gpt pro 5.4 attempt on Math-785

Erdős Problem 373

Problem statement.

Show that the equation \[ n! = a_{1}!a_{2}!\cdots a_{k}!, \qquad n-1>a_{1}\geq a_{2}\geq \cdots \geq a_{k}\geq 2, \] has only finitely many solutions.

Tools in this round.

Web browsing: YES. Computation: YES.

1) ROUND-2 OBJECTIVE

I am not pursuing path (A), because the full finiteness statement is still open. The Round-2 objective is strict advancement beyond Round 1 by turning the qualitative prime-gap obstruction into a quantitative theorem.

Round 1 only proved $a_{1}>n/2$. In this round I will prove the substantially sharper statement \[ a_{1}>n\left(1-\frac{\eta}{(\log n)^{2}}\right) \] for every fixed $\eta>1/25$ and all sufficiently large solutions. Equivalently, \[ n-a_{1}=O\!\left(\frac{n}{(\log n)^{2}}\right). \]

2) Round-1 FOUNDATION USED

I rely on the main structural lemma proved in Round 1.

If \[ n! = a_{1}!a_{2}!\cdots a_{k}!, \qquad n-1>a_{1}\geq a_{2}\geq \cdots \geq a_{k}\geq 2, \] then the interval $(a_{1},n]$ contains no prime numbers. Equivalently, every prime $p\leq n$ satisfies $p\leq a_{1}$.

Round 1 also extracted the easy corollary $a_{1}>n/2$ from Bertrand's postulate. I will now replace that by a much sharper asymptotic bound.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is Dusart's explicit short-interval prime theorem:

for every $x>396738$, there exists a prime $p$ with \[ x<p\leq x\left(1+\frac{1}{25(\log x)^{2}}\right). \]

Combining this with the Round-1 prime-free interval lemma shows that $(a_{1},n]$ cannot be much longer than a Dusart interval ending at $n$. This converts the qualitative prime-gap obstruction into a quantitative restriction on $a_{1}$.

4) ATTACK PLAN (ROUND-2)

The exact gap after Round 1 was clear: knowing merely that $(a_{1},n]$ is prime-free does not rule out infinitely many solutions, because large prime gaps do occur.

To advance beyond that, I will:

1. choose a point $x$ just below $n$;
2. prove that Dusart's interval starting from $x$ still ends below $n$;
3. conclude that there must be a prime in $(x,n]$;
4. invoke the Round-1 prime-free interval lemma to force $a_{1}>x$.

This yields the new theorem $a_{1}>n(1-\eta/(\log n)^{2})$.

5) WORK (ROUND-2)

Fix a real number $\eta>1/25$. I will prove that every sufficiently large solution satisfies \[ a_{1}>n\left(1-\frac{\eta}{(\log n)^{2}}\right). \]

Let \[ L=\log n, \qquad x=n\left(1-\frac{\eta}{L^{2}}\right). \] For all sufficiently large $n$ we have $x>396738$, so Dusart's theorem applies.

It remains to check that the corresponding Dusart interval stays below $n$.

Since $\eta/L^{2}\to 0$, the Taylor expansion $\log(1-z)=-z+O(z^{2})$ gives \[ \log x =\log n+\log\left(1-\frac{\eta}{L^{2}}\right) =L-\frac{\eta}{L^{2}}+O\!\left(\frac{1}{L^{4}}\right). \] Hence \[ \frac{1}{(\log x)^{2}} =\frac{1}{L^{2}}+O\!\left(\frac{1}{L^{5}}\right), \] because replacing $L$ by $L+O(L^{-2})$ changes its reciprocal square by $O(L^{-5})$.

Therefore \[ x\left(1+\frac{1}{25(\log x)^{2}}\right) =n\left(1-\frac{\eta}{L^{2}}\right) \left(1+\frac{1}{25L^{2}}+O\!\left(\frac{1}{L^{5}}\right)\right). \] Multiplying out, \[ x\left(1+\frac{1}{25(\log x)^{2}}\right) =n\left(1-\frac{\eta-1/25}{L^{2}}+O\!\left(\frac{1}{L^{4}}\right)\right). \] Since $\eta>1/25$, the coefficient of $1/L^{2}$ is negative. Thus for all sufficiently large $n$, \[ x\left(1+\frac{1}{25(\log x)^{2}}\right)<n. \]

Now apply Dusart. There exists a prime $p$ such that \[ x<p\leq x\left(1+\frac{1}{25(\log x)^{2}}\right)<n. \] So in fact \[ p\in (x,n]. \]

Suppose for contradiction that $a_{1}\leq x$. Then $p>x\geq a_{1}$, hence \[ p\in (a_{1},n], \] which contradicts the Round-1 lemma that $(a_{1},n]$ contains no prime.

Therefore $a_{1}>x$, i.e. \[ a_{1}>n\left(1-\frac{\eta}{(\log n)^{2}}\right) \] for all sufficiently large solutions.

This proves the claimed theorem.

Corollary.

Choosing, for example, $\eta=1/5$, one obtains the concrete bound \[ n-a_{1}<\frac{n}{5(\log n)^{2}} \] for every sufficiently large solution.

6) ADVERSARIAL VERIFICATION

The delicate points are as follows.

• Validity of the Taylor expansions. Here $z=\eta/L^{2}\to 0$, so both $\log(1-z)=-z+O(z^{2})$ and the reciprocal-square expansion are legitimate.
• Was the Dusart interval really forced below $n$? Yes. The main term is $1-(\eta-1/25)/L^{2}$, and because $\eta>1/25$ this is strictly less than $1$ for large $n$; the $O(L^{-4})$ term cannot overturn that sign eventually.
• Did I use the Round-1 lemma correctly? Yes. If $a_{1}\leq x$ then the prime $p$ found above satisfies $p>x\geq a_{1}$ and also $p\leq n$, so $p\in (a_{1},n]$, impossible.
• Does this solve finiteness? No. The new theorem only shows that $a_{1}$ is very close to $n$ on the scale $n/(\log n)^{2}$. That is much stronger than Round 1, but it still does not rule out infinitely many solutions.

No hidden conflict with the Round-1 results appears; the new theorem is a direct strengthening of the previous prime-gap argument.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The strict new advance is the theorem \[ \forall \eta>1/25,\qquad a_{1}>n\left(1-\frac{\eta}{(\log n)^{2}}\right) \] for every sufficiently large solution. In particular, \[ n-a_{1}=O\!\left(\frac{n}{(\log n)^{2}}\right). \] This is substantially stronger than the Round-1 bound $a_{1}>n/2$, and it cleanly identifies the remaining weakness of the prime-gap approach: even such a strong closeness estimate does not by itself force finiteness.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 58%

9) REFERENCES

• P. Dusart, Estimates of some functions over primes without R.H., arXiv:1002.0442, especially Proposition 6.8.
• T. F. Bloom, Erdős Problem #373, accessed 2026-03-07.

10) OUTPUT FORMAT

Erdős problem number: $x=373$. This section is written as a drop-in LaTeX fragment and is included in the combined file 369_371_373_round2_solutions.tex.