Is there a product of $k$ consecutive integers which is one more (or less) than a perfect cube?

Asked Mar 28 '23 at 07:43

Active Mar 28 '23 at 07:43

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I've seen this question with numerous proofs on the product of four consecutive integers being one less than a perfect square.

Then I've started to wonder: how many consecutive integers does one need to multiply in order to get a number one more or one less than a perfect cube? If it's not possible, I'd like to know a proof of that.

asked Mar 28 '23 at 07:43

Rusurano

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$18 \cdot 19 =342 = 341+1=7^3+1.$ – coffeemath Mar 28 '23 at 08:00
@coffeemath is it generalizeable? – Rusurano Mar 28 '23 at 08:05
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Please note that I did a typo. Should be $18\cdot 19 = 342 = 343-1= 7^3-1.$ My glaring error was to think 7^3=341. It's 343. I haven-t come up with say a one parameter family of such examples. – coffeemath Mar 28 '23 at 08:26
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Found another: $36 \cdot 37 = 11^3+1.$ – coffeemath Mar 28 '23 at 09:08
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@Rusurano, $35 \cdot 36 \cdot 37 \cdot 38= 121^3-1$ – Tomita Mar 28 '23 at 10:45
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A cube cannot be one more or one less than a product of three consecutive positive integers since $(n-1)n(n+1)=n^3-n$ and for all $n>1$, $(n-1)^3+1<n^3-n<n^3-1$. – Adam Bailey Mar 28 '23 at 11:27

Is there a product of $k$ consecutive integers which is one more (or less) than a perfect cube?

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