Reused writeup on Math-6162 (gpt pro 5.2)

Erdős Problem 1108 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1108

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1108.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1108.tex.

FORMAL RESTATEMENT

Let $\mathbb{N}=\{1,2,3,\dots\}$ and define \[A := \left\{ \sum_{n\in S} n! \ :\ S\subset \mathbb{N}\ \text{finite}\right\} \subseteq \mathbb{Z}_{\ge 0},\] where the empty sum is $0$.

Fix an integer $k\ge 2$.

• Question 1: Does $A$ contain only finitely many $k$th powers, i.e. integers of the form $m^k$ with $m\in\mathbb{Z}_{\ge 0}$?
• Question 2: Does $A$ contain only finitely many powerful numbers? Here I use the standard convention that a (positive) integer $n$ is powerful (squarefull) if for every prime $p\mid n$ we have $p^2\mid n$.

QUICK LITERATURE/CONTEXT CHECK

The problem text notes that it is open even whether there are infinitely many squares of the form $1+n!$. It also cites a result of Brindza and Erdős: for any fixed $r$, if $n_1!+\cdots+n_r!$ is powerful then the smallest factorial index $n_1$ is bounded in terms of $r$. I do not use any results not explicitly stated in the problem file.

ATTACK PLAN

Proof-track strategies.

• Use $p$-adic valuation: compare $v_p\left(\sum_{n\in S} n!\right)$ with constraints forced by being a $k$th power or powerful.
• Work modulo $(m+1)!$: because all terms $n!$ with $n\ge m+1$ vanish mod $(m+1)!$, the residue of an element of $A$ mod $(m+1)!$ is controlled by finitely many "digits" $\in\{0,1\}$.
• Attempt to show that if $\sum_{n\in S} n!$ is a large $k$th power then there must be a "carry" phenomenon in factorial base that is incompatible with having only $0/1$ digits.

Disproof-track strategies.

• Search computationally for large squares/cubes/powerful values in $A$ and attempt to detect a pattern that could generate infinitely many.
• Try to engineer identities of the shape $(x\pm y)^2 = \sum_{n\in S} n!$ by forcing the right-hand side to approximate a square with controlled error using factorial growth.

WORK

FAST REALITY CHECK (computation).

I enumerated all subset sums of $\{1!,2!,\dots,m!\}$ for $m\le 12$ (so $|A\cap[0,(13)! )|=2^{12}=4096$ elements) and checked for perfect powers and powerful (squarefull) numbers.

Up to $\sum_{n=1}^{12} n! = 522956313 < 13!$, the set $A$ contains exactly $15$ squares: \[ 0,1,9,25,121,144,729,841,5041,5184,45369,46225,363609,403225,3674889. \] Their $0/1$ factorial representations (as a set of indices $S$) are:

• $9=1!+2!+3!$;
• $25=1!+4!$;
• $121=1!+5!$;
• $144=4!+5!$;
• $729=1!+2!+3!+6!$;
• $841=1!+5!+6!$;
• $5041=1!+7!$;
• $5184=4!+5!+7!$;
• $45369=1!+2!+3!+7!+8!$;
• $46225=1!+4!+5!+6!+7!+8!$;
• $363609=1!+2!+3!+6!+9!$;
• $403225=1!+4!+8!+9!$;
• $3674889=1!+2!+3!+6!+7!+8!+10!$.

For cubes, up to the same bound $A$ contains exactly $5$ cubes: \[0,1,8,27,729\] with $8=2!+3!$, $27=1!+2!+4!$, and $729=1!+2!+3!+6!$. For fifth powers up to the same bound, $A$ contains $0,1,32$ with $32=2!+3!+4!$.

These computations do not settle finiteness; they only show that there are already many small squares/powers in $A$.

Lemma 1108.1 (a basic factorial-sum inequality).

For every integer $m\ge 2$, \[\sum_{j=1}^{m-1} j! < m!.\]

Proof.

For $1\le j\le m-1$ we have \[(m-1)! = j!\cdot (j+1)(j+2)\cdots(m-1).\] If $j\le m-2$ then each factor in $(j+1)(j+2)\cdots(m-1)$ is at least $2$, and there are $(m-1)-(j+1)+1 = m-j-1$ such factors. Therefore \[j!\le \frac{(m-1)!}{2^{m-j-1}}\qquad (1\le j\le m-2),\] and trivially $(m-1)!\le (m-1)!$ for $j=m-1$. Summing gives \[\sum_{j=1}^{m-1} j! \le (m-1)!\left(1+\sum_{t=1}^{m-2} 2^{-t}\right) < (m-1)!\left(1+\sum_{t=1}^{\infty} 2^{-t}\right) = 2(m-1)!.\] If $m=2$ then $2(m-1)!=2=2!$ and the left-hand side equals $1! =1$, so the strict inequality holds. If $m\ge 3$ then $2(m-1)!\le m(m-1)! = m!$, and since the inequality above is strict we get $\sum_{j=1}^{m-1} j! < m!$. $\square$

Lemma 1108.2 (injectivity of $0/1$ factorial sums).

If $S,T\subset\mathbb{N}$ are finite and \[\sum_{n\in S} n! = \sum_{n\in T} n!,\] then $S=T$. Equivalently, the map $S\mapsto \sum_{n\in S} n!$ from finite subsets of $\mathbb{N}$ to $\mathbb{Z}_{\ge 0}$ is injective.

Proof.

Assume for contradiction that $S\ne T$. Let \[m:=\max(S\triangle T),\] where $S\triangle T$ is the symmetric difference. Then exactly one of $S,T$ contains $m$. Without loss of generality assume $m\in S\setminus T$. Rewrite the difference of the sums as \[0=\sum_{n\in S} n! - \sum_{n\in T} n! = m! + \sum_{n\in S\cap\{1,\dots,m-1\}} n! - \sum_{n\in T\cap\{1,\dots,m-1\}} n!.\] Thus \[m! = \sum_{n\in T\cap\{1,\dots,m-1\}} n! - \sum_{n\in S\cap\{1,\dots,m-1\}} n!. \] Taking absolute values and using the triangle inequality gives \[m! \le \sum_{n\in T\cap\{1,\dots,m-1\}} n! + \sum_{n\in S\cap\{1,\dots,m-1\}} n! \le 2\sum_{j=1}^{m-1} j!. \] By Lemma 1108.1, $\sum_{j=1}^{m-1} j! < m!$, so the right-hand side is strictly less than $2m!$. However, we can argue more sharply: each of the two sums on the right is at most $\sum_{j=1}^{m-1} j! < m!$, hence their difference has absolute value strictly less than $m!$. But the equation above asserts that this difference equals $m!$, a contradiction. Therefore $S=T$. $\square$

Lemma 1108.3 (counting $A$ below $(m+1)!$; density zero).

For every integer $m\ge 1$, \[\left|A\cap\{0,1,\dots,(m+1)!-1\}\right| = 2^m. \] In particular, $A$ has natural density $0$ in $\mathbb{Z}_{\ge 0}$.

Proof.

Any subset $S\subseteq\{1,2,\dots,m\}$ produces an element $\sum_{n\in S} n!\in A$. By Lemma 1108.2 these $2^m$ sums are all distinct. Also, each such sum is $< (m+1)!$ because it is at most $\sum_{n=1}^{m} n!$ and $\sum_{n=1}^{m} n! < (m+1)!$ follows from Lemma 1108.1 applied to $m+1$. So we get at least $2^m$ elements of $A$ in $[0,(m+1)!-1]$.

Conversely, if $S$ contains an index $\ge m+1$ then $\sum_{n\in S} n!\ge (m+1)!$. Therefore every element of $A$ lying in $[0,(m+1)!-1]$ must come from a subset of $\{1,\dots,m\}$, giving at most $2^m$ possibilities. Hence the count is exactly $2^m$.

For density: take $X=(m+1)!-1$. Then \[\frac{|A\cap[0,X]|}{X+1} = \frac{2^m}{(m+1)!} \to 0\qquad (m\to\infty),\] so the natural density is $0$. $\square$

VERIFICATION

• The definition of $A$ includes $0$ (empty subset). This contributes a single trivial $k$th power ($0^k$) and does not affect finiteness questions.
• Lemma 1108.2 uses Lemma 1108.1 via the inequality $\sum_{j=1}^{m-1} j! < m!$; Lemma 1108.1 was proved with an explicit geometric-series bound.
• The computation enumerated all $2^m$ subset sums for $m\le 12$, so within the range $[0,13!)$ the reported lists are exhaustive.

FINAL

UNRESOLVED

(i) Strongest proved partial result. Lemmas 1108.1--1108.3 give a complete injective "$0/1$ factorial-base" description of $A$ and show $|A\cap[0,(m+1)! )|=2^m$, hence $A$ has density $0$.

(ii) First gap. Prove (or refute) that for each fixed $k\ge 2$, only finitely many elements of $A$ are perfect $k$th powers; and similarly, only finitely many elements of $A$ are powerful (squarefull) numbers.

(iii) Top 3 next moves. 1. Develop a $p$-adic obstruction: for a large element $a=\sum_{n\in S}n!$ with $m=\max S$, analyze $v_p(a)$ for primes $p\le m$ and compare with the constraints $v_p(a)\equiv 0\pmod{k}$ for $k$th powers. 2. Work modulo $(m+1)!$ and smaller factorial moduli to restrict possible residues of $k$th powers within the $0/1$ digit set; attempt to show that for large $m$ no residue class can match. 3. Extend the brute-force search to larger $m$ (e.g. $m\le 16$ gives $65536$ elements) to look for new squares/cubes/powerful numbers beyond those found for $m\le 12$, and record their factorial-digit patterns.

(iv) What a minimal counterexample would likely look like. If there are infinitely many $k$th powers in $A$, then for arbitrarily large $m$ there exists a subset $S\subseteq\{1,\dots,m\}$ with $m\in S$ such that $\sum_{n\in S} n!$ lands exactly on a $k$th power. By Lemma 1108.2 this forces a highly rigid pattern of $0/1$ digits in factorial base; a "minimal" large counterexample would likely have $m\in S$ and then a carefully tuned choice of lower factorials to match the gap between consecutive $k$th powers near $m!$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.