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:

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:

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

Case $2$: a number is divisible by $7^3$
The only multiples of $7^3$ less than $1377$ are $343, 686, 1029, 1372$, and:

Hence none of these numbers is in a triplet.

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:

Hence none of these numbers is in a triplet.