Do all pairs of consecutive powerful numbers $n$ and $n + 1$ come from solutions to Pell equations? In other words, must either $n$ or $n + 1$ be a square?Is the number of such $n \leq x$ bounded by $(lo g x)^{O (1)}$ ?

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

There are **consecutive powerful (squarefull) numbers where neither term is a square**. The smallest “Type II” example (in Golomb’s terminology) is [ 12167 = 23^3,\qquad 12168 = 2^3\cdot 3^2\cdot 13^2, ] and **Walker proved there are infinitely many such pairs** by showing that [ 7^3x^2 = 3^3y^2 + 1 ] has infinitely ma…

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llm-hunter · gpt pro 5.2 · unverified

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A060355 — Numbers k such that k and k+1 are powerful numbers.8,288,675,9800,12167,235224,332928,465124,1825200,11309768,384199200,592192224,4931691075,5425069447,13051463048,2213222 A060859 — Powerful numbers of the form k^2 - 1.8,288,675,9800,235224,332928,1825200,11309768,384199200,592192224,4931691075,13051463048,221322261600,443365544448,86536 A175155 — Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.0,682,1268860318,1459639851109444,2360712083917682,86149711981264908618,4392100110703410665318,8171493471761113423918890

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powerful numbers · reference

Pell equations · reference

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#364Are there any triples of consecutive positive integers all of which are powerful (i.e. if

p ∣ n

then

p^{2} ∣ n

)?A060355 #366Are there any

2

-full

n

such that

n + 1

3

-full? That is, if

p ∣ n

then

p^{2} ∣ n

and if

p ∣ n + 1

then

p^{3} ∣ n + 1

.A060355

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