Problems / erdos-1094 / erdos-1094--h3--h1--h1--h1--h2
Residue form, 6 | k, 5 ∤ k, no carry at 5
erdos-1094--h3--h1--h1--h1--h2
open
Hole h_k6rest of erdos-1094--h3--h1--h1--h1 (split by 20261015T145000Z-t1015c-a12-partial.lean).
$\displaystyle \exists K_2\ \forall n,k\in\mathbb{N}:\quad K_2<k,\ \ 2k\le n<k^2,\ \ 2\mid k,\ \ 3\mid k,\ \ 5\nmid k,\ \ k \bmod 5 \le n \bmod 5 \implies \exists\,p\le k \text{ prime with } n \bmod p < k \bmod p$
For all large $k$ divisible by $6$ but not by $5$, every $n$ with $2k\le n<k^2$ whose residue mod $5$ is at least that of $k$ (so $p=5$ itself gives no carry) has a prime $p\le k$ whose residue of $n$ is smaller than its residue of $k$. Lean writes $5\nmid k$ as 0 < k % 5 and $k^2$ as k * k. n % p is the least non-negative residue, so the condition reads $n \bmod p < k \bmod p$, a carry at the last base-$p$ digit when $k$ is subtracted from $n$ (hence $p \mid \binom{n}{k}$ by Kummer's theorem, though the statement names no binomial coefficient).
Lean
theorem erdos_1094__h3__h1__h1__h1__h2 : ∃ (K₂ : ℕ),
∀ (n k : ℕ),
K₂ < k →
(2 : ℕ) * k ≤ n →
n < k * k →
k % (2 : ℕ) = (0 : ℕ) →
k % (3 : ℕ) = (0 : ℕ) →
(0 : ℕ) < k % (5 : ℕ) → k % (5 : ℕ) ≤ n % (5 : ℕ) → ∃ (p : ℕ), Nat.Prime p ∧ p ≤ k ∧ n % p < k % p := by
sorryWhy does it end in sorry?
A statement is written := by sorry: the sorry holds the place of a proof, because a statement file only says what is claimed. A proof the gate checked is its own file, Proof.lean. How to read a proof page →
How it is being proved
Open. A direct proof closes it.
What is known
Work on it
How to contribute → · Pull requests in flight ↗ · See it in the problem’s graph →
Record
Statement: the Lean file and its words (1 version)
Statement not audited — only a problem’s root statement is audited for fidelity (D-9); the kernel checks a proof against this Lean as written.
import Mathlib
import Nodes.«erdos-1094--h3--h1--h1--h1--h2».Context
open scoped Nat
/-! Hole `h_k6rest` of a merged partial proof, as a node (D-12 #5, D-29).
Statement generated by the gate from the assembly; immutable (D-3). -/
theorem erdos_1094__h3__h1__h1__h1__h2 : ∃ (K₂ : ℕ),
∀ (n k : ℕ),
K₂ < k →
(2 : ℕ) * k ≤ n →
n < k * k →
k % (2 : ℕ) = (0 : ℕ) →
k % (3 : ℕ) = (0 : ℕ) →
(0 : ℕ) < k % (5 : ℕ) → k % (5 : ℕ) ≤ n % (5 : ℕ) → ∃ (p : ℕ), Nat.Prime p ∧ p ≤ k ∧ n % p < k % p := by
sorryA statement is written := by sorry: the sorry holds the place of a proof, because a statement file only says what is claimed. A proof the gate checked is its own file, Proof.lean. How to read a proof page →
In words, unverified — what this statement says; written by f29-words, drafted with Claude Fable 5.1 (claude-fable-5-1), 2026-10-18; no one has read it against the Lean and signed it.
Draft · drafted, by the model named above
$\displaystyle \exists K_2\ \forall n,k\in\mathbb{N}:\quad K_2<k,\ \ 2k\le n<k^2,\ \ 2\mid k,\ \ 3\mid k,\ \ 5\nmid k,\ \ k \bmod 5 \le n \bmod 5 \implies \exists\,p\le k \text{ prime with } n \bmod p < k \bmod p$
For all large $k$ divisible by $6$ but not by $5$, every $n$ with $2k\le n<k^2$ whose residue mod $5$ is at least that of $k$ (so $p=5$ itself gives no carry) has a prime $p\le k$ whose residue of $n$ is smaller than its residue of $k$. Lean writes $5\nmid k$ as 0 < k % 5 and $k^2$ as k * k. n % p is the least non-negative residue, so the condition reads $n \bmod p < k \bmod p$, a carry at the last base-$p$ digit when $k$ is subtracted from $n$ (hence $p \mid \binom{n}{k}$ by Kummer's theorem, though the statement names no binomial coefficient).
Rendered from gloss/a77434bc0d8b….md · Improve these words →
Rendered from targets/erdos-1094/nodes/erdos-1094--h3--h1--h1--h1--h2/Statement.lean. Origin: compiler-derived.
Witness
Not checked — no run on this page records step 7 passing for it.
1aa8a1c1b9f3.import Mathlib open scoped Nat theorem witness : True := trivial
No gloss yet: no one has written in words what this witness says (D-3 v3.30); the Lean above is the witness. How to write one →
How to read this page
No explainer yet. An explainer is a merged proof written as a journal would print it, so there is nothing to explain until this statement is proved (D-3).