Erdős Problem #1056

Erdős Problem 1056 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1056

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_5.2/1056.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1056.tex.

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 $<k$.

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.

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 $<k$.