Finiteness conjecture for equal radicals of consecutive products

Let kk and \ell be integers greater than 11, and let mm and nn be integers. Define

Rk(m)=rad(m(m+1)(m+k1)),R_{k}(m)=\operatorname{rad}\bigl(m(m+1)\cdots(m+k-1)\bigr),

where rad(r)\operatorname{rad}(r) is the largest squarefree divisor of rr. Finiteness conjecture. If (k,)(2,2)(k,\ell)\neq(2,2), then the equation

rad(m(m+1)(m+k1))=rad(n(n+1)(n+1))\operatorname{rad}\bigl(m(m+1)\cdots(m+k-1)\bigr)=\operatorname{rad}\bigl(n(n+1)\cdots(n+\ell-1)\bigr)

has only finitely many solutions. The conjecture concerns the expected rarity of pairs of consecutive products with equal radicals; the exceptional case (k,)=(2,2)(k,\ell)=(2,2) is explicitly excluded, and the source gives no resolution.

Sources & referencesView supporting material

Primary source

Noah Lebowitz-Lockard, “On pairs of consecutive sequences with the same radicals”, arXiv:2507.09899 (2025).

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