Reused writeup on Math-980 (gpt pro 5.2)

Erdős Problem 913 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/913

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_5.2/913.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/913.tex.

FORMAL RESTATEMENT.
For $n\in\mathbb{N}$ write the prime factorisation of $n(n+1)$ as \[ n(n+1) = \prod_i p_i^{k_i} \] with distinct primes $p_i$. The question asks whether there exist infinitely many $n$ such that the exponents $k_i$ are pairwise distinct (i.e. $k_i\ne k_j$ for $i\ne j$).

QUICK LITERATURE/CONTEXT CHECK.
The problem text gives a conditional infinite family: if there are infinitely many primes $p$ for which $8p^2-1$ is prime, then taking $n=8p^2-1$ yields exponent set $\{1,2,3\}$. I will not assume any unproved distributional statements about such primes.

ATTACK PLAN.
Construction track. Seek explicit parametric families of $n$ forcing a controlled factorisation of $n$ and $n+1$ (e.g. one of them a small-prime power times a prime). Use coprimality of $n$ and $n+1$.

Computation track. Search small $n$ for which the distinct-exponent property holds, to guess patterns/families.

WORK.
Phase 1: fast reality check (computation). I checked the condition "all exponents in $n(n+1)$ are distinct" by factoring $n$ and $n+1$ separately (they are coprime) and comparing the multiset of exponents.

• Up to $n\le 5000$, there are $57$ solutions. The first few are \[ 1,3,4,7,8,16,24,27,31,48,63,71,72,107,108,124,127,199,242,243,256,\dots \]
• Up to $n\le 10^5$, there are $140$ solutions.
• Up to $n\le 10^6$, there are $299$ solutions.

(These are exact counts from a direct scan using integer factorisation.)

Lemma 913.1 (exponents come from $n$ and $n+1$ separately). Let $n\ge 1$. Since $\gcd(n,n+1)=1$, every prime divisor of $n(n+1)$ divides exactly one of $n$ and $n+1$. Hence the multiset of exponents in the prime factorisation of $n(n+1)$ is the disjoint union of the exponent multisets for $n$ and for $n+1$.

Proof. The coprimality $\gcd(n,n+1)=1$ implies no prime divides both. If $p^a\parallel n$ (i.e. $p^a\mid n$ but $p^{a+1}\nmid n$), then $p\nmid n+1$ and therefore $p^a\parallel n(n+1)$. Similarly for primes dividing $n+1$. Thus exponents in $n(n+1)$ are exactly those appearing in $n$ together with those appearing in $n+1$, with no overlap of primes. $\square$

Lemma 913.2 (the $8p^2-1$ pattern gives exponents $\{1,2,3\}$). Let $p$ be prime and suppose $q:=8p^2-1$ is also prime. Set $n=q$. Then the exponents in the prime factorisation of $n(n+1)$ are $\{1,2,3\}$ (hence all distinct).

Proof. By definition, $n=q$ is prime so $n=q^1$. Also \[ n+1 = q+1 = 8p^2 = 2^3\,p^2. \] Since $q$ is odd and $p$ is prime, $q,2,p$ are three distinct primes. Therefore \[ n(n+1)= q\cdot 2^3\,p^2, \] so the exponent multiset is $\{1,3,2\}$, which has three distinct elements. $\square$

VERIFICATION.

• Lemma 913.1 uses only $\gcd(n,n+1)=1$ and the uniqueness of prime factorisation.
• Lemma 913.2 verifies line-by-line the factorisation and distinctness of primes.
• The computation does not prove infinitude; it only confirms that solutions exist and are not extremely rare at small scales.

FINAL.
UNRESOLVED.

(i) Strongest proved partial result: structural reduction Lemma 913.1, and an explicit conditional infinite family Lemma 913.2 (conditional only on the existence of infinitely many primes $p$ with $8p^2-1$ prime). Computationally, there are $140$ solutions up to $10^5$ and $299$ up to $10^6$.

(ii) First gap (crisp): Prove unconditionally that there are infinitely many $n$ such that the exponents in $n(n+1)$ are all distinct, or else prove that only finitely many exist.

(iii) Top 3 next moves:

1. Search for other parametric patterns forcing a bounded set of primes with prescribed exponents, e.g. $n$ prime and $n+1=2^a p^b$ with $a,b$ distinct and both $\ne 1$; attempt to make such patterns accessible to sieve methods.
2. Relax primality conditions and try to prove infinitude of solutions with "almost primes" (e.g. $n$ having two prime factors with distinct exponents) as a stepping stone; then attempt to upgrade.
3. Extend computation to much larger ranges and classify solutions by the exponent pattern (e.g. $\{1,2,3\}$, $\{1,4\}$, $\{2,3,1\}$, etc.) to identify the dominant families.

(iv) Minimal counterexample structure (if infinitude fails): beyond some $N$, every $n\ge N$ would force at least one repeated exponent among the prime factors of $n$ and $n+1$ (by Lemma 913.1 this repetition could occur within $n$, within $n+1$, or as the same exponent appearing once in $n$ and once in $n+1$).

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.