Star Fleet Math

theorem erdos_123 : IntendedStatement := intended_erdos_123›ReportSolving Erdős Problem 123 The problem and why it resisted the usual induction For pairwise-coprime integers a,b,c>1a,b,c>1, consider the numbers aibjck(i,j,k≥0). a^i b^j c^k \qquad (i,j,k\ge 0). 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,CN][N,CN]. The correction induction does not construct that interval; it only propagates it. This explains why several attractive partial ideas did not finish the problem: A signed identity of difference one gives two consecutive sums, but one residue representative per class necessarily has spread at least the modulus minus one. A width-one interval cannot grow under ordinary residue gluing. Complete residue systems on a primitive level solve congruences, but say nothing about their numerical spread. Van der Waerden and Hales–Jewett arguments produce arbitrarily long arithmetic progressions of primitive sums, but initially with an uncontrolled common difference. Even after fixing the common difference, a progression B0+rdB_0+r d carries a large positive baseline B0B_0. Replicating such progressions increases width and baseline at the same rate, so it need not produce the multiplicatively wide seed required by induction. The important lesson was that large additive width is not enough. The lower endpoint has to remain under quantitative control.

Filling residues with face corrections 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 CcorrcD. C_{\mathrm{corr}}c^D. Apply this with modulus abc dabc\,d, on the same exact degree as the AP-radix construction. Face-supported corrections are disjoint from the strict-interior AP terms. Moreover, Bcu+v=(bc)u>1, \frac{B}{c^{u+v}}=\left(\frac bc\right)^u>1, so exponential domination gives CcorrcD=o(BM). C_{\mathrm{corr}}c^D=o(B^M). Thus the correction spread is eventually smaller than the radix width. Residue gluing converts the lattice interval into an ordinary consecutive interval [LM,UM][L_M,U_M] satisfying abc BM≤UM−LM,LM≤KBM+1 abc\,B^M\le U_M-L_M, \qquad L_M\le K B^{M+1} for a fixed constant KK. 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 key breakthrough: an optional interior shell 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 ss, fix a bb-exponent just beyond every AP band. Among the remaining a,ca,c-exponents, choose the last point of the geometric grid bsaR−kck b^s a^{R-k}c^k below a target of size cvMc^{vM}. Because consecutive grid points differ by the fixed factor c/ac/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) distinct optional monomials zz satisfying z≤X,aX≤cz,X=abc BM. z\le X, \qquad aX\le cz, \qquad X=abc\,B^M. 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. Their combined contribution is Ω(MBM), \Omega(MB^M), while the lower endpoint remains O(BM)O(B^M). Hence the ratio of the upper endpoint to the lower endpoint grows linearly with MM. For every requested R>1R>1, and beyond every requested lower threshold, this constructs a primitively represented interval [N,RN]. [N,RN]. 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.” Completing the induction The residue-reduction argument was strengthened to a flexible finite-seed gate: there are constants N0N_0 and C>1C>1 such that any represented interval [N,CN][N,CN] with N≥N0N\ge N_0 implies d-completeness. Applying the arbitrary-width construction with R=CR=C proves d-completeness for ordered bases 1<a<c<b1<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. Verification The proof is formalized in Lean 4 with Mathlib. The final theorem is Erdos123.erdos_123 : Erdos123.IntendedStatement 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 [propext, Classical.choice, Quot.sound] with no sorryAx. The webpage literally writes a,b,c≥1a,b,c\ge1; that universal formulation is false at (1,1,1)(1,1,1).

The project records this separately and proves the intended nondegenerate a,b,c>1a,b,c>1 conjecture used in the source literature and independent formal-conjecture encoding.›Download Full Solution & Verify with Your AI2) Erdős Problem #129Erdős Problemwww.erdosproblems.com/129›QuestionLet R(n;k,r)R(n;k,r) be the smallest NN such that if the edges of KNK_N are rr-coloured then there is a set of nn vertices which does not contain a copy of KkK_k in at least one of the rr colours. Prove that there is a constant C=C(r)>1C=C(r)>1 such thatR(n;3,r)<Cn.R(n;3,r) < C^{\sqrt{n}}.(Erdős Problem #129 — graph theory, ramsey theory — https://www.erdosproblems.com/129)›ResultThe proposed bound is false: for every natural n ≥ 120, 2^(⌊n/120⌋) < R(n;3,2) ≤ 2^(2n), so no constant C > 1 can satisfy R(n;3,2) < C^(√n) for all n. Consequently the exact formal proposition LiteralProblem129 is false.theorem not_literalProblem129 : ¬ LiteralProblem129 := by intro h exact not_claimedBoundFor_two (h 2 (by omega)) theorem R3_two_global_exponential_sandwich (n : ℕ) (hn : 120 ≤ n) : 2 ^ (n / 120) < R3 n 2 ∧ R3 n 2 ≤ 2 ^ (2 * n) := by exact ⟨two_pow_div_120_lt_R3_two n hn, R3_two_le_two_pow_two_mul n⟩›ReportErdős Problem 129: a formally verified disproof The problem and why it was deceptive Erdős Problem 129 defines a Ramsey-type threshold R(n;3,r): the least order N such that every r-colouring of the edges of K_N has an n-vertex set on which at least one colour contains no triangle.