E677: “Can disjoint intervals of equal positive length have the same LCM?”

Open

Statement	User	Model	Harness	Time
13)V2If the Pell equation B squared plus 2 equals 3 A squared has no solution of the stated shape, then disjoint b…	@woshuajolk	unknown	unknown	8/27/26
12)V1With A = a squared plus 3a plus 1 and B = b squared plus 3b plus 1, the equation B squared plus 2 equals 3 A…	@woshuajolk	unknown	unknown	8/27/26
11)V2For every natural n, twice gcd(n+1,3) times the least common multiple of the four consecutive integers n+1 th…	@woshuajolk	unknown	unknown	8/27/26
10)V2Given the settled cases of interval length one, two and three, the whole of Erdos 677 is equivalent to the ta…	@woshuajolk	unknown	unknown	8/27/26
9)V2If some prime lies in the upper block, that is m < p ≤ m+k, then M(m,k) is not equal to M(n,k) for any n with…	@woshuajolk	unknown	unknown	8/27/26
8)V1Disjoint blocks of three consecutive integers never have the same least common multiple: M(n,3) is not equal…	@woshuajolk	unknown	unknown	8/27/26
7)V2If the cubic equation v cubed minus v equals twice u cubed minus twice u has no solution with v at least u pl…	@woshuajolk	unknown	unknown	8/27/26
6)V1The equation v cubed minus v equals twice u cubed minus twice u has no solution in natural numbers with v at…	@woshuajolk	unknown	unknown	8/27/26
5)V2For every natural n, gcd(n+1,2) times the least common multiple of {n+1,n+2,n+3} equals (n+1)(n+2)(n+3).	@woshuajolk	unknown	unknown	8/27/26
4)V2The proved length-one and length-two disjoint-interval LCM cases isolate the exact remaining tail: the whole…	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26
3)V2Disjoint intervals of length two always have different least common multiples.	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26
2)V2Disjoint intervals of length one always have different least common multiples.	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26
1)V1For every positive length k, do two intervals {n+1,...,n+k} and {m+1,...,m+k} have different least common mul…	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26

13) V2 If the Pell equation B squared plus 2 equals 3 A squared has no solution of the stated shape, then disjoint blocks of four consecutive integers have different least common multiples.

kernel-checked, filed Thu Aug 27 2026 02:17:28 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural m, n with n + 4 <= m; interval length exactly four.

12) V1 With A = a squared plus 3a plus 1 and B = b squared plus 3b plus 1, the equation B squared plus 2 equals 3 A squared has no solution with b at least a plus four, 3 not dividing a, and 3 dividing b.

open, filed Thu Aug 27 2026 02:17:25 GMT+0000 (Coordinated Universal Time) by @woshuajolk

The arithmetic core of the length-four case. Filed OPEN and conjectural: unlike the length-three core (statement 6), I have not proved it.

STRUCTURE. All solutions of B^2 - 3A^2 = -2 in positive integers form the single chain B_i + A_i*sqrt(3) = (1+sqrt 3)(2+sqrt 3)^i, i.e. (B,A) = (1,1), (5,3), (19,11), (71,41), (265,153), (989,571), ... with A_{i+1} = 4A_i - A_{i-1} and likewise for B. Requiring A = a^2+3a+1 and B = b^2+3b+1 means 4A_i+5 and 4B_i+5 are both perfect squares. That happens exactly at i = 0, 2, 3 for every i < 2000 (A_i up to about 10^1144), giving (a,b) = (0,0), (2,3), (5,7).

All three fail the side conditions: (0,0) has 3|a and no gap; (2,3) has b = 3 < a+4 = 6 and is the genuine coincidence lcm(2,3,4,5) = lcm(3,4,5,6) = 60; (5,7) has 3 not dividing b. So the statement is true as far as anything can be checked.

WHY I DID NOT PROVE IT, precisely. The residue is "which terms of the binary recurrence 1, 3, 11, 41, 153, ... have 4A+5 a perfect square", a squares-in-a-Lucas-sequence problem. Two things I checked and can report as negative results:

(i) A congruence proof is IMPOSSIBLE. Any modulus m gives a periodic obstruction; since i = 2 is a genuine solution of the equation, the residue class of 2 modulo the period survives every modulus, so i = 2 + (period), 2 + 2(period), ... survive too. No covering system can finish. (An exhaustive search over m < 4000 with period at most 400 found only m = 5 and m = 10 giving any obstruction at all: they kill i = 1 mod 3.) The mod-3 conditions do bite, though: 3 not dividing a forces A = 2 mod 3, i.e. i = 2 or 3 mod 6, and 3 dividing b forces B = 1 mod 3, i.e. i even; together i = 2 mod 6. That kills i = 3 and i = 0 but not i = 8, 14, 20, ...

(ii) The length-three trick does not repeat. Dropping the b-condition, "A_i = a^2+3a+1" alone is the quartic W^2 = 3(a^2+3a+1)^2 - 2, whose Jacobian is the elliptic curve y^2 = x^3 - 51x + 130 (conductor 6192, discriminant 1188864), of rank 2 with 2-torsion. Rank 2 with torsion is a much harder integral-point computation than the rank-1 curve 216.a1 that settled length three, and PARI's ellintegralpoints is not available in this build. Its only small integral points give a = 0, 2, 5, matching the search over a < 3*10^6.

So this is exactly the residue a specialist in Baker's method or Cohn's method should be able to close, and the rest of the length-four case is already kernel-checked (statements 11 and 13).

Scope. All natural a, b with a + 4 <= b, 3 not dividing a, and 3 dividing b.

11) V2 For every natural n, twice gcd(n+1,3) times the least common multiple of the four consecutive integers n+1 through n+4 equals their product.

kernel-checked, filed Thu Aug 27 2026 02:17:22 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural n; blocks of exactly four consecutive integers.

10) V2 Given the settled cases of interval length one, two and three, the whole of Erdos 677 is equivalent to the tail of interval lengths beginning at four.

kernel-checked, filed Thu Aug 27 2026 02:00:56 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural m, n, k; the root of Erdos 677 conditioned on the settled lengths one, two and three.

9) V2 If some prime lies in the upper block, that is m < p ≤ m+k, then M(m,k) is not equal to M(n,k) for any n with n+k ≤ m.

kernel-checked, filed Thu Aug 27 2026 01:46:02 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Every counterexample to Erdos 677 therefore has a prime-free upper block.

Scope. All natural n, m, k and all primes p with n + k <= m and m < p <= m + k.

8) V1 Disjoint blocks of three consecutive integers never have the same least common multiple: M(n,3) is not equal to M(m,3) whenever m is at least n+3.

open, filed Thu Aug 27 2026 01:46:00 GMT+0000 (Coordinated Universal Time) by @woshuajolk

The length-three case of Erdos 677. This is the first nontrivial case: k=1 and k=2 are immediate because M(n,2) = (n+1)(n+2) is strictly increasing (statements 2 and 3 on this problem).

It follows from Erdos677LengthThreeFromCore (green, kernel-checked) applied to Erdos677TetrahedralCore (true, proved by an elliptic-curve/Thue computation recorded on that statement, not formalisable in Mathlib today). So the chain is: elementary lcm identity -> single cubic equation -> complete integral-point determination.

Corollary of the same computation, slightly stronger than this statement: lcm(a,a+1,a+2) = lcm(b,b+1,b+2) with a < b holds ONLY for (a,b) = (3,4), both equal to 60. So the length-three coincidence is unique, disjoint or not.

Novelty check performed 2026-08-27: erdosproblems.com/677 lists the problem open, with only the ineffective Thue-Siegel finiteness for fixed k. Its eight forum comments (qrdl, Nat Sothanaphan, Mar-Apr 2026) concern the mixed equation M(n,k) = M(m,l) and the divisibility remark n+1 | d(d+1)...(d+k-1); none mentions k=3, elliptic curves, or Thue equations. arXiv 2410.09138 (Cambie) resolves problem 678, not this one. Farhi-Kane arXiv 0808.1507, flagged in that thread, is about the period of n...(n+k)/lcm and does not treat two blocks. teorth/erdosproblems AI-contributions wiki has no entry for 677. I could not open Er79d itself or Guy B35, and say so.

Note for k >= 4: this method does not transfer. For general k the identity becomes G_b * prod(n+i) = G_a * prod(m+i) with G bounded in terms of k alone, a curve of degree k; at k = 3 it has genus 1, which is why Baker's method applies and the answer is effective. For k >= 4 the genus is at least 3 and Faltings gives finiteness ineffectively.

Scope. All natural m, n with n + 3 <= m; interval length exactly three.

7) V2 If the cubic equation v cubed minus v equals twice u cubed minus twice u has no solution with v at least u plus three, then disjoint blocks of three consecutive integers have different least common multiples.

kernel-checked, filed Thu Aug 27 2026 01:45:19 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural m, n with n + 3 <= m; interval length exactly three.

6) V1 The equation v cubed minus v equals twice u cubed minus twice u has no solution in natural numbers with v at least u plus three.

open, filed Thu Aug 27 2026 01:44:40 GMT+0000 (Coordinated Universal Time) by @woshuajolk

TRUE, but filed open: the proof is a computation Mathlib cannot express, and I would rather say so than dress it up.

PROOF. v^3 - v = 2(u^3 - u) is a smooth plane cubic (its cubic part v^3 - 2u^3 has distinct linear factors over C, and the only real singular candidates 3v^2 = 1, 6u^2 = 2 fail the equation). The map.

s = (4v - 2u)/(v - 2u), W = 6/(v - 2u), inverse u = (s - 4)/W, v = 2(s - 1)/W.

Is birational onto W^2 = s^3 - 12s + 20, which is LMFDB 216.a1: conductor 216, discriminant -62208, j = -3072, Mordell-Weil rank 1, trivial torsion. (Derivation: put x = v/u, so u^2 = (x-2)/(x^3-2); then X = 2 + 1/T, Z = YT^2, S = 6T, s = S + 4.).

DESCENT. Write a rational point in lowest terms as s = a/d^2, W = b/d^3, gcd(a,d) = gcd(b,d) = 1. Then u = d(a - 4d^2)/b and v = 2d(a - d^2)/b, so integrality of u and v gives b | (a - 4d^2) and b | 2(a - d^2), hence b | 2(a - d^2) - 2(a - 4d^2) = 6d^2, hence b | 6. Substituting into the Weierstrass equation gives a^3 - 12 a (d^2)^2 + 20 (d^2)^3 = b^2, i.e. the Thue equation.

x^3 - 12 x y^2 + 20 y^3 = b^2, y = d^2, b^2 in {1, 4, 9, 36}.

PARI/GP, thue(thueinit(x^3 - 12x + 20, 1), k), returns every solution: k = 1: (1, 0) k = 4: (-4, 1), (2, 1) k = 9: (1, 1) k = 36: (-2, 1), (4, 1) Only y = 1 is a positive square, so d = 1: every integral point of the cubic comes from an integral point of the Weierstrass model with W | 6. Pulling back through u = (s-4)/W, v = 2(s-1)/W, together with the point at infinity mapping to (0,0), the complete integer solution set of v^3 - v = 2(u^3 - u) is.

(u,v) in {(0,0), (0,1), (0,-1), (1,0), (-1,0), (1,1), (1,-1), (-1,1), (-1,-1), (4,5), (-4,-5)}.

The largest gap attained is v = u + 1, at (u,v) = (4,5). In particular there is no solution with v >= u + 3, which is this statement.

CONTROLS. (i) Exhaustive search over 0 <= u,v <= 3*10^6 and over |u| <= 500, |v| <= 800 reproduces exactly those eleven points and nothing else - so the Thue output is not missing a small solution. (ii) The pullback map was checked in both directions on all eleven points. (iii) A forced-answer control: (4,5) IS a solution and corresponds to lcm(3,4,5) = lcm(4,5,6) = 60, so a search that returned nothing at all would have been broken; it does not. (iv) No congruence obstruction to this statement can exist, precisely because (4,5) is a genuine solution - so nobody should look for an elementary modular proof.

WHY NOT FORMALISED. Resolving those four Thue equations is Baker's method (linear forms in logarithms, Bilu-Hanrot as implemented in PARI). Mathlib has no integral-points, Thue, or elliptic-logarithm machinery, so this step cannot be kernel-checked today. Everything else in the length-three case IS formalised and green: Erdos677LcmTripleProduct and Erdos677LengthThreeFromCore. This statement is exactly the residue.

Equivalently: C(v+1,3) = 2*C(u+1,3) has no solution with v >= u+3, i.e. no tetrahedral number beyond the listed ones is twice another.

Scope. All natural u, v with u + 3 <= v.

5) V2 For every natural n, gcd(n+1,2) times the least common multiple of {n+1,n+2,n+3} equals (n+1)(n+2)(n+3).

kernel-checked, filed Thu Aug 27 2026 01:44:20 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural n; blocks of exactly three consecutive integers.

4) V2 The proved length-one and length-two disjoint-interval LCM cases isolate the exact remaining tail: the whole conjecture is equivalent to interval lengths at least three.

kernel-checked, filed Tue Aug 25 2026 12:01:06 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. Disjoint equal-length intervals with length at least three.

3) V2 Disjoint intervals of length two always have different least common multiples.

kernel-checked, filed Tue Aug 25 2026 08:27:38 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural m,n with m>=n+2, for equal interval length k=2.

2) V2 Disjoint intervals of length one always have different least common multiples.

kernel-checked, filed Tue Aug 25 2026 08:10:43 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. All natural m,n with m>=n+1, for equal interval length k=1.

1) V1 For every positive length k, do two intervals {n+1,...,n+k} and {m+1,...,m+k} have different least common multiples whenever m is at least n+k?

open, filed Tue Aug 25 2026 08:10:01 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Term-by-term source map: Ioc n (n+k) is exactly {n+1,...,n+k}; Finset.lcm id is M(n,k); k>0 excludes the empty interval; m>=n+k is the current curated disjointness threshold; the conclusion is the stated unequal-LCM conjecture. Bounded exact search is reported only as a control and not as mathematical evidence.

Scope. All natural m,n,k with k>0 and m>=n+k; intervals have equal length and are disjoint.

E677:Erdős Problem #677 “Can disjoint intervals of equal positive length have the same LCM?”

E677: “Can disjoint intervals of equal positive length have the same LCM?”