# Smallest Triplet of Consecutive Integers Divisible by Cube

## Contents

## 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

We will show that:

\(\displaystyle 1375\) | \(=\) | \(\displaystyle 11 \times 5^3\) | |||||||||||

\(\displaystyle 1376\) | \(=\) | \(\displaystyle 172 \times 2^3\) | |||||||||||

\(\displaystyle 1377\) | \(=\) | \(\displaystyle 51 \times 3^3\) |

is the smallest such triplet.

Each number in such triplets of consecutive integers is divisible by a cube of some prime number.

Only $2, 3, 5, 7, 11$ are less than $\sqrt [3] {1377}$.

Since the numbers involved are small, we can check the result by brute force.

For general results one is encouraged to use the Chinese Remainder Theorem.

### Case $1$: a number is divisible by $11^3$

The only multiple of $11^3$ less than $1377$ is $1331$, and:

\(\displaystyle 1330\) | \(=\) | \(\displaystyle 2 \times 5 \times 7 \times 19\) | |||||||||||

\(\displaystyle 1332\) | \(=\) | \(\displaystyle 2^2 \times 3^2 \times 37\) |

Since neither $1330$ nor $1332$ are divisible by a cube of some prime number, $1331$ is not in a triplet.

$\Box$

### Case $2$: a number is divisible by $7^3$

The only multiples of $7^3$ less than $1377$ are $343, 686, 1029, 1372$, and:

\(\displaystyle 342\) | \(=\) | \(\displaystyle 2 \times 3^2 \times 19\) | |||||||||||

\(\displaystyle 344\) | \(=\) | \(\displaystyle 2^3 \times 43\) | |||||||||||

\(\displaystyle 345\) | \(=\) | \(\displaystyle 3 \times 5 \times 23\) | |||||||||||

\(\displaystyle 685\) | \(=\) | \(\displaystyle 5 \times 137\) | |||||||||||

\(\displaystyle 687\) | \(=\) | \(\displaystyle 3 \times 229\) | |||||||||||

\(\displaystyle 1028\) | \(=\) | \(\displaystyle 2^2 \times 257\) | |||||||||||

\(\displaystyle 1030\) | \(=\) | \(\displaystyle 2 \times 5 \times 103\) | |||||||||||

\(\displaystyle 1371\) | \(=\) | \(\displaystyle 3 \times 457\) | |||||||||||

\(\displaystyle 1373\) | \(\text {is}\) | \(\displaystyle \text {prime}\) |

Hence none of these numbers is in a triplet.

$\Box$

### Case $3$: the numbers are divisible by $2^3, 3^3, 5^3$ respectively

Let $n = k \times 5^3$.

We show that $k$ cannot be divisible by $3$ or $4$.

Suppose $3 \divides k$.

Then none of $n \pm 1, n \pm 2$ are divisible by $3$, and consequently $3^3$.

Suppose $4 \divides k$.

Then none of $n \pm 1, n \pm 2$ are divisible by $4$, and consequently $2^3$.

The only multiples of $5^3$ less than $1377$ are $125, 250, 375, 500, 625, 750, 875, 1000, 1125, 1250, 1375$, and we eliminate $375, 500, 750, 1000, 1125$ due to the reasons above.

Now:

\(\displaystyle 124\) | \(=\) | \(\displaystyle 2^2 \times 31\) | |||||||||||

\(\displaystyle 126\) | \(=\) | \(\displaystyle 2 \times 3^2 \times 7\) | |||||||||||

\(\displaystyle 249\) | \(=\) | \(\displaystyle 3 \times 83\) | |||||||||||

\(\displaystyle 251\) | \(\text {is}\) | \(\displaystyle \text {prime}\) | |||||||||||

\(\displaystyle 623\) | \(=\) | \(\displaystyle 7 \times 89\) | |||||||||||

\(\displaystyle 624\) | \(=\) | \(\displaystyle 2^4 \times 3 \times 13\) | |||||||||||

\(\displaystyle 626\) | \(=\) | \(\displaystyle 2 \times 313\) | |||||||||||

\(\displaystyle 874\) | \(=\) | \(\displaystyle 2 \times 19 \times 23\) | |||||||||||

\(\displaystyle 876\) | \(=\) | \(\displaystyle 2^2 \times 3 \times 73\) | |||||||||||

\(\displaystyle 1030\) | \(=\) | \(\displaystyle 2 \times 5 \times 103\) | |||||||||||

\(\displaystyle 1249\) | \(\text {is}\) | \(\displaystyle \text {prime}\) | |||||||||||

\(\displaystyle 1251\) | \(=\) | \(\displaystyle 3^2 \times 139\) |

Hence none of these numbers is in a triplet.

$\blacksquare$

## Historical Note

This result is reported by David Wells in his *Curious and Interesting Numbers* of $1986$ as appearing in *Eureka* in $1982$, but it has been impossible to corroborate this.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $1375$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $1375$