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import FormalConjecturesUtilErdős Problem 698
References:
[ErSz78] Erdős, P. and Szekeres, G., Some number theoretic problems on binomial coefficients. Austral. Math. Soc. Gaz. (1978), 97-99.
[Be11] Bergman, George M., On common divisors of multinomial coefficients. Bull. Aust. Math. Soc. (2011), 138--157.
namespace Erdos698open FilterIs there some $h(n)\to \infty$ such that for all $2\leq i<j\leq n/2$ $$\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \geq h(n)?$$
This was resolved by Bergman [Be11], who proved that for any $2\leq i<j\leq n/2$ $$\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \gg n^{1/2}\frac{2^i}{i^{3/2}},$$ where the implied constant is absolute.
@[category research solved, AMS 5 11]
theorem erdos_698 : answer(True) ↔
∃ h : ℕ → ℕ, Tendsto h atTop atTop ∧
∀ n i j : ℕ, 2 ≤ i → i < j → j ≤ n / 2 →
h n ≤ Nat.gcd (n.choose i) (n.choose j) := ⊢ True ↔ ∃ h, Tendsto h atTop atTop ∧ ∀ (n i j : ℕ), 2 ≤ i → i < j → j ≤ n / 2 → h n ≤ (n.choose i).gcd (n.choose j)
All goals completed! 🐙A problem of Erdős and Szekeres, who observed that $$\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \geq \frac{\binom{n}{i}}{\binom{j}{i}} \geq 2^i$$ (in particular the greatest common divisor is always $>1$).
@[category research solved, AMS 5 11]
theorem erdos_698.variants.erdos_szekeres (n i j : ℕ) (hi : 1 ≤ i) (hij : i < j)
(hj : j ≤ n / 2) :
(n.choose i : ℝ) / (j.choose i : ℝ) ≤ (Nat.gcd (n.choose i) (n.choose j) : ℝ) ∧
(2 : ℝ) ^ i ≤ (n.choose i : ℝ) / (j.choose i : ℝ) := n:ℕi:ℕj:ℕhi:1 ≤ ihij:i < jhj:j ≤ n / 2⊢ ↑(n.choose i) / ↑(j.choose i) ≤ ↑((n.choose i).gcd (n.choose j)) ∧ 2 ^ i ≤ ↑(n.choose i) / ↑(j.choose i)
All goals completed! 🐙This inequality is sharp for $i=1$, $j=p$, and $n=2p$.
@[category research solved, AMS 5 11]
theorem erdos_698.variants.erdos_szekeres_sharp (p : ℕ) (hp : p.Prime) (hp2 : 2 < p) :
(Nat.gcd ((2 * p).choose 1) ((2 * p).choose p) : ℝ) =
((2 * p).choose 1 : ℝ) / (p.choose 1 : ℝ) ∧
((2 * p).choose 1 : ℝ) / (p.choose 1 : ℝ) = (2 : ℝ) ^ 1 := p:ℕhp:Nat.Prime php2:2 < p⊢ ↑(((2 * p).choose 1).gcd ((2 * p).choose p)) = ↑((2 * p).choose 1) / ↑(p.choose 1) ∧
↑((2 * p).choose 1) / ↑(p.choose 1) = 2 ^ 1
All goals completed! 🐙This was resolved by Bergman [Be11], who proved that for any $2\leq i<j\leq n/2$ $$\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \gg n^{1/2}\frac{2^i}{i^{3/2}},$$ where the implied constant is absolute.
@[category research solved, AMS 5 11, formal_proof using lean4 at "https://github.com/plby/lean-proofs/blob/main/src/v4.29.1/ErdosProblems/Erdos698.lean"]
theorem erdos_698.variants.bergman :
∃ c : ℝ, 0 < c ∧ ∀ n i j : ℕ, 2 ≤ i → i < j → j ≤ n / 2 →
c * (Real.sqrt (n : ℝ) * (2 : ℝ) ^ i / ((i : ℝ) * Real.sqrt (i : ℝ))) ≤
(Nat.gcd (n.choose i) (n.choose j) : ℝ) := ⊢ ∃ c,
0 < c ∧ ∀ (n i j : ℕ), 2 ≤ i → i < j → j ≤ n / 2 → c * (√↑n * 2 ^ i / (↑i * √↑i)) ≤ ↑((n.choose i).gcd (n.choose j))
All goals completed! 🐙end Erdos698