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): $1375$