Reused writeup on Math-6012 (gpt pro 5.2)

Erdős Problem 951 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/951

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_5.2/951.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/951.tex.

Erdős Problem #951

FORMAL RESTATEMENT

Let $(a_i)_{i\ge1}$ be a strictly increasing sequence of real numbers with $a_1>1$. Assume that for every pair of distinct finitely supported tuples $k=(k_i)_{i\ge1}$ and $\ell=(\ell_i)_{i\ge1}$ of nonnegative integers, \[ \left|\prod_i a_i^{k_i} - \prod_i a_i^{\ell_i}\right|\ge 1. \] Let $A(x):=\#\{i: a_i\le x\}$. Question: is it true that $A(x)\le \pi(x)$ for all $x\ge1$?

QUICK LITERATURE/CONTEXT CHECK

The problem text calls such a sequence a set of Beurling prime numbers and mentions (without proof here) a conjecture of Beurling about the counting of multiplicative semigroup elements. I will not use any external results.

ATTACK PLAN

• Reality check: understand immediate consequences of the spacing condition, in particular constraints on the $a_i$ themselves.
• Prove unconditional lemmas: (a) injectivity of the exponent map; (b) $a_i\ge 2$ and $|a_i-a_j|\ge1$; (c) trivial bound $A(x)\le x-1$.
• Special-case proof attempt: show the conjectured bound $A(x)\le \pi(x)$ holds if the $a_i$ are integers (or rationals), using prime-factor exponent vectors and linear independence.
• Stop: the full real-valued problem seems to require new ideas.

WORK

Fast reality check (tiny examples).

• The usual primes $a_i=p_i$ satisfy the condition because distinct prime-power products are distinct integers, hence differ by at least $1$.
• The sequence $a_1=2$, $a_2=4$ fails: $4=2^2$ gives difference $0$ between tuples $(0,1,0,\dots)$ and $(2,0,0,\dots)$.

Lemma 951.1 (basic consequences of the spacing hypothesis). Under the hypothesis:

1. For each $i$, $a_i\ge 2$.
2. For $i\ne j$, $|a_i-a_j|\ge 1$.
3. The map $\Phi$ from finitely supported $k\in\mathbb Z_{\ge0}^{(\mathbb N)}$ to $\prod_i a_i^{k_i}$ is injective.

Proof. $1$ Compare the tuple $k$ with $k_i=1$ and $k_{m\ne i}=0$ to the zero tuple $\ell\equiv0$. Then the inequality gives $|a_i-1|\ge1$. Since $a_i>1$, it follows $a_i\ge2$.

$2$ Compare the tuple with one $1$ at position $i$ to the tuple with one $1$ at position $j$. Then $|a_i-a_j|\ge1$.

$3$ If $\Phi(k)=\Phi(\ell)$ for distinct $k\ne\ell$, then the left-hand side of the hypothesis is $0$, contradicting $\ge1$. $\square$

Lemma 951.2 (trivial growth bound). For every $x\ge2$, one has $A(x)\le \lfloor x\rfloor-1$.

Proof. By Lemma 951.1$1$, every $a_i\ge2$. By Lemma 951.1$2$, the set $\{a_i\}$ is $1$-separated. Thus in the interval $[2,x]$ one can fit at most $\lfloor x\rfloor-1$ many points with pairwise distance at least $1$. Formally, order the $a_i\le x$ as $b_1<\cdots<b_m$. Then $b_m-b_1\ge (m-1)\cdot 1$; since $b_1\ge2$ and $b_m\le x$, we have $x-2\ge m-1$, so $m\le x-1$ and hence $m\le\lfloor x\rfloor-1$. $\square$

Lemma 951.3 (the conjectured bound holds for integer sequences). Assume in addition that each $a_i$ is an integer $\ge 2$. Then for every real $x\ge2$, \[ A(x)=\#\{i: a_i\le x\}\le \pi(x). \]

Proof. Fix $x\ge2$ and let $I:=\{i: a_i\le x\}$, so $|I|=A(x)$. For each $i\in I$, write the prime factorization \[ a_i=\prod_{p\le x} p^{v_p(a_i)}, \] where $v_p(a_i)\in\mathbb Z_{\ge0}$ is the usual $p$-adic valuation. Define the exponent vector \[ v(i):=(v_p(a_i))_{p\le x}\in \mathbb Z_{\ge0}^{\pi(x)}. \]

Claim: The vectors $\{v(i): i\in I\}$ are linearly independent over $\mathbb Z$. Indeed, suppose there is an integer relation \[ \sum_{i\in I} c_i\, v(i)=0\quad\text{in }\mathbb Z^{\pi(x)}, \] with not all $c_i$ zero. Write $c_i=c_i^+-c_i^-$ with $c_i^+,c_i^-\in\mathbb Z_{\ge0}$ and $c_i^+c_i^-=0$. Then the vector equation implies (componentwise for each prime $p\le x$) \[ \sum_i c_i^+ v_p(a_i)=\sum_i c_i^- v_p(a_i), \] which in turn implies equality of integers \[ \prod_{i\in I} a_i^{c_i^+}=\prod_{i\in I} a_i^{c_i^-}. \] The exponent tuples $(c_i^+)_{i\in I}$ and $(c_i^-)_{i\in I}$ are distinct because some $c_i\ne0$. This contradicts Lemma 951.1$3$, which says distinct exponent tuples give distinct products. So no nontrivial integer relation exists, proving the claim.

Now the vectors $v(i)$ lie in the free abelian group $\mathbb Z^{\pi(x)}$ of rank $\pi(x)$. A set of $\mathbb Z$-linearly independent vectors in $\mathbb Z^{\pi(x)}$ has size at most the rank, i.e. at most $\pi(x)$. Therefore $A(x)=|I|\le \pi(x)$. $\square$

VERIFICATION

• Lemma 951.1: checked that comparing exponent tuples with single $1$ vs all $0$ forces $a_i\ge2$; comparing different singletons forces $|a_i-a_j|\ge1$.
• Lemma 951.2: verified inequality $b_m-b_1\ge m-1$ for ordered $1$-separated points.
• Lemma 951.3: the key step is that any nontrivial integer dependence among prime-exponent vectors yields an equality between two distinct monomials in the $a_i$, forbidden by the spacing hypothesis.

FINAL

UNRESOLVED

(i) Strongest proved partial result. We proved $a_i\ge2$ and the $a_i$ are pairwise $1$-separated, giving the trivial bound $A(x)\le x-1$ (Lemmas 951.1--951.2). If the $a_i$ are integers, then the desired inequality $A(x)\le\pi(x)$ holds (Lemma 951.3).

(ii) First gap. Extend Lemma 951.3 from integer sequences to arbitrary real sequences satisfying the spacing condition, i.e. find an invariant replacing prime-factor exponent vectors that forces $A(x)\le\pi(x)$.

(iii) Top 3 next moves.

1. Translate the condition into a packing problem in $\mathbb R$ for the multiplicative semigroup $\{\prod a_i^{k_i}\}$: it is $1$-separated in the Euclidean metric. Seek entropy/geometry-of-numbers bounds relating the number of generators $\le x$ to how many semigroup elements fit below $x$.
2. Examine whether one can force many semigroup elements $\le x$ from having too many generators $\le x$, contradicting the $1$-separation.
3. Search for explicit constructions with $A(x) > \pi(x)$ (to disprove) by choosing reals with carefully spaced logarithms and checking the $1$-separation numerically for initial segments.

(iv) What a minimal counterexample would likely look like. A counterexample would require a real sequence $(a_i)$ with many terms $\le x$ (significantly more than $x/\log x$) while keeping all distinct products $\prod a_i^{k_i}$ at distance $\ge1$ from each other. This would likely involve very sparse multiplicative interactions (strong multiplicative independence) and rapidly growing gaps to prevent near-collisions of large products.

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.