2026-08-29 · article unsolved problemsmethodrecords

Drawing unsolved problems at random — a record of the smaller problems

The method was fixed in advance.

Problems are drawn at random. When it feels stuck, move to another problem and switch attention. Record the thinking at the point of getting stuck, together with the order and the intervals of the draws (measured from the machine). Measure whatever mathematical software can measure.

A lottery box was built and drawn from with random numbers. Seven problems drawn, six taken up.

Four of the problems drawn here have articles of their own. What stays in this article is the record of the smaller problems — Sierpiński numbers, Riesel numbers, distinct distances — and the method itself. Five problems around Erdős problems taken up afterwards are in §05 Small records.

※ Not drawn in this lottery, but the same day's work also took in the Collatz conjecture and the Riemann hypothesis (those were taken up by choice, not by random draw).

paperproved, not yet machine-checked computationchecked on this machine, within the range stated knowna restatement, a known theorem, or a check of the literature. Nothing in this article is closed in Lean.

※ The square peg problem, drawn seventh, was redrawn 8 seconds later. Why it was put back is not recorded. What the record holds is only that it was 8 seconds.

The largest thing to come out of this round was that three entirely different problems had the same shape.

"The provable part" is already closed.
What remains is only the part that method cannot reach in principle.

Sierpiński numbers (#7)

Is 78557 the smallest Sierpiński number? That 78557 is a Sierpiński number at all is settled by a finite check.

The least common multiple of the orders of 2 for the covering set {3,5,7,13,19,37,73} is L=36, so checking the 36 cases of n mod 36 settles it for every n.—no uncovered residue. Proof complete.

What about candidates below 78557? For the 39,278 odd k < 78557, we searched for the smallest n making k·2ⁿ+1 prime.

The 5 for which no prime has yet been found (21181, 22699, 24737, 55459, 67607) are all inside that 351. The other 346 are ones resolved at larger n (the ones PrimeGrid has ground down over decades). These 5 have not been shown to lack a covering set; they have not been proved to be Sierpiński numbers at all — all that is established is that there is no prime for n < 45,543,700, and the expectation runs the other way (a prime will eventually appear). known

And here is the real point. Whether k has a covering set of period L is decided by a finite computation. (If p divides k·2ⁿ+1 then the order of 2 must divide L, so the candidate primes are restricted to the prime factors of 2^L−1.)

Below 78557 there is not one k with a covering set.
That is, the road of "proof by covering set" is completely closed.
What remains is a quite different question: is there a Sierpiński number with no covering set? — posed as Erdős problem #1113, and open. It cannot be settled by a finite computation. known

It is not only the periods of primes that cover

The count above restricts "covering" to the periods of primes. That is not the only thing that covers a whole residue class — algebraic factorisations cover too. The Sophie Germain identity:

Take k = m⁴ (m odd) and write n = 4j+2, so that k·2ⁿ = 4·(m·2^j)⁴; then for n ≡ 2 (mod 4) compositeness follows without using a single prime (81·2⁶+1 = 5,185 = 71 × 73; 625·2⁶+1 = 40,001 = 199 × 201). There is a quarter less to cover.

There are 7 fourth powers of odd numbers below 78557 (81, 625, 2401, 6561, 14641, 28561, 50625). Checking for each whether n ≢ 2 (mod 4) can be covered by primes up to period 96, none of the 7 can be covered, and even the closest leave 3 classes (2401 = 7⁴, 14641 = 11⁴ and 28561 = 13⁴ leave 3 classes at period 12). computation

The odd m for which an algebraic factorisation is available are completely classified by Capelli's criterion — for m > 1, only m = c^q (q an odd prime; the class n ≡ 0 mod q) and m = c⁴ (the class n ≡ 2 mod 4). So for m > 1 the set covered by algebra is always periodic and of density below 1. known The core of #1113 is whether a third mechanism, neither covering nor algebraic factorisation, can exist; no theorem rules it out.

"Has a covering set or not" becomes a continuous quantity through the sieve weight (the Proth weight) W_k — it is 0 when a finite period covers, and settles at a positive value for ordinary k. The weights of the remaining 5 are between 0.07 and 0.27 (the population mean is 2), and the expected number of k surviving to depth 4.5×10⁷ is 4 to 7 — there is nothing unnatural in 5 being still unresolved. known

Riesel numbers (same machine, straight after)

The k·2ⁿ−1 side. The conjectured smallest is 509203, covering set {3,5,7,13,17,241}, L=24. By the same procedure—the only odd k ≤ 509203 with a covering set is 509203 itself.

The Erdős arithmetic-progression conjecture (#22)

The one that became its own article had the same shape. The road of "raising the exponent of the logarithm above 1" goes through for k=3 and does not for k=4. And Green–Tao themselves write that this "is the limit of our method". The position at which the method closes is clearly known.

g(n) is the minimum number of distinct distances determined by n points in the plane. In 2015 Guth–Katz proved g(n) ≥ c·n/log n, effectively settling it, but Erdős's conjecture is n/√(log n). The gap that remains is a factor of √log n.

We counted the distinct distances of an m×m grid directly.

Divided by n/√log n, the ratio levels off around 1.11—you can see numerically that the shape of Erdős's conjecture is right.
Divided by n/log n, the ratio keeps rising, 3.46 → 4.35—the gap between the Guth–Katz lower bound and the true value is plainly visible even at this size.

But this problem is different in character from the previous two. Nobody knows the exact values beyond n=13. The configuration space is continuous, so combinatorial search cannot reach it—it is not "closed" but "entered from a different door".

Five problems around Erdős problems, taken up after the draw. In none of them is the problem itself solved. Each either narrows where to look for the answer, shows that a certain route of construction is closed, or re-measures a known quantity. The numbers are those of the Erdős problems site (erdosproblems.com), numbered differently from the "#" in the tables above.

#1087 — A vertex-transitive point set lies on one circle

The problem is to estimate the maximum f(n), over n points in the plane, of the number of 4-point sets in which some two distances are equal. The Erdős–Purdy estimate is n3 log n ≪ f(n) ≪ n7/2, and the question is whether f(n) ≤ n3+o(1). known Highly symmetric configurations — those with a group of congruences acting transitively on the points (vertex-transitive point sets) — are natural candidates, but a finite vertex-transitive planar point set always lies on one circle.paper, one line

Conclusion: the family of vertex-transitive configurations does not yield the factor log n in the lower bound.

#104 — 3-rich circles of a fixed radius inside a rational lattice

The problem asks whether, for n points in the plane, the number of unit circles containing 3 or more of them is o(n2). The known lower bound is Elekes's, of order n3/2. known Next door, in #101, a conjecture of the same shape is broken by a Behrend-type construction, so we checked whether the same transfer works for unit circles. When the points are in a rational lattice, the circumcentre of three lattice points is a rational point, and the squared distances D of pairs that can lie on a circle of radius ρ (ρ2 = a/b) are limited to those satisfying 4ρ2 = D(1+s2) (s rational). Their kinds are determined by the divisors of 4a, and there are only ao(1) of them, so there are at most n·ao(1) 3-rich circles of radius ρ, and at most n1+o(1) if a ≤ nO(1).paper For ρ = 5, the predicted allowed squared distances {2, 10, 20, 36, 50, 64, 80, 90, 98, 100} agreed with a measurement over all triples of the 13×13 lattice, and the number of circles was 641 on the 19×19 lattice (n = 361), growing almost in proportion to n. computation

Conclusion: a Behrend-type transfer from a single rational lattice does not reach n3/2. It says nothing about the body of the prize (whether it is o(n2)).

#99 — Optimal configurations without a unit equilateral triangle, and the pairs realising the diameter

The problem asks whether, among n points with minimum distance 1, the configuration of minimum diameter contains an equilateral triangle of side 1 when n is large. We put two known facts side by side. One is Eppstein 2018's upper bound 2n − Ω(√n) on the number of edges of a triangle-free penny graph. The other is the count that an infinitesimally rigid planar bar framework has at least 2n − 3 edges. known Under the explicitly stated assumption (J) that the framework whose bars are the pairs at distance 1 and the pairs realising the diameter (t of them) is infinitesimally rigid, the two give t ≥ c√n − 3 (c being Eppstein's constant). paper, conditional On the other hand, the perimeter of the convex hull gives t ≤ 3.30√n or so, so the contest comes down to a comparison of constants. The assumption (J) is not a theorem, and the value of Eppstein's constant has not been checked either.

Conclusion: a conditional reduction under the assumption (J). If (J) falls, the inequality falls with it.

#40 — g(N) = Nε cannot be the answer

The problem asks for which g(N) → ∞ it holds that "if |A ∩ [1,N]| ≫ N1/2/g(N) then limsup 1A*1A(n) = ∞". The Erdős–Rényi construction in the notes to #39 gives, for every ε > 0, a set A with |A ∩ [1,N]| ≫ N1/2−ε whose representation function is bounded. known This satisfies the hypothesis for g(N) = Nε and breaks the conclusion. paper What this rests on is the description of Erdős–Rényi written on the site; the original paper has not been consulted.

Conclusion: the meaningful range is squeezed into 1 ≪ g(N) ≤ No(1) (for example (log N)C).

#30 — Re-measuring the error term for Sidon sets with a convex program

For the largest size h(N) of a Sidon set in [1,N], what #30 asks is whether h(N) = N1/2 + Oε(Nε). What was dealt with here is the neighbouring quantity, the coefficient γ in h(N) ≤ N1/2 + γN1/4 + O(1). The computer-assisted part of Carter–Hunter–O'Bryant discretises the function in the form "fix the values and move the breakpoints", and so needs an enumeration of cells. Switching to the form "fix the breakpoints and move the values" turns the same problem into a convex program, and the certificate for the upper bound is a single rational tangent plane.paper By this method b∞ ≤ 1.89715, that is γ ≤ 0.94858, was confirmed with an independent checker that uses rational numbers only. computation But this is worse than the best known. The γ0 = 0.94349… of Hou–Zhao (arXiv:2607.01169) is smaller, and even the ceiling of this method (γ ≈ 0.9466 by extrapolating to finer grids) does not reach it. known

Conclusion: as a value, it does not reach the best known. What remains is only the side of the method: the cell enumeration is replaced by a convex program and the certificate becomes a single tangent plane. It does not touch #30 itself (Nε).

All three were of the type "the name did not match the contents". And all three were saved by numbers that were being printed.

The three BSD errors overlapped to produce the value |Ш| = 1/18. The cause was findable because the discrepancy had the simple shape of 2 and 6; had it stayed at 1/18 I think it would have gone unexplained for a long time.

"Searched up to 10⁶ and did not find it" and "does not exist" are different. In the same way, "examined A₅" and "examined a variable named A₅" are different.
In this round the latter happened three times. Had the order, the height and the shape of the discrepancy not been printed, all three would have passed.

A problem drawn twice. χ(ℝ²) ∈ {5, 6, 7} has not moved. The record of entering from the side of the fractional chromatic number, and the quantity measured from there — f(α), the largest unit-distance graph with independence number at most α; f(3) = 10, and f(4) is 14 or 15 — is in The Hadwiger–Nelson problem.

Seven problems drawn, six taken up, and none of them solved. Even so, some things about the method worked.

Some things did not work. None of the problems got more than one round's depth. Only #22 was carried on and became an article; the rest stopped at "measured the terrain at the entrance".

This method, together with the discipline of checking and the standard for claims put outside, is summarised on one page in How the work is done.

Sources and reproduction

Apart from §05, there is no new mathematics in this article. Every theorem used and every known record is public. The five in §05 are paper arguments and computations connecting known results — small observations that may be obvious to specialists, although no description of the same form was found in the literature searched. What this article did was check them again on its own machine, and record which positions could be checked and which could not.

The number that stays most from this record is the 8 seconds. The square peg problem was drawn and redrawn 8 seconds later. Why it was put back is not in the record. What remains is only the fact that it was 8 seconds. The record of an action can be more accurate than the introspection of the one who acted—recording the order and intervals by measurement is there for that.