Smallest Triplet of Consecutive Integers Divisible by Cube

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Theorem

The smallest sequence of triplets of consecutive integers each of which is divisible by a cube greater than $1$ is:

$\tuple {1375, 1376, 1377}$


Proof

\(\displaystyle 1375\) \(=\) \(\displaystyle 11 \times 5^3\)
\(\displaystyle 1376\) \(=\) \(\displaystyle 172 \times 2^3\)
\(\displaystyle 1377\) \(=\) \(\displaystyle 51 \times 3^3\)



Historical Note

This result is reported by David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$ as appearing in Eureka in $1982$, but it has been impossible to corroborate this.


Sources