Built by Colin Snyder · colin@colinsnyder.com：mailto:colin@colinsnyder.com

Advised by Mike Kim · proposed solutions ↓：#solutions

Inspired by Ignis · previously built by Myself, Dhruv Agarwal, & Nitin Kesarwani at the New Turing Institute

These are “proposed solutions”, so if you take issue with any of them, please let us know.

Star Fleet is an AI system that solves the world's hardest open mathematics problems using Lean 4. It's a Mac desktop app that controls up to 20 custom agentic harnesses called “starships” in parallel, each running its own GPT-5.6 instance on a dedicated 60-vCPU server and working on a separate math problem. Everything is built from scratch in TypeScript & Bun.

Many problems listed as “open” carry informal or partial answers already available online; we tried extremely hard to avoid working on any such problems (but most likely failed in certain cases). If there are attribution mistakes, please let us know.

Let a , b , c ≥ 1 a,b,c\geq 1 a , b , c ≥ 1 be three integers which are pairwise coprime. Is every large integer the sum of distinct integers of the form a k b l c m a^kb^lc^m a k b l c m ( k , l , m ≥ 0 k,l,m\geq 0 k , l , m ≥ 0 ), none of which divide any other?

(Erdős Problem #123 — prize: $250 — number theory — https://www.erdosproblems.com/123)

For every pairwise-coprime triple of integers a,b,c>1, every sufficiently large integer is a sum of distinct terms a^i b^j c^k such that no selected term divides another. In Lean, this is the theorem Erdos123.erdos_123 : Erdos123.IntendedStatement.

For pairwise-coprime integers a , b , c > 1 a,b,c>1 a , b , c > 1 , consider the numbers

The question asks whether every sufficiently large integer is a sum of distinct such numbers, with the additional requirement that no chosen summand divides another .

The divisibility condition is the real source of difficulty. Ordinary completeness arguments can use many terms from different scales, but terms from different scales tend to be comparable by divisibility. Conversely, a set chosen to be a divisibility antichain can be too arithmetically sparse to fill consecutive integers.

Earlier work had developed a powerful reduction scheme: choose a correction with the required residue modulo one base, subtract it, divide by that base, and induct. For particular triples this succeeds after a finite computer check. In general, however, it leaves a stubborn finite-seed problem : one must first represent every integer in a multiplicatively wide interval [ N , C N ] [N,CN] [ N , C N ] . The correction induction does not construct that interval; it only propagates it.

This explains why several attractive partial ideas did not finish the problem:

The important lesson was that large additive width is not enough . The lower endpoint has to remain under quantitative control.

The first structural simplification is to work on one homogeneous exponent level

For pairwise-coprime bases greater than one, divisibility of monomials is coordinatewise comparison of their exponents. Therefore two distinct monomials on the same level can never divide one another. Every subset of a homogeneous level is automatically primitive.

This turns the problem into an additive question about subset sums while making primitiveness essentially free—as long as all pieces of the construction can be placed on the same exact degree.

An edge-code construction supplies c n c^n c n primitive subset sums on one level with distinct residues modulo c n c^n c n and a bounded carry. Coloring by that carry and applying finite van der Waerden, itself obtained from Mathlib’s Hales–Jewett theorem, gives arbitrarily long exact arithmetic progressions of primitive homogeneous subset sums.

and then choose v > 0 v>0 v > 0 so that

Define two coprime homogeneous translation weights

Copies of one AP digit family are translated by the weights

The choice u > H + 1 u>H+1 u > H + 1 places different copies in disjoint bands of the b b b -exponent. Multiplying every term by a b c abc ab c makes every AP term strict-interior. All copies then lie on one exact exponent degree.

A bounded homogeneous-radix lemma proves that coefficient sums

contain a full interval of width at least 2 A B M + 1 2AB^{M+1} 2 A B M + 1 . Replacing each coefficient by the corresponding AP digit set realizes this as an interval on a lattice of step a b c d abc\,d ab c d , where d d d is the AP difference.

The next ingredient constructs, on every sufficiently high exact degree , a primitive correction for each residue modulo any prescribed modulus. The corrections are supported on the three coordinate faces and, in the ordered case, have total size bounded by

Apply this with modulus a b c d abc\,d ab c d , on the same exact degree as the AP-radix construction. Face-supported corrections are disjoint from the strict-interior AP terms. Moreover,

Thus the correction spread is eventually smaller than the radix width. Residue gluing converts the lattice interval into an ordinary consecutive interval [ L M , U M ] [L_M,U_M] [ L M ​ , U M ​ ] satisfying

At this stage there is a genuine interval, but its multiplicative width is still only bounded by a constant. This is exactly where the earlier baseline problem remained.

The decisive idea was to exploit monomials that had not yet been used, on the same exact homogeneous level .

For each of linearly many indices s s s , fix a b b b -exponent just beyond every AP band. Among the remaining a , c a,c a , c -exponents, choose the last point of the geometric grid

below a target of size c v M c^{vM} c v M . Because consecutive grid points differ by the fixed factor c / a c/a c / a , the selected monomial lies in a controlled multiplicative window. After restoring the common factors, this produces at least M − O ( 1 ) M-O(1) M − O ( 1 ) distinct optional monomials z z z satisfying

Every optional term is therefore no larger than the already available interval width. Adding such a term optionally—either use it or do not—extends a consecutive interval without changing its lower endpoint. Since all optional terms remain on the same exact level and lie beyond the AP exponent bands, primitiveness and disjointness are preserved.

while the lower endpoint remains O ( B M ) O(B^M) O ( B M ) . Hence the ratio of the upper endpoint to the lower endpoint grows linearly with M M M . For every requested R > 1 R>1 R > 1 , and beyond every requested lower threshold, this constructs a primitively represented interval

This interior-shell amplification is what removes the finite-seed obstruction. The successful coordinate change was not merely “work on a homogeneous level,” but “place the main growth along an interior homogeneous ray, then use the unused transverse strip as optional mass.”

The residue-reduction argument was strengthened to a flexible finite-seed gate: there are constants N 0 N_0 N 0 ​ and C > 1 C>1 C > 1 such that any represented interval [ N , C N ] [N,CN] [ N , C N ] with N ≥ N 0 N\ge N_0 N ≥ N 0 ​ implies d-completeness.

Applying the arbitrary-width construction with R = C R=C R = C proves d-completeness for ordered bases 1 a c b 1 1 a c b . Pairwise-coprime bases greater than one are distinct, so every triple has one of six strict orderings. Explicit permutations of the exponents show that permuting the bases leaves the smooth set unchanged, completing all cases.

The proof is formalized in Lean 4 with Mathlib. The final theorem is

where IntendedStatement quantifies over all pairwise-coprime natural bases greater than one and asserts an explicit eventual threshold for primitive representations.

The complete project builds successfully, and a source scan finds no proof placeholders. Lean’s axiom report for the final theorem is exactly

with no sorryAx . The webpage literally writes a , b , c ≥ 1 a,b,c\ge1 a , b , c ≥ 1 ; that universal formulation is false at ( 1 , 1 , 1 ) (1,1,1) ( 1 , 1 , 1 ) . The project records this separately and proves the intended nondegenerate a , b , c > 1 a,b,c>1 a , b , c > 1 conjecture used in the source literature and independent formal-conjecture encoding.

The download contains everything needed to verify this solution independently: the original problem, the formal statement, the complete pinned Lean project, and the verifier script — plus step-by-step instructions for your agent. About 20 minutes, mostly downloading Mathlib.

↓ Download full solution raw：/downloads/verify/erdos-123/erdos-123-solution.zip

Let A ⊆ N A\subseteq \mathbb{N} A ⊆ N be such that ∣ A ∩ [ 1 , 2 x ] ∣ − ∣ A ∩ [ 1 , x ] ∣ → ∞ as x → ∞ \lvert A\cap [1,2x]\rvert -\lvert A\cap [1,x]\rvert \to \infty\textrm{ as }x\to \infty ∣ A ∩ [ 1 , 2 x ]∣ − ∣ A ∩ [ 1 , x ]∣ → ∞ as x → ∞ and ∑ n ∈ A { θ n } = ∞ \sum_{n\in A} \{ \theta n\}=\infty n ∈ A ∑ ​ { θ n } = ∞ for every θ ∈ ( 0 , 1 ) \theta\in (0,1) θ ∈ ( 0 , 1 ) , where { x } \{x\} { x } is the distance of x x x from the nearest integer. Then every sufficiently large integer is the sum of distinct elements of A A A .

(Erdős Problem #254 — number theory — https://www.erdosproblems.com/254)

Erdős Problem #254 is true: if A ⊆ ℕ has dyadic shell counts tending to infinity and ∑_{n∈A} ‖θn‖ = ∞ for every real 0

Let A ⊆ ℕ . Assume that the number of elements of A in every dyadic shell (x,2x] tends to infinity, and that

for every 0 , where ‖x‖ is distance to the nearest integer. Erdős asked whether these two hypotheses force every sufficiently large integer to be a sum of distinct elements of A .

The hypotheses control two very different phenomena:

Neither condition alone is close to sufficient. The real difficulty is to assign disjoint elements of A to several roles—two piecewise-Bohr classes, a phase-correction class, and a final syndetic class—without destroying the phase hypothesis or reusing a summand.

Several tempting shortcuts are false.

The key was therefore to solve the allocation problem globally and replace the missing ergodic machinery by a finite-cyclic spectral proof that could be kernel-checked from first principles.

The decisive structural observation is that dyadic abundance makes the set of phases with finite total mass countable .

two sufficiently close phases cannot both satisfy that bound. Indeed, if their difference is δ , look at a dyadic shell at scale about 1/(8δ) . Every element of that shell contributes a controlled positive amount to the difference phase, and the growing shell cardinality contradicts the assumed bound. Hence every bounded phase sublevel is finite, and the union of those sublevels is countable.

This converts an uncountable allocation problem into a countable diagonalization. We split off a shell-abundant seed, enumerate only its countably many convergent phases, and balance reserves against those phases. The resulting reserve has syndetic distinct subset sums, while its actual complement still has divergent phase mass for every nonzero phase. Splitting the reserve by rank produces three pairwise-disjoint syndetic finite-sum classes and a disjoint universally phase-divergent correction class.

That resolves the support-allocation obstruction completely.

The remaining input was the Bergelson–Furstenberg–Weiss theorem: the sum of two syndetic subsets of ℕ contains a piecewise-Bohr set.

Instead of formalizing an abstract Kronecker factor, we built the spectral argument from finite cyclic groups.

A syndetic set has a uniformly positive number of points in long finite blocks. Given dense blocks from two syndetic sets, finite cyclic averaging finds a large fiber on which all pairs have the same exact sum. A dense subblock of that fiber has the property that each positive internal difference, after one common translation, belongs to the original sumset.

For a signal Φ : ZMod N → ℂ , we proved Parseval’s identity for Mathlib’s unnormalized DFT: