We say that $N$ is powerful if whenever $p ∣ N$ we also have $p^{2} ∣ N$ . Let $k \geq 3$ . Can the product of any $k$ consecutive positive integers ever be powerful?

Worked, still open.

number theory · open · formalized (Lean) · 0 attempts

machinery: consecutive-integer-window,powerful-numbers,S-unit-equation,perfect-power-in-products,abc/kernel-bound,Pell-equation,prime-distribution

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

evidence

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

A **powerful** (also called **squarefull**) integer is one where every prime divisor shows up with exponent at least $2$: if (p\mid N) then (p^2\mid N). ([Wikipedia][1])

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 11 · open (literature)

theorem erdos_137 : answer(sorry) ↔ ∀ k ≥ 3, ∀ n, ¬ (∏ x ∈ Finset.Ioc n (n + k), x).Powerful

formal-conjectures/137.lean ↗

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open

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formal

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