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Updated Oct 8

Problem 137

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

Status. Open.

Source. erdosproblems.com/137, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #137, https://www.erdosproblems.com/137.

References.

[Er82c] Erdős, P., Miscellaneous problems in number theory. Congr. Numer. (1982), 25-45.
[Er97c] Erdős, Paul, Some of my favorite problems and results. The mathematics of Paul Erdős, I (1997), 47-67.
[ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathematique (1980).
[ErSe75] Erdős, P. and Selfridge, J. L., The product of consecutive integers is never a power. Illinois J. Math. (1975), 292-301.

Formalization. Statement in formal-conjectures.

Progress

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Linked library material

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Linked from (10)

Diophantine Problems and Powers Diophantine Problems and Powers diophantine_problems/erdos_1975_product_consecutive_integers_is_never_power Conjecture (pp. 292-293): a prime greater than k divides the product exactly once Theorem 2: a prime at least k divides the product to a power not divisible by l diophantine_problems/pandey_2024_squarefree_numbers_short_intervals Theorem 1.1: the squarefree count in intervals of length X^{1/5-eta}factorials_binomials/erdos_1982_miscellaneous_problems_number_theory Conjecture (p. 27): many exponents equal to 1 in a long product of consecutive integers Conjecture (p. 28): some exponent equals 1 in a product of consecutive integers

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