Problems / erdos-1094 / erdos-1094--h3--h1--h2--h1--h1
Residue form, k and n odd, 15 | k
erdos-1094--h3--h1--h2--h1--h1
open
Hole h_k15 of erdos-1094--h3--h1--h2--h1 (split by 20261015T145000Z-t1015c-a10-partial.lean).
$\displaystyle \exists K_1\ \forall n,k\in\mathbb{N}:\quad K_1<k,\ \ 2k\le n<k^2,\ \ k \text{ odd},\ \ n \text{ odd},\ \ 3\mid k,\ \ 5\mid k \implies \exists\,p\le k \text{ prime with } n \bmod p < k \bmod p$
For all large odd $k$ divisible by $3$ and by $5$, every odd $n$ with $2k\le n<k^2$ has a prime $p\le k$ whose residue of $n$ is smaller than its residue of $k$. Lean writes oddness as k % 2 = 1 and n % 2 = 1, the divisibilities as k % 3 = 0 and k % 5 = 0, 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__h2__h1__h1 : ∃ (K₁ : ℕ),
∀ (n k : ℕ),
K₁ < k →
(2 : ℕ) * k ≤ n →
n < k * k →
k % (2 : ℕ) = (1 : ℕ) →
n % (2 : ℕ) = (1 : ℕ) →
k % (3 : ℕ) = (0 : ℕ) → k % (5 : ℕ) = (0 : 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--h2--h1--h1».Context
open scoped Nat
/-! Hole `h_k15` 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__h2__h1__h1 : ∃ (K₁ : ℕ),
∀ (n k : ℕ),
K₁ < k →
(2 : ℕ) * k ≤ n →
n < k * k →
k % (2 : ℕ) = (1 : ℕ) →
n % (2 : ℕ) = (1 : ℕ) →
k % (3 : ℕ) = (0 : ℕ) → k % (5 : ℕ) = (0 : 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
$\displaystyle \exists K_1\ \forall n,k\in\mathbb{N}:\quad K_1<k,\ \ 2k\le n<k^2,\ \ k \text{ odd},\ \ n \text{ odd},\ \ 3\mid k,\ \ 5\mid k \implies \exists\,p\le k \text{ prime with } n \bmod p < k \bmod p$
For all large odd $k$ divisible by $3$ and by $5$, every odd $n$ with $2k\le n<k^2$ has a prime $p\le k$ whose residue of $n$ is smaller than its residue of $k$. Lean writes oddness as k % 2 = 1 and n % 2 = 1, the divisibilities as k % 3 = 0 and k % 5 = 0, 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/cece1162c34a….md · Improve these words →
Rendered from targets/erdos-1094/nodes/erdos-1094--h3--h1--h2--h1--h1/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).