Erdős Problem 699
TheoremErdős Problem 699#
Reference: erdosproblems.com/699
Quoted from FormalConjectures/ErdosProblems/699.lean in Formal Conjectures (Google DeepMind) at commit 2055e5ad, Apache-2.0, as are the docstrings and Lean statements below.
The problem's page on erdosproblems.com: https://www.erdosproblems.com/699.
The file states 3 research or textbook statements: 2 open and 1 solved.
Erdos699.sylvester_schur
Solved: stated here; a Lean proof is at https://github.com/AllenGrahamHart/FormalConjectures-Bench/blob/482dacc4d9335240f26218cdc62032da3100392b/formalizations/erdos699/Erdos699Formalization.lean#L7679.
Sylvester and Schur: for \(1 \le i \le n/2\) there is a prime \(p > i\) dividing
n.choose i.
theorem sylvester_schur (n i : ℕ) (hi : 1 ≤ i) (hi_half : i ≤ n / 2) :
∃ p : ℕ, p.Prime ∧ i < p ∧ p ∣ Nat.choose n i
Category research solved; AMS 11 Number theory. Declared in FormalConjectures/ErdosProblems/699.lean, line 33.
Erdos699.erdos_699
Open: stated without proof.
Erdős Problem 699. Is it true that for every \(1 \le i < j \le n / 2\) there exists a prime \(p \ge i\) with \(p \mid \gcd\big(\binom{n}{i}, \binom{n}{j}\big)\)?
theorem erdos_699 :
∀ n i j : ℕ,
1 ≤ i →
i < j →
j ≤ n / 2 →
∃ p : ℕ, p.Prime ∧ i ≤ p ∧ p ∣ Nat.gcd (Nat.choose n i) (Nat.choose n j)
Category research open; AMS 11 Number theory. Declared in FormalConjectures/ErdosProblems/699.lean, line 42.
Erdos699.erdos_szekeres_strengthening
Open: stated without proof.
Erdős and Szekeres conjectured that, apart from a finite exceptional set of triples
(n, i, j), one can always takep > iin the prime divisor statement.
theorem erdos_szekeres_strengthening :
∃ E : Finset (ℕ × ℕ × ℕ), ∀ n i j : ℕ,
1 ≤ i →
i < j →
j ≤ n / 2 →
(n, i, j) ∉ E →
∃ p : ℕ, p.Prime ∧ i < p ∧ p ∣ Nat.gcd (Nat.choose n i) (Nat.choose n j)
Category research open; AMS 11 Number theory. Declared in FormalConjectures/ErdosProblems/699.lean, line 53.
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