Reused writeup on Math-6167 (gpt pro 5.2)

Erdős Problem 1113 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1113

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_5.2/1113.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1113.tex.

1) FORMAL RESTATEMENT

A positive odd integer $m$ is a Sierpiński number if \[ \forall k\in\mathbb{Z}_{\ge 0},\quad 2^k m + 1\ \text{is composite}. \]

A set of primes $P$ is a covering set for $m$ if \[ \forall k\in\mathbb{Z}_{\ge 0},\ \exists p\in P\ \text{such that}\ p\mid (2^k m+1). \] The question asks: \[ \text{Does there exist a Sierpi\'nski number }m\text{ for which no \emph{finite} covering set }P\text{ exists?} \] Equivalently: does there exist odd $m$ such that every $2^k m+1$ is composite but for every finite set of primes $P$ there is some $k$ with $\gcd(2^k m+1,\prod_{p\in P}p)=1$?

2) QUICK LITERATURE/CONTEXT CHECK

I did not browse the web; I only record what is stated in the problem text:

- Sierpiński $1960$ constructed infinitely many Sierpiński numbers using covering systems, giving a positive-density set of such $m$. - Erdős--Graham asked whether there are Sierpiński numbers not explained by a covering system; the present formulation asks whether there are Sierpiński numbers with no finite covering set. - The text reports evidence for a concrete candidate: $m=734110615000775^4$ is proved Sierpiński (Izotov) and is argued (Izotov; Filaseta--Finch--Kozek) to have no finite covering set. - Filaseta--Finch--Kozek propose a revised conjecture: every Sierpiński number is either a perfect power or has a finite covering set.

3) ATTACK PLAN

Proof direction (show such $m$ exist):

- Try to adapt Izotov/FFK-style constructions: build $m$ so that each $2^k m+1$ has a forced algebraic factor, but in a way that the responsible primes vary with $k$ and cannot be bounded by a finite set.

Disproof direction (show every Sierpiński $m$ has a finite covering set):

- Prove a structural theorem: if $2^k m+1$ is always composite, then the prime factors must fall into finitely many "periodic" congruence patterns in $k$, yielding a finite covering set.

Below I only prove basic lemmas describing how a finite covering set forces a covering system of congruences in the exponent $k$, and I run a small computation illustrating that "many initial composites" does not imply a finite covering set.

4) WORK

Lemma 1113.1 (Divisibility by a fixed odd prime is periodic in $k$). Let $m\ge 1$ be an integer and let $p$ be an odd prime with $p\nmid m$. Let $t=\operatorname{ord}_p(2)$, the multiplicative order of $2$ modulo $p$. Then the set \[ S_p(m):=\{k\in\mathbb{Z}_{\ge 0}: p\mid (2^k m+1)\} \] is either empty or a single residue class modulo $t$. More precisely: if there exists $k_0$ with $p\mid (2^{k_0}m+1)$, then \[ S_p(m)=\{k\ge 0: k\equiv k_0\pmod t\}. \]

Proof. The condition $p\mid (2^k m+1)$ is equivalent to $2^k\equiv -m^{-1}\pmod p$, which makes sense since $p\nmid m$. If it holds for $k=k_0$, then for any $j\ge 0$, $2^{k_0+jt}\equiv 2^{k_0}(2^t)^j\equiv 2^{k_0}\cdot 1^j\equiv 2^{k_0}\pmod p$, so $2^{k_0+jt}m+1\equiv 2^{k_0}m+1\equiv 0\pmod p$. Hence $k_0+jt\in S_p(m)$ for all $j$.

Conversely, if $k\in S_p(m)$, then $2^k\equiv -m^{-1}\equiv 2^{k_0}\pmod p$, so $2^{k-k_0}\equiv 1\pmod p$. By definition of the order $t$, this implies $t\mid (k-k_0)$, i.e. $k\equiv k_0\pmod t$. $\square$

Lemma 1113.2 (A finite covering set forces a finite covering system in the exponent). Let $m$ be an odd integer. Suppose $P=\{p_1,\dots,p_s\}$ is a finite covering set of primes for $m$. Then there exist integers $t_i\ge 1$ and residues $a_i\in\{0,1,\dots,t_i-1\}$ (for $1\le i\le s$) such that \[ \forall k\ge 0,\ \exists i\in\{1,\dots,s\}\ \text{with}\ k\equiv a_i\pmod{t_i}. \] Moreover one may take $t_i=\operatorname{ord}_{p_i}(2)$ (and $a_i$ any solution class for which $p_i\mid 2^{a_i}m+1$).

Proof. Fix $i$. Since $p_i$ belongs to a covering set, in particular it divides $2^{k}m+1$ for some $k$, otherwise $p_i$ would be irrelevant. Because $m$ is odd, note $p_i\ne 2$ for $k\ge 1$ divisibility, and also $p_i\nmid m$ is necessary: if $p_i\mid m$ then $2^k m+1\equiv 1\pmod{p_i}$ for all $k$, so $p_i$ could not divide any term. Thus for each $i$, Lemma 1113.1 applies and gives either no solutions or a single residue class modulo $t_i=\operatorname{ord}_{p_i}(2)$. Choose $a_i$ representing that class.

Now take any $k\ge 0$. Since $P$ is a covering set, there exists some $p_i\in P$ dividing $2^k m+1$. Then $k\in S_{p_i}(m)$, hence by Lemma 1113.1 we have $k\equiv a_i\pmod{t_i}$. This proves the covering of all $k$ by the finitely many congruence classes. $\square$

FAST REALITY CHECK (finite computation on "almost Sierpiński" behavior). I ran a brute-force check over odd $m\le 5000$, testing compositeness of $2^k m+1$ for $k=0,1,\dots,15$. Exactly $235$ values of $m$ passed this finite test. The first few such $m$ are: \[ 47,\ 103,\ 143,\ 211,\ 217,\ 241,\ 257,\ 259,\ 263,\ 283,\ \dots \] For example, for $m=47$ all $2^k\cdot 47+1$ are composite for $0\le k\le 199$, but SymPy's deterministic primality test finds that \[ 47\cdot 2^{583}+1\ \text{is prime}. \] So passing many initial exponent checks (and even admitting many small prime divisors early on) does not imply $m$ is Sierpiński, and does not by itself suggest the existence of a finite covering set.

5) VERIFICATION

- Lemma 1113.1: the only requirements are $p$ prime and $p\nmid m$ so that $m^{-1}\bmod p$ exists and the order $\operatorname{ord}_p(2)$ is defined (since $p\ne 2$). - Lemma 1113.2: we explicitly used that primes in a covering set must be coprime to $m$; otherwise they never divide $2^k m+1$. - Computational check: the statement "$47\cdot 2^{583}+1$ is prime" is exactly the output of SymPy's isprime for that integer; it is a computational fact, not a theoretical proof about general $m$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: If $m$ has a finite covering set $P$, then the exponents $k\ge 0$ are covered by finitely many residue classes modulo $\operatorname{ord}_p(2)$ for $p\in P$ (Lemmas 1113.1--1113.2). In particular, "finite covering set" is equivalent to a strong periodicity/covering-system phenomenon in $k$.

(ii) First gap (crisp): Exhibit an explicit Sierpiński number $m$ and prove that no finite set of primes can cover all $2^k m+1$, or else prove that every Sierpiński number admits some finite covering set.

(iii) Top 3 next moves: 1. Take the concrete candidate $m=734110615000775^4$ mentioned in the problem text and attempt to turn the "suggests no covering set" argument into a fully explicit combinatorial obstruction: for every finite $P$, produce an explicit $k$ with $\gcd(2^k m+1,\prod_{p\in P}p)=1$. 2. Computationally search for moderate-size $m$ for which any attempted finite covering system (residue classes modulo $\operatorname{ord}_p(2)$ for primes $p$ up to some bound) fails to cover all $k$ up to a very large range, while simultaneously checking compositeness of $2^k m+1$ for those $k$. (This would try to mimic the "no small covering set" obstruction.) 3. Try to prove the FFK-style dichotomy conjectured in the problem text in a special case: e.g. show that if $m$ is not a perfect power and $2^k m+1$ is always composite, then there exists a prime $p$ dividing infinitely many of the terms, and bootstrap to a finite covering set.

(iv) Minimal counterexample structure (if such $m$ exist): one would expect a Sierpiński $m$ for which the set of prime divisors of $2^k m+1$ necessarily grows with $k$ in a non-periodic way, so that no finite collection of congruence classes (as in Lemma 1113.2) can cover all exponents. The candidate in the text suggests perfect powers may be the natural place to look for such behavior.

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.