Erdős Problem #373

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.

Erdős Problem #373: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

Are there only finitely many solutions of $n!=a_1!\cdots a_k!$ with $n-1>a_1\ge\cdots\ge a_k\ge2$?

All supplied versions considered: 373.tex. The source forces $a_1$ close to $n$ through a prime-gap theorem. I derive complementary restrictions on the remaining factorials using elementary size and valuation arguments; no stronger prime-gap result is assumed.

2) WORK

Put $h=n-a_1\ge2$. Cancelling $a_1!$ gives \[ \prod_{i=2}^k a_i!=(a_1+1)\cdots n\le n^h. \] Each factor on the left is at least two, so \[ k-1\le h\log_2 n. \tag{1} \] Also for $a_2\ge2$ the last $\lfloor a_2/2\rfloor$ factors in $a_2!$ are all at least $a_2/2$, giving \[ \lfloor a_2/2\rfloor\log(a_2/2)\le h\log n. \tag{2} \] For any fixed $h$, inequality $2$ implies $a_2=O_h(\log n/\log\log n)$: inserting $a_2\ge C h\log n/\log\log n$ makes its left side greater than $h\log n$ for a sufficiently large absolute $C$ and large $n$.

There is also an exact prime restriction. If $p$ is prime with $n/2<p\le a_1$, then $v_p(n!)=v_p(a_1!)=1$. Every other factorial must have $p$-valuation zero, so $a_2<p$. If $a_1<p\le n$ instead, the equation is impossible. Thus a hypothetical solution with a fixed short terminal block must express that block's product using factorials whose arguments are only logarithmic in $n$, despite the block entries being of order $n$.

3) ADVERSARIAL VERIFICATION

Repeated factorial arguments are permitted in the problem and in $1$. The bound for fixed $h$ is not uniform in a growing $h$ unless its dependence is retained; the source only gives a growing upper bound on $h$. Smooth products of large consecutive integers are not excluded merely by these inequalities.

4) FINAL

UNRESOLVED. No contradiction for all large $n$ is obtained, even when $h$ is fixed. The remaining issue is the arithmetic feasibility of the residual factorial product, not just the closeness of $a_1$ to $n$.

5) SOURCES CHECKED

No external theorem is used in the new argument. The supplied source was read in full; the saved Tao status snapshot is dated 2026-09-22.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 10%

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

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

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%.