gpt pro 5.2 attempt on Math-6110

FORMAL RESTATEMENT.

Fix an integer $k\ge 2$. Does there exist a prime $p$ and consecutive nonempty integer intervals $I_1,\dots,I_k$ such that \[ \prod_{n\in I_i} n \equiv 1 \pmod p\qquad\text{for all }1\le i\le k? \] Here “consecutive intervals” means there exist integers $a_1\le b_1 < a_2\le b_2 < \cdots < a_k\le b_k$ with $a_{i+1}=b_i+1$ and $I_i=\{a_i,a_i+1,\dots,b_i\}$.

QUICK LITERATURE/CONTEXT CHECK.

I did not use external sources. I only use the examples stated in the problem text and perform a local computation search for small $k$.

ATTACK PLAN.

Proof-track: reduce the condition to factorial congruences in $\mathbb{F}_p$ and try to force repeated factorial values; relate to Wilson's theorem and possible cycle structure of $t!\bmod p$. Disproof-track: search for an obstruction (e.g. show that for each fixed $p$ the maximum $k$ is bounded). Best path here: obtain equivalent formulations (lemmas) and do computational exploration; no general proof is found.

WORK.

Lemma 1 (localization away from multiples of $p$). If $(p;I_1,\dots,I_k)$ is a solution, then translating all intervals by a multiple of $p$ (replacing every $n$ by $n+tp$ for a fixed integer $t$) yields another solution. Moreover, in any solution, none of the intervals may contain a multiple of $p$.

Proof. For any integer $t$, $n+tp\equiv n\pmod p$, so $\prod_{n\in I_i}(n+tp)\equiv \prod_{n\in I_i}n\equiv 1\pmod p$. Consecutiveness is preserved under translation. If some interval contained a multiple of $p$, then its product would be $0\pmod p$, contradicting the requirement that it be $1\pmod p$. $\square$

Lemma 2 (factorial-chain formulation). Assume $I_1,\dots,I_k$ lie inside a block of consecutive integers with no multiples of $p$. After translation (Lemma 1), we may assume $I_1\cup\cdots\cup I_k \subseteq \{1,2,\dots,p-1\}$. Write $I_i=[a_i,b_i]$ with $1\le a_i\le b_i\le p-1$ and $a_{i+1}=b_i+1$. Define $a_0:=a_1-1$ and $a_i:=b_i$ for $1\le i\le k$. Then the condition $\prod_{n\in I_i} n \equiv 1\pmod p$ for all $i$ is equivalent to \[ a_{i}! \equiv a_{i-1}! \pmod p\qquad\text{for all }1\le i\le k, \] where $0!=1$.

Proof. For $1\le a\le b\le p-1$, all numbers $1,2,\dots,p-1$ are invertible modulo $p$, so \[ \prod_{n=a}^{b} n \equiv \frac{b!}{(a-1)!}\pmod p. \] Thus $\prod_{n=a}^{b}n\equiv 1$ iff $b!\equiv (a-1)!$. Applying this to $I_i=[a_i,b_i]$ gives $b_i!\equiv (a_i-1)!$. Since $a_i-1=b_{i-1}$, this is exactly $a_i!\equiv a_{i-1}!$ in the notation above. $\square$

FAST REALITY CHECK (computation). Using Lemma 2, I searched primes $p$ for increasing chains $a_0<a_1<\cdots<a_k\le p-1$ with $a_i!\equiv a_{i-1}!\pmod p$. Within the searched range, the smallest prime $p$ found for each $k\in\{2,\dots,8\}$ (together with one solution as consecutive intervals in $[1,p-1]$) was:

k=2:  p=5,   intervals: [1,1], [2,3]
k=3:  p=17,  intervals: [1,1], [2,5], [6,11]
k=4:  p=17,  intervals: [1,1], [2,5], [6,11], [12,15]
k=5:  p=23,  intervals: [1,1], [2,4], [5,8], [9,11], [12,21]
k=6:  p=71,  intervals: [8,9], [10,19], [20,51], [52,61], [62,63], [64,70]
k=7:  p=599, intervals: [29,50], [51,122], [123,183], [184,250], [251,289],
               [290,500], [501,539]
k=8:  p=599, same first 7 intervals plus [540,555]

(Each list is consecutive and lies in $[1,p-1]$, hence can be translated to an integer solution by Lemma 1.)

VERIFICATION.

For $k=2$, $p=5$ gives $[1,1]$ product $1$, and $[2,3]$ product $6\equiv 1\pmod 5$. For $k=3$, $p=17$ and the reported intervals satisfy the factorial equalities $1!\equiv 0!$, $5!\equiv 1!$, and $11!\equiv 5!$ (verified by the program).

FINAL: UNRESOLVED.

(i) Strongest proved partial result here. The problem is equivalent to finding long chains of repeated factorial values modulo $p$ (Lemma 2), and explicit solutions exist at least for $2\le k\le 8$ with primes and intervals as above. (ii) First gap. Prove that for every $k$ there exists some prime $p$ with a factorial repetition chain of length $k$, or else exhibit some $k$ and prove that no such prime $p$ exists. (iii) Top 3 next moves. $1$ Study the directed graph on $\{0,1,\dots,p-1\}$ with edges $a\to b$ iff $a<b$ and $b!\equiv a!\pmod p$; show that its longest path length can be made arbitrarily large by choosing $p$. $2$ Use Wilson's theorem $(p-1)!\equiv -1\pmod p$ to relate factorial collisions near $p-1$ to multiplicative inverses and possibly force repetitions. $3$ Extend computation to larger $k$ to guess the growth of the minimal prime $p(k)$ and look for constructive patterns in the interval endpoints. (iv) What a minimal counterexample would look like. A counterexample would be a specific $k$ such that for every prime $p$, the factorial sequence $0!,1!,\dots,(p-1)!$ has fewer than $k+1$ indices with equal factorial values along an increasing chain, i.e. the graph described in (iii) has maximum path length $<k$.