Smallest Quadruplet of Consecutive Integers Divisible by Cube

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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

\(\displaystyle 22 \, 624\) \(=\) \(\displaystyle 2828 \times 2^3\)
\(\displaystyle 22 \, 625\) \(=\) \(\displaystyle 181 \times 5^3\)
\(\displaystyle 22 \, 626\) \(=\) \(\displaystyle 838 \times 3^3\)
\(\displaystyle 22 \, 627\) \(=\) \(\displaystyle 17 \times 11^3\)



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