gpt pro 5.4 attempt on Math-5618

Erdős Problem #387

1) ROUND-2 OBJECTIVE

I pursue (C) obstruction/partial reduction. Round-1 produced only the weak divisor $n/\gcd(n,k)$ in $[n/k,n]$. The most promising direction now is to show that any genuine obstruction to a uniform linear divisor cannot come from the interior of Pascal's triangle. In this round I prove exactly that: on every fixed interior strip \[ \alpha n\leq k\leq (1-\alpha)n, \] the conjectured conclusion already holds with $c=\alpha$. Consequently, any counterexample sequence must live in the boundary regime \[ \min(k,n-k)=o(n), \] and, when combined with the Round-1 divisor $n/\gcd(n,k)$, must also satisfy \[ \gcd(n,k)\to\infty. \]

2) Round-1 FOUNDATION USED

I use the following Round-1 result.

1. Round-1 Lemma 1: $n/\gcd(n,k)$ divides $\binom{n}{k}$, hence $\binom{n}{k}$ has a divisor in $[n/k,n]$.

I also use the symmetry $\binom{n}{k}=\binom{n}{n-k}$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is the classical Sylvester--Schur theorem on prime divisors of products of consecutive integers. Applied to the numerator block \[ n(n-1)\cdots(n-r+1),\qquad r:=\min(k,n-k), \] it yields a prime divisor strictly larger than $r$, and this prime cannot be cancelled by the denominator $r!$. This produces a divisor of $\binom{n}{k}$ that is already linear in $n$ throughout every fixed interior strip.

The real gain over Round-1 is conceptual: it proves that the open difficulty is not spread across all $k$, but is confined to a thin boundary layer.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was the lack of any divisor of size $cn$ when $k$ is large.

The new plan is:

1. use Sylvester--Schur to show that $\binom{n}{k}$ always has a prime divisor exceeding $r=\min(k,n-k)$;
2. deduce that every fixed interior strip $\alpha n\leq k\leq (1-\alpha)n$ already satisfies the conjectured conclusion with $c=\alpha$;
3. combine this with the Round-1 divisor $n/\gcd(n,k)$ to isolate the only possible obstruction class.

This directly answers the question "where could a counterexample sequence hide?"

5) WORK (ROUND-2)

Define \[ M(n,k):=\max\{d\leq n: d\mid \binom{n}{k}\}. \] The Erdős problem asks whether there is a constant $c>0$ such that \[ M(n,k)>cn \] for all $1\leq k\leq n-1$.

By symmetry we may assume $1\leq k\leq n/2$. Set $r:=k$ in this range; more invariantly, $r=\min(k,n-k)$.

Lemma 387.1. For every $n\geq 2$ and every $1\leq k\leq n-1$, the binomial coefficient $\binom{n}{k}$ has a prime divisor $p$ with \[ p>\min(k,n-k). \] In particular, \[ M(n,k)>\min(k,n-k). \]

Proof. Let $r:=\min(k,n-k)$ and write \[ \binom{n}{k}=\binom{n}{r}=\frac{n(n-1)\cdots(n-r+1)}{r!}. \] The numerator is a product of $r$ consecutive integers, all strictly larger than $r$. By the Sylvester--Schur theorem, such a product has a prime divisor $p>r$. Since every prime divisor of $r!$ is at most $r$, this prime $p$ cannot be cancelled by the denominator. Hence $p\mid \binom{n}{r}=\binom{n}{k}$ and $p>r=\min(k,n-k)$. Because $p\leq n$, it contributes to $M(n,k)$, proving \[ M(n,k)\geq p>\min(k,n-k). \] This proves the lemma. $\square$

Theorem 387.2. Fix any real $\alpha$ with $0<\alpha\leq 1/2$. Then for every $n\geq 2$ and every integer $k$ with \[ \alpha n\leq k\leq (1-\alpha)n, \] one has \[ M(n,k)>\alpha n. \] Equivalently, the Erdős--387 conclusion is already true on every fixed interior strip, with the explicit constant $c=\alpha$.

Proof. Let $r=\min(k,n-k)$. The strip condition implies $r\geq \alpha n$. By Lemma 387.1, \[ M(n,k)>r\geq \alpha n. \] This is exactly the required conclusion. $\square$

Corollary 387.3. If the original Erdős problem has a negative answer, then any counterexample sequence $(n_j,k_j)$ must satisfy \[ \min(k_j,n_j-k_j)=o(n_j). \]

Proof. If not, then some $\alpha>0$ and an infinite subsequence would satisfy \[ \alpha n_j\leq k_j\leq (1-\alpha)n_j \] after symmetry. Theorem 387.2 would then give \[ M(n_j,k_j)>\alpha n_j \] on that subsequence, contradicting the assumption that no fixed linear bound works. $\square$

Corollary 387.4. If there exists a sequence $(n_j,k_j)$ with \[ \frac{M(n_j,k_j)}{n_j}\to 0, \] then necessarily \[ \min(k_j,n_j-k_j)=o(n_j) \qquad\text{and}\qquad \gcd(n_j,k_j)\to\infty. \]

Proof. The first assertion is exactly Corollary 387.3. For the second, Round-1 Lemma 1 gives the divisor \[ \frac{n_j}{\gcd(n_j,k_j)}\mid \binom{n_j}{k_j}. \] Since this divisor is at most $M(n_j,k_j)$, we get \[ \frac{1}{\gcd(n_j,k_j)} \leq \frac{M(n_j,k_j)}{n_j} \to 0. \] Hence $\gcd(n_j,k_j)\to\infty$. $\square$

Corollary 387.5. For each fixed integer $K\geq 1$, the Erdős--387 conclusion is true uniformly on the boundary strip \[ 1\leq k\leq K \qquad\text{or}\qquad n-K\leq k\leq n-1, \] with any constant $c<1/K$ (for instance $c=1/(K+1)$).

Proof. If $1\leq k\leq K$, then Round-1 gives a divisor at least $n/k\geq n/K$, and therefore strictly larger than $cn$ for every fixed $c<1/K$. By symmetry the same holds for $n-k\leq K$. $\square$

Thus the only unresolved regime is a moving boundary layer in which \[ \min(k,n-k)\to\infty, \qquad \min(k,n-k)=o(n), \qquad \gcd(n,k)\to\infty. \]

6) ADVERSARIAL VERIFICATION

The potentially delicate steps are these.

1. Does the Sylvester--Schur input apply at the endpoint $n=2r$? Yes. The numerator block is then $r+1,r+2,\dots,2r$, a product of $r$ consecutive integers all greater than $r$.
2. Could the prime $p>r$ from Sylvester--Schur be cancelled by the denominator? No. Every prime divisor of $r!$ is at most $r$, so cancellation is impossible.
3. Did I accidentally prove the full conjecture? No. The divisor $p>r$ is linear in $n$ only when $r$ itself is a positive proportion of $n$. When $r=o(n)$ this gives only $o(n)$, so the main open problem survives exactly in that edge regime.
4. Is Corollary 387.4 logically stronger than what was proved? No. It is just the conjunction of the new interior-strip theorem and the Round-1 divisor $n/\gcd(n,k)$.

The reduction is therefore logically sound.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advance beyond Round-1 is the structural reduction:

1. every fixed interior strip $\alpha n\leq k\leq (1-\alpha)n$ already satisfies the conjectured conclusion with $c=\alpha$;
2. hence any hypothetical counterexample sequence must satisfy \[ \min(k,n-k)=o(n); \]
3. combining with the Round-1 divisor $n/\gcd(n,k)$, any such sequence must also satisfy \[ \gcd(n,k)\to\infty. \]

So the unresolved part of Erdős #387 is now sharply localized to a thin moving boundary layer with large common divisors.

8) COMPLETION ESTIMATE

COMPLETION: 54%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #387, erdosproblems.com/387, accessed 2026-03-07.
2. P. Erdős, A theorem of Sylvester and Schur, J. London Math. Soc. 9 $1934$, 282--288.