$ and $n-(m-k)^2$ be $n^{0.4}$-smooth. Three hundred and nine holes remain on $[2,80000]$, the last at $79922$. Making the first summand a nearby square of a smooth integer does not buy an exponent below $1/2$.

Even reduction. If $m=a+b$ then $2m=(2a)+(2b)$, so $F(n)\le\max(2,F(\mathrm{odd\ part}(n)))$. The identity is correct and useless: the hard $n$ are odd, and the negative-pseudosquare construction lives in the class $7\pmod 8$.

Short intervals of length $n^\alpha$ as a proof of $F(n)\le n^\alpha$. This would lift if every interval $(n-n^\alpha,n)$ contained an $n^\alpha$-smooth integer. For $\alpha=1/2$ the only genuine holes on $[2,10^6]$ are $n=7,8,23,24$, and $F$ itself exceeds $\sqrt n$ only at $3,7,23$. For $\alpha=0.4$ the short-interval holes run up to $3496$; two large summands still save those $n$ as soon as one computes $F$, but the short-interval template does not cover them. Unconditionally, the literature (Balog 1987, Harman, Matomäki) gives $x^\varepsilon$-smooth numbers in intervals of length $\sqrt x$, which again returns $F(n)\le\sqrt n$, not a new exponent.

2. The useful failure

The residue-class search dying at a single class modulo $8$ looked, for a moment, like a near-cover. It taught the opposite lesson. The seven live classes were live because a short list of small smooth offsets happened to hit an $n^{0.4}$-smooth complement on a short prefix. Class $5$ is not arithmetically special; it is the first place the same finite list ran out of luck. Lengthening the list, or the modulus, only postpones the first death. Smoothness is a multiplicative monoid. A finite set of additive residue templates cannot manufacture two large smooth summands for every $n$.

The same failure, read against Balog's reduction, is sharper. Balog splits off the $N^\alpha$-smooth kernel of $N$ and reduces to $N^\alpha$-rough $N$ — primes and products of large primes. Those are exactly the inputs a small-shift cover never sees: the complement $n-a$ is then comparable to $n$, and asking it to be $n^\alpha$-smooth is the original problem.

3. The click

The click was not a cover. It was the realisation that the negative-pseudosquare obstruction, which is the only infinite local obstruction we can write down, is too weak to produce $F(n)>n^{2/5}$ past a few hundred, and too strong to let any cover we can certify go below the Burgess exponent.

If $m\equiv 7\pmod 8$ and $(-m/q)=1$ for every odd prime $q\le y$, then $m$ is not a sum of two positive $y$-smooth integers. The proof is Jacobi symbols plus quadratic reciprocity on the even and odd summands; it does not use Dirichlet or Linnik. Consequently $G(y)\le A045535(\pi(y)-1)$, and for primes $p\equiv 3\pmod 4$ the least quadratic non-residue satisfies $n(p)\le F(p)$. A uniform bound $F(n)\le n^{0.1}$ would beat Burgess on that prime class. That is Green's / Wooley's wall, and it is real.

But the same construction, evaluated at the certified points, gives $F(131486759)=83$ while $131486759^{2/5}\approx 1585$. The obstruction forbids $y=79$, not $y=n^{2/5}$. Along the A062241 sequence the inequality $F(G(y))>G(y)^{2/5}$ holds for the first few terms and then flips: $F(1559)=17<1559^{2/5}\approx 18.93$. After that the obstruction no longer produces counterexamples to $F(n)\le n^{2/5}$. That is why a C sweep of $[2,10^6]$ finds exactly sixteen exceptions to $F(n)\le n^{2/5}$, the last at $n=479=G(11)$, and why the same sweep at exponent $0.27$ still has exceptions at the right endpoint of $[2,2\cdot 10^5]$.

So the cover we wanted — a finite residue recipe for some $\varepsilon<1/2$ — is blocked from below by the trivial square template and from above by Fouvry. The obstruction explains why $\varepsilon=0.1$ is hopeless and why $\varepsilon=2/5$ can look finite-exception on a long prefix without being a theorem we proved.

4. The argument, in the order it was found

Green #59 and Bloom #334 were fetched first. The live Bloom page (edited 03 April 2026) still states Balog's exponent and the $n^{o(1)}$ conjecture; two comments add the A045535 comparison and the Vinogradov wall. Balog's five-page paper was fetched next. The first paragraph contains the two sentences that organised the night: $N^{1/2}$ is almost trivial, and the method cannot reach $N^\varepsilon$.

The trivial covering was written down and certified so it could not be “discovered” later. Then the Jacobi obstruction was proved from scratch and run against every small negative pseudosquare, and against $M=131486759$ with displayed roots and a two-pointer non-representation test. That gave the exact value $F(M)=83$.

Only then did the cover search run. Fixed shifts, square adjustments, short intervals, and the even reduction were each given a verifier and each produced holes. The useful picture was the exception sweep with integer tests $F^2>n$, $F^5>n^2$, $F^3>n$: floating-point $\mathtt{pow}$ had falsely flagged the cubes $8,27,125$ as $n^{1/3}$ exceptions. The exact lists are finite on the computed prefixes and stop at values of $G(y)$.

In parallel the C bitset independently reproduced A062241 through $y=79$. The 73-smooth and 79-smooth counts through $M$ agreed with the Python generator to the integer. That is a finite certificate, not a new bound.

5. Computer search

Stored under compute/certs/.

• trivial_cover.json — square template, zero failures on the stated range.
• obstruction.json — A045535 prefix, roots of $-M$, $F(M)=83$, smooth counts.
• g_certified.json — $G(y)$ through $y=79$, C bitset plus two-pointer.
• covering_search.json and covering_refined.json — holes for every finite template that was tried.
• f_exceptions_exact.json — the sixteen $n^{2/5}$ exceptions and the seventy-six $n^{1/3}$ exceptions, each with an explicit splitting that realises $F(n)$.

Figures: figures/g_of_y.png, figures/exceptions_two_fifths.png.

The bitset that certifies $G(73)=131486759$ ORs the 73-smooth characteristic vector, shifted by each of the $22176$ 73-smooth integers up to $200000$, and finds no hole in $[2,M-1]$. The same $M$ has no 79-smooth splitting.

6. What is proved vs still open

Proved tonight, all checkable:

• $F(n)<2\sqrt n+1$ for every $n\ge 2$ (elementary; the computer only audits).
• The negative-pseudosquare lemma, and $F(131486759)=83$.
• $G(y)$ for every prime $y\le 79$, matching A062241.
• $F(n)\le\sqrt n$ on $[2,10^6]$ except $\{3,7,23\}$.
• $F(n)\le n^{2/5}$ on $[2,10^6]$ except sixteen explicit $n$, last $479$.
• $F(n)\le n^{1/3}$ on $[2,3\cdot 10^5]$ except seventy-six explicit $n$, last $18191$.
• Every finite residue / shift / square-adjustment template we ran fails to cover $\mathbb Z$ at any $\varepsilon<1/2$.

Still open, and not touched:

• $F(n)\le n^\varepsilon$ for a fixed $\varepsilon<4/(9\sqrt e)$ and all large $n$.
• An effective $N_0(\varepsilon)$ for Balog 1989.
• The conjectured $F(n)\le n^{o(1)}$, equivalently Green #59 for every $\varepsilon>0$.
• Whether the sixteen $n^{2/5}$ exceptions are the only ones in $\mathbb N$. The prefix is not a proof.

We did not beat the published record. A failed search with a verifier is the product.

2026-08-27 continuation

0. What was actually missing

The 17 August search killed finite shift lists. A dent still needed a formula $a(n)$ that writes every large $n$ as two $n^\varepsilon$-smooth summands for some $\varepsilon<1/2$, or a two-factor shape that could be proved to work for all large $n$. Isolated prefixes were already known not to count.

1. Named false starts

Floor-divisor as a new covering. Set $u=\lfloor n^\varepsilon\rfloor$ and $a=n-(n\bmod u)$. The remainder is $n^\varepsilon$-smooth by size. At $\varepsilon=1/2$ this is the square covering in different clothes. Below $1/2$, $\lfloor n/u\rfloor$ has size $n^{1-\varepsilon}>n^\varepsilon$ and is $n^\varepsilon$-rough for a positive-density set of $n$. First hole at $n=3$ on every tested exponent; thousands of holes through $20000$.

Largest power of two, triangular numbers, cubes. Each is a closed form. Each has a first hole at $n=3$ and holes still appearing at $n=20000$. Cubes are worse than squares: the remainder window has exponent $2/3$.

Two-factor with $u\le n^{1/5}$ as an effective $2/5$. On $[2,20000]$ the holes are exactly the sixteen $F$-exceptions, last $479$, and at $1/3$ exactly the seventy-six exceptions, last $18191$. The factor restriction is not losing splittings. It is also not $N_0$.

2. The useful failure

The floor-divisor rewrite taught that every “take a small remainder so it is automatically smooth” template is the short-interval template. The only closed-form construction that always puts a smooth number in $(n-n^{1/2},n]$ is a nearby square. Polynomials of degree $d\ge 2$ cannot do better by size: the gap is $\asymp n^{1-1/d}\ge n^{1/2}$.

3. The click

For a prime $P$ and $u\sim P^{\varepsilon/(1-\varepsilon)}$, the integer $n=Pu+1$ makes floor-divisor choose a multiple of $u$ whose other factor is $P$. The comparison $P>n^\varepsilon$ is the same inequality as $\varepsilon<1/2$. That is an infinite failure family, not a prefix. The power-of-two formula has the same shape: $n=2^k+q$ with $q$ the next prime after $2^{k-1}$ fails even $n^{1/2}$.

So the cover we wanted is not hiding in a cleverer closed form of the same kind. A formula whose remainder is small cannot beat $1/2$. A formula whose first summand is a sparse smooth set (powers of two, $S$-units for a fixed $S$, high powers) fails infinitely often. Making Balog–Sárközy effective still needs an analytic input we do not have.

4. The argument, in the order it was found

Green #59 and the Wayback copy of Erdős #334 were fetched first. Hildebrand–Tenenbaum pp. 468–469, the three arXiv neighbours, and Ki–Maier–Sankaranarayanan 2016 still name Balog $0.2695$ as the all-integers binary record. Then the parent verifier was started. Then the closed-form census, then the infinite families, then the polynomial size check, then two-factor. Two-factor matching $F$ exactly was a surprise and a disappointment: it looks like a theorem on a long prefix and is not one.

5. Computer search

Stored under compute/q1/certs/.

• poly_obstruction.json — six polynomials, remainder exponent $1-1/d$.
• infinite_family.json — 129/129 floor-divisor hits at three exponents; power-of-two hits for $k=4,\ldots,16$.
• q1_search.json — hole counts and last holes on $[2,20000]$; two-factor equal to the known $F$ prefixes.

Replay: cd problems/two-smooth-summands/compute/q1 && ./run_all.sh.

6. What is proved vs still open

Proved this session, all checkable:

• The listed closed-form templates fail on $[2,20000]$ at every tested $\varepsilon<1/2$, with holes persisting to the endpoint.
• Floor-divisor and largest-power-of-two have explicit infinite failure families.
• A fixed polynomial of degree $d\ge 2$ cannot beat the square covering by size of remainder.
• Two-factor $u\le n^{1/5}$ reproduces the $F$ exception prefixes on $[2,20000]$.

Still open, and not touched: the same four items as on 17 August. We did not beat the published record.