Smallest Triplet of Consecutive Integers Divisible by Cube

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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