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Math-686 · mathematics · Mathematics · Number theory

Erdős Problem #137

Status Open · erdosproblems.comLLM status Open unverified

Statement

A positive integer $N$ is powerful if for every prime $p\mid N$ we have $p^2\mid N$ (equivalently, $v_p(N)\ge 2$ for all primes dividing $N$).

Fix an integer $k\ge 3$. For $m\ge 1$ define the product of $k$ consecutive integers \[ N_{m,k}:= m(m+1)\cdots (m+k-1). \]

Question. Can $N_{m,k}$ be powerful for some $m$ when $k\ge 3$? (The case $k=2$ has the example $8\cdot 9=72$.)

Restatement from Math-686-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-686-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-686

Erdős Problem 137: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

An integer is powerful if each prime dividing it occurs to exponent at least two. Can $P_{m,k}=m(m+1)\cdots(m+k-1)$ be powerful for any $k\ge3$? Source: https://www.erdosproblems.com/137. We give exact valuation criteria, exclude the range $m\le k+1$, and describe the two parity cases for three consecutive factors. The general question remains unresolved.

Valuations and large primes

Counting multiples of every prime power gives the finite exact formula \[ v_p(P_{m,k})=\sum_{j\ge1}\left( \left\lfloor\frac{m+k-1}{p^j}\right\rfloor- \left\lfloor\frac{m-1}{p^j}\right\rfloor\right). \] The product is powerful if and only if none of these valuations equals one. If $p\ge k$, it divides at most one factor: two multiples would differ by at least $p$, while the span of the factors is $k-1$. Thus every prime $p\ge k$ occurring in any factor must already have exponent at least two in that factor. Equivalently, a necessary condition is that every factor have a representation \[ m+i=a_i b_i^2c_i^3, \] where $a_i$ is squarefree and supported on primes less than $k$. To see this, put each exponent-one prime in $a_i$; every even exponent is supplied by $b_i^2$, and every odd exponent at least three is supplied by $c_i^3b_i^2$. Such a representation excludes exponent-one large primes. After imposing it, powerfulness is decided by the remaining finitely many valuations at primes $p<k$; the representation alone does not check those primes.

A whole initial range is impossible

Suppose $m\le k+1$ and put $U=m+k-1\ge3$. Bertrand's postulate gives a prime \[ U/2<p\le U. \] For even $U=2t\ge4$, use a prime $t<p<2t$; for odd $U=2t+1\ge5$, the same interval has integer $p\ge t+1>U/2$. The case $U=3$ uses $p=2$. Since $U\ge2m-2$, integrality gives $p\ge m$. Its only multiple in $[m,U]$ is $p$ itself, so its exponent in the product is exactly one. Hence no product with $m\le k+1$ is powerful. Bertrand's postulate is the explicit classical input to this exclusion.

The exact three-factor distinction

Let $k=3$. If $m$ is odd, the three factors are pairwise coprime: the only possible common divisor of the outer two divides two, and they are odd. Therefore their product is powerful if and only if each factor individually is powerful. If $m$ is even, every odd prime divides at most one factor. The odd parts of $m$ and $m+2$, and the entire odd integer $m+1$, must consequently be powerful. Conversely these three conditions suffice: one of $m,m+2$ has 2-adic valuation one and the other at least two, so their total 2-adic valuation is at least three. This reduces the even case to three specified odd powerful parts; it does not solve their simultaneous existence.

A warning about the perfect-power theorem

Being powerful is weaker than being a perfect power: for example $72=2^3\cdot3^2$ is powerful but not a perfect power. Thus the known exclusion of perfect powers among consecutive products does not settle this question. The earlier GPT-Pro-5.2 draft's large-prime obstruction is retained and strengthened above; its unrecorded searches are not used as proof. The missing step is a uniform source of an exponent-one prime when $m>k+1$, or an actual powerful counterexample.

Completion rate estimate: 20%.

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Math-686-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-686

1) FORMAL RESTATEMENT

A positive integer $N$ is powerful if for every prime $p\mid N$ we have $p^2\mid N$ (equivalently, $v_p(N)\ge 2$ for all primes dividing $N$).

Fix an integer $k\ge 3$. For $m\ge 1$ define the product of $k$ consecutive integers \[ N_{m,k}:= m(m+1)\cdots (m+k-1). \]

Question. Can $N_{m,k}$ be powerful for some $m$ when $k\ge 3$? (The case $k=2$ has the example $8\cdot 9=72$.)

2) QUICK LITERATURE/CONTEXT CHECK

(Restricted to statements explicitly present in 123-137.tex.) - The problem is open for all $k\ge 3$ as stated.

3) ATTACK PLAN

- Prove elementary constraints on prime exponents in products of consecutive integers. - Use computation to search for small examples (if any exist).

4) WORK

Lemma 4.1 (Large primes occur in at most one factor). Let $p$ be a prime and $k\ge 1$. If $p>k$, then among the $k$ consecutive integers $m,m+1,\dots,m+k-1$, at most one is divisible by $p$.

Proof. If $p$ divides both $m+i$ and $m+j$ with $0\le i<j\le k-1$, then $p$ divides their difference $(m+j)-(m+i)=j-i$. But $1\le j-i\le k-1<p$, impossible. $\square$

Lemma 4.2 (A sufficient condition for non-powerfulness). If there exists a prime $p>k$ such that for some $0\le i\le k-1$ we have $p\mid (m+i)$ but $p^2\nmid (m+i)$, then $N_{m,k}$ is not powerful. In particular, if the interval $[m,m+k-1]$ contains a prime $p>k$, then $N_{m,k}$ is not powerful.

Proof. By Lemma 4.1, such a prime $p>k$ divides exactly one factor among the $k$ consecutive integers. Therefore $v_p(N_{m,k})=v_p(m+i)=1$, which violates powerfulness. If $p$ itself lies in $[m,m+k-1]$ and $p>k$, then $p$ divides the product exactly once because $p^2\nmid p$. $\square$

FAST REALITY CHECK (computation). A brute-force search found no powerful products in the tested ranges: - For $k=3$, no $m\le 50000$ has $m(m+1)(m+2)$ powerful. - For $k=4$, no $m\le 5000$ has $m(m+1)(m+2)(m+3)$ powerful. - For $k=5$, no $m\le 5000$ has $m\cdots(m+4)$ powerful.

5) VERIFICATION

- Lemma 4.1 is a standard "difference less than $p$" argument. - Lemma 4.2 correctly translates the exponent condition $v_p(N)\ge 2$. - The computation is a finite search; it provides evidence but not a proof.

6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result: - If a prime $p>k$ appears to exponent exactly 1 in some factor of the interval, then the product is not powerful (Lemmas 4.1--4.2). - No examples were found for $k=3$ up to $m=50000$.

(ii) Exact first gap: - Either construct an explicit example $(m,k)$ with $k\ge 3$ for which $N_{m,k}$ is powerful, or prove such examples cannot exist.

(iii) Top 3 next moves: 1. Use results on primes in short intervals to guarantee existence of a prime $p>k$ in $[m,m+k-1]$ infinitely often (or always for large $m$), which would rule out powerfulness. 2. Analyze squarefull constraints: if $p>k$ divides some $m+i$, then $p^2$ must divide it, forcing very rigid factorization patterns across the interval. 3. For fixed small $k$ (e.g. $k=3$), attempt a Diophantine descent using the near-coprimality of consecutive integers and valuation constraints.

(iv) What a minimal counterexample would likely look like: - A short interval of length $k$ consisting entirely of integers each having all its "large" prime factors squared, arranged so that every prime dividing the product appears at least twice overall.

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