Reused writeup on Math-6000 (gpt pro 5.2)

Erdős Problem 930 --- collection record

PROVENANCE AND STATUS

TeX syntax corrected for display; mathematical content unchanged. Official problem: https://www.erdosproblems.com/930

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 with TeX syntax corrections for display.

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_5.2/930.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/930.tex.

1) FORMAL RESTATEMENT

Ambiguity check: the statement says "intervals of consecutive integers". If $0\in I_i$ for some $i$ then the product is $0$, which is a perfect power ($0=0^t$ for any $t\ge 2$). So the literal statement is false unless one restricts to (strictly) positive integers.

Corrected (standard) formulation. Fix an integer $r\ge 1$. An interval of consecutive positive integers means a set \[ I=[a,b]\cap\mathbb Z=\{a,a+1,\dots,b\}\subset\mathbb Z_{\ge 1} \] with $1\le a\le b$, and its length is $|I|=b-a+1$. Call a positive integer $P$ a perfect power if $P=x^t$ for some integers $x\ge 2$ and $t\ge 2$.

The conjecture is: \[ \forall r\ge 1\ \exists k\ge 1\ \forall\text{ pairwise disjoint intervals }I_1,\dots,I_r\subset\mathbb Z_{\ge1} \ \left(\min_i |I_i|\ge k\ \Rightarrow\ \prod_{i=1}^r\prod_{m\in I_i} m\ \text{is not a perfect power}\right). \]

2) QUICK LITERATURE/CONTEXT CHECK

From the provided problem statement: Erdős--Selfridge proved the case $r=1$ (the product of consecutive positive integers is never a perfect power). For $r=2$ there are explicit constructions (referenced as [363] in the source) showing that some largeness condition on the intervals in terms of $r$ is necessary. No further results are stated in the source file.

3) ATTACK PLAN

• Disproof track (literal statement): exploit the $0$ issue to give a counterexample if intervals are allowed to include $0$.
• Proof track (corrected statement): try to force a prime to appear to exponent $1$ in the total product. A standard sufficient condition is: if the union contains a prime $q$ exceeding half the maximum element, then $q$ appears exactly once.
• Computational track: for small $r$ (especially $r=2$), brute force small intervals to see what minimum lengths still allow a perfect power.

4) WORK

Counterexample 930.0 (literal statement if $0$ is allowed). Let $r\ge 1$ be arbitrary. Let $k\ge 1$ be any proposed bound. Take \[ I_1=[0,k-1]\cap\mathbb Z,\qquad I_2=[k,k+(k-1)]\cap\mathbb Z,\quad\dots,\quad I_r=[(r-1)k,rk-1]\cap\mathbb Z. \] Then the $I_i$ are pairwise disjoint consecutive-integer intervals, each of length $k$, and $0\in I_1$. Hence \[ \prod_{i=1}^r\prod_{m\in I_i} m = 0, \] which is a perfect power (e.g. $0=0^2$). Therefore no $k$ can work in the literal "integers" formulation. $\square$

Lemma 930.1 (a large prime forces non-power). Let $S\subset\mathbb Z_{\ge 1}$ be a finite nonempty set of positive integers, and let $M:=\max S$. If $S$ contains a prime $q$ with $M/2<q\le M$, then the product $\prod_{m\in S} m$ is not a perfect power.

Proof. Because $q$ is prime and $q\in S$, the exponent of $q$ in the prime factorization of $\prod_{m\in S} m$ is at least $1$. If $m\in S$ and $q\mid m$, then $m\ge q$. Since $m\le M$ and $2q>M$, the only multiple of $q$ in $\{1,2,\dots,M\}$ is $q$ itself. Hence $q\mid m$ implies $m=q$. Therefore $q$ divides exactly one element of $S$, namely $q$, and contributes exponent exactly $1$ to the total product. A perfect power $x^t$ with $t\ge 2$ has every prime exponent divisible by $t$, in particular no prime can appear to exponent $1$. Hence $\prod_{m\in S} m$ is not a perfect power. $\square$

Proposition 930.2 (small $r=2$ constructions exist; $k(2)\ge 4$ is forced). There exist disjoint intervals $I_1,I_2$ of consecutive positive integers with $|I_1|\ge 3$ and $|I_2|\ge 3$ such that $(\prod_{m\in I_1}m)(\prod_{m\in I_2}m)$ is a perfect power. In particular, for the corrected statement with $r=2$, no choice $k\le 3$ can work.

Proof (explicit verification). Take \[ I_1=\{1,2,3,4,5,6\}=[1,6],\qquad I_2=\{8,9,10\}=[8,10]. \] These are disjoint intervals, with $|I_1|=6\ge 3$ and $|I_2|=3\ge 3$. Their product equals \[ (1\cdot2\cdot3\cdot4\cdot5\cdot6)\,(8\cdot9\cdot10)=720\cdot720=720^2, \] which is a perfect square. Thus any valid $k(2)$ must satisfy $k(2)\ge 4$. $\square$

FAST REALITY CHECK (local brute force, $r=2$). I brute-forced pairs of disjoint intervals contained in $\{1,2,\dots,60\}$ with each interval length $\le 8$ and checked whether the product of the union is a perfect power.

• For $k=2$ and $k=3$ (minimum interval length), many examples occur; one example for $k=3$ is the square above.
• For $k=4$ (minimum length $4$) and $k=5$, I found no examples in this search window.

This is only a sanity check on small ranges and does not prove anything asymptotic.

5) VERIFICATION

• Counterexample 930.0: all intervals have length $k$ and are disjoint; $0$ is included, so the total product is exactly $0$, which is a perfect power. Checked.
• Lemma 930.1: the only subtle point is "$q$ divides no other element of $S$". This follows from $2q>M$ and $m\le M$ for all $m\in S$.
• Proposition 930.2: the numerical identity is exact: $1\cdot2\cdot3\cdot4\cdot5\cdot6=720$ and $8\cdot9\cdot10=720$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result here. For the corrected positive-integer formulation, if the union of the intervals contains a prime $q>\tfrac12\max(\cup I_i)$, then the total product is not a perfect power (Lemma 930.1). Also, for $r=2$ one must have $k(2)\ge 4$ (Proposition 930.2).

(ii) First gap (crisp). Show that for each fixed $r$ there exists $k=k(r)$ such that every union of $r$ disjoint intervals of length at least $k$ necessarily contains a prime $q$ whose exponent in the total product is not divisible by any $t\ge 2$ (in particular, a prime $q>\tfrac12\max$ would suffice by Lemma 930.1).

(iii) Top 3 next moves.

1. Try to force a prime $q>\tfrac12\max$ inside (or as a factor of) the topmost interval using quantitative bounds on prime gaps (would require an explicit theorem strong enough to guarantee a prime in short intervals).
2. For $r=2$, do a targeted computation to guess the true smallest $k(2)$ (search larger windows and allow longer intervals), and then look for a structural description of all perfect-power examples.
3. Attempt to adapt the Erdős--Selfridge argument for a single interval to unions of intervals by tracking $p$-adic valuations prime-by-prime and showing an unavoidable "parity defect" for some prime.

(iv) What a minimal counterexample would look like. A minimal counterexample (for fixed $r$) would be $r$ disjoint long intervals whose union avoids every prime in $(\tfrac12M,M]$ (where $M$ is the maximum element), and for which every prime exponent in the product is simultaneously divisible by some $t\ge 2$. In particular, every number in the union near $M$ would have to share its prime factors extensively with other numbers in the union so that no prime occurs to exponent $1$.

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.