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Math-785-A2 · revision r1 · on Math-785

gpt pro 5.2 attempt on Math-785

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem 373
  2. 1) FORMAL RESTATEMENT
  3. 2) KEY DEFINITIONS AND KNOWN FACTS
  4. Quick literature/context check (browsing available).
  5. 3) ATTEMPTED SOLUTION
  6. 4) DETAILED WORK (with complete reasoning)
  7. 4.1 A proved structural constraint: $(a_1,n]$ is prime-free
  8. 4.2 A fully proved numerical consequence: $a_1>n/2$ for $n\ge 4$
  9. 4.3 Why this does not yet imply finiteness
  10. 5) VERIFICATION
  11. 6) FINAL
  12. 7) COMPLETION ESTIMATE (MANDATORY)

Problem 373

1) FORMAL RESTATEMENT

For integers $n\ge 1$ and $k\ge 2$, consider solutions in integers \[ n-1>a_1\ge a_2\ge \cdots \ge a_k\ge 2 \] of the factorial Diophantine equation \[ n! = a_1!\,a_2!\cdots a_k!. \] The problem asks to prove that there are only finitely many such solutions $(n; a_1,\dots,a_k)$.

2) KEY DEFINITIONS AND KNOWN FACTS

  • Factorial: $n!=1\cdot2\cdots n$.
  • For any prime $p$, if $p\le n$ then $p\mid n!$.
  • If $a<n$ and $p$ is a prime with $a<p\le n$, then $p$ divides $n!$ but does not divide $a!$.
Quick literature/context check (browsing available).
The problem is listed as open as of late 2025. It is conjectured that the only solutions are the "Hickerson list" $9!=2!\,3!\,3!\,7!$, $10!=6!\,7!$, $10!=3!\,5!\,7!$, $16!=14!\,5!\,2!$. Conditional on the ABC conjecture, Luca proved there are only finitely many solutions. Unconditionally, Luca proved that the set of $n\le x$ admitting a nontrivial solution is very sparse (subexponential in $x$ in an explicit sense). In the case $k=2$ (and in fact for general $k$), there are strong bounds forcing $a_1$ to be extremely close to $n$ (e.g. $a_1\ge n - C\log\log n$ with an explicit constant $C$), and numerical work has ruled out additional $k=2$ solutions up to astronomically large $n$.

3) ATTEMPTED SOLUTION

PHASE 1 (computational reality check). A brute-force search (by recursive factor selection with pruning) for nontrivial solutions with $n\le 30$ found exactly the classical solutions \[ \begin{aligned} 9! &= 7!\,3!\,3!\,2!,\\ 10! &= 7!\,6!,\\ 10! &= 7!\,5!\,3!,\\ 16! &= 14!\,5!\,2!, \end{aligned} \] and no others in that range.

PHASE 2 (proof attempt). I attempted to prove finiteness by forcing $a_1$ to be extremely close to $n$ via prime divisibility considerations. One can show that the interval $(a_1,n]$ contains no primes, hence $a_1$ must be at least the largest prime $\le n$. This gives a nontrivial constraint (e.g. $a_1>n/2$ by Bertrand), but it does not by itself rule out infinitely many $n$ because large prime gaps exist.

4) DETAILED WORK (with complete reasoning)

4.1 A proved structural constraint: $(a_1,n]$ is prime-free

Lemma: Assume $n! = a_1!\cdots a_k!$ with $n-1>a_1\ge a_2\ge\cdots\ge a_k\ge2$. Then there is no prime $p$ with $a_1<p\le n$.
Proof: Suppose for contradiction that there exists a prime $p$ with $a_1<p\le n$.

Because $p\le n$, we have $p\mid n!$.

On the other hand, for each $i\in\{1,\dots,k\}$ we have $a_i\le a_1<p$, so $p$ does not divide $a_i!$ (since all factors of $a_i!$ are at most $a_i<p$). Hence $p\nmid a_i!$ for all $i$, so $p\nmid a_1!\cdots a_k!$.

This contradicts $n! = a_1!\cdots a_k!$, because the left-hand side is divisible by $p$ while the right-hand side is not. ∎

Corollary: Let $p^*(n)$ denote the largest prime $\le n$ (for $n\ge2$). Any solution must satisfy $a_1\ge p^*(n)$.
Proof: If $a_1<p^*(n)$, then the prime $p^*(n)$ lies in $(a_1,n]$, contradicting Lemma Lemma. ∎

4.2 A fully proved numerical consequence: $a_1>n/2$ for $n\ge 4$

Corollary: For any solution with $n\ge 4$, one has $a_1>\frac{n}{2}$.
Proof: By Bertrand's postulate, for every integer $m\ge 2$ there exists a prime $p$ with $m<p<2m$. Apply this with $m=\lfloor n/2\rfloor\ge 2$ (since $n\ge4$). Then there exists a prime $p$ with \[ \lfloor n/2\rfloor < p < 2\lfloor n/2\rfloor \le n. \] In particular $p\le n$ and $p>n/2$. Hence $p^*(n)\ge p>n/2$. By Corollary Corollary, $a_1\ge p^*(n)>n/2$. ∎

4.3 Why this does not yet imply finiteness

Lemma Lemma shows $n-a_1$ is at most the length of the prime gap immediately preceding $n$. Prime gaps are unbounded, so this constraint alone does not rule out infinitely many solutions.

To finish a proof of finiteness one seemingly needs an additional ingredient connecting the factorial-product structure to large prime factors of products of short intervals (as suggested by known reductions to bounds on $P(n(n\pm 1))$), or to show that the prime-free interval constraint forces $n-a_1$ into a regime incompatible with balancing the remaining factorial factors.

5) VERIFICATION

  • Lemma Lemma uses only basic divisibility and is correct: a prime $p$ in $(a_1,n]$ divides $n!$ but cannot divide any $a_i!$ because all $a_i<p$.
  • Corollary Corollary follows from Bertrand's postulate; the inequality chain is checked explicitly.
  • The computational solutions listed in Phase 1 satisfy $n-1>a_1$ and $a_k\ge2$.

6) FINAL

UNRESOLVED.

(i) Strongest fully proved partial result obtained here. Lemma Lemma and its corollaries: any solution forces the interval $(a_1,n]$ to contain no primes, hence $a_1\ge p^*(n)$ and in particular $a_1>n/2$ for $n\ge4$.

(ii) Weakest specific gap preventing a full proof. Even though $(a_1,n]$ must be prime-free, prime gaps can be large; I do not have an argument that combines this with the remaining factorial-product constraints to preclude infinitely many solutions.

(iii) What lemmas would be needed to complete the proof. One plausible route would be to prove that for all sufficiently large $n$, the product $\prod_{j=a_1+1}^{n} j$ necessarily has a prime factor exceeding $a_1$ unless $n-a_1$ is extremely small in a way incompatible with the factorial decomposition. Any theorem of the form \[ \max_{a<j\le n} P(j) > a\quad\text{whenever } n-a \text{ lies in some explicit range} \] would begin to convert Lemma Lemma into finiteness.

(iv) Minimal explicit computation supporting finiteness. The search up to $n\le 30$ finds only the four classical solutions listed in PHASE 1.

7) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%.