Smallest Quadruplet of Consecutive Integers Divisible by Cube
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This theorem requires a proof.
It remains to be shown there is no smaller such quadruplet.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
Contents
Theorem
The smallest sequence of quadruplets of consecutive integers each of which is divisible by a cube greater than $1$ is:
- $\tuple {22 \, 624, 22 \, 625, 22 \, 626, 22 \, 627}$
Proof
| \(\displaystyle 22 \, 624\) | \(=\) | \(\displaystyle 2828 \times 2^3\) | |||||||||||
| \(\displaystyle 22 \, 625\) | \(=\) | \(\displaystyle 181 \times 5^3\) | |||||||||||
| \(\displaystyle 22 \, 626\) | \(=\) | \(\displaystyle 838 \times 3^3\) | |||||||||||
| \(\displaystyle 22 \, 627\) | \(=\) | \(\displaystyle 17 \times 11^3\) |
It remains to be shown there is no smaller such quadruplet.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
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Historical Note
This result is reported by David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$ as the work of Stephane Vandemergel, but details are lacking.
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1375$
