# Smallest Triplet of Consecutive Integers Divisible by Cube

## 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 $\sqrt  {1377}$.

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.