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:

$\ds 1375$

$=$

$\ds 11 \times 5^3$

$\ds 1376$

$=$

$\ds 172 \times 2^3$

$\ds 1377$

$=$

$\ds 51 \times 3^3$

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:

$\ds 1330$

$=$

$\ds 2 \times 5 \times 7 \times 19$

$\ds 1332$

$=$

$\ds 2^2 \times 3^2 \times 37$

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:

$\ds 342$

$=$

$\ds 2 \times 3^2 \times 19$

$\ds 344$

$=$

$\ds 2^3 \times 43$

$\ds 345$

$=$

$\ds 3 \times 5 \times 23$

$\ds 685$

$=$

$\ds 5 \times 137$

$\ds 687$

$=$

$\ds 3 \times 229$

$\ds 1028$

$=$

$\ds 2^2 \times 257$

$\ds 1030$

$=$

$\ds 2 \times 5 \times 103$

$\ds 1371$

$=$

$\ds 3 \times 457$

$\ds 1373$

$\text {is}$

$\ds \text {prime}$

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:

$\ds 124$

$=$

$\ds 2^2 \times 31$

$\ds 126$

$=$

$\ds 2 \times 3^2 \times 7$

$\ds 249$

$=$

$\ds 3 \times 83$

$\ds 251$

$\text {is}$

$\ds \text {prime}$

$\ds 623$

$=$

$\ds 7 \times 89$

$\ds 624$

$=$

$\ds 2^4 \times 3 \times 13$

$\ds 626$

$=$

$\ds 2 \times 313$

$\ds 874$

$=$

$\ds 2 \times 19 \times 23$

$\ds 876$

$=$

$\ds 2^2 \times 3 \times 73$

$\ds 1030$

$=$

$\ds 2 \times 5 \times 103$

$\ds 1249$

$\text {is}$

$\ds \text {prime}$

$\ds 1251$

$=$

$\ds 3^2 \times 139$

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): $1375$