Smallest Triplet of Consecutive Integers each Divisible by Fourth Power

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Theorem

This triplet of consecutive integers has the property that each of them is divisible by a fourth power:

$33 \, 614, 33 \, 615, 33 \, 616$

This is the smallest such triplet.


Proof

\(\ds 33 \, 614\) \(=\) \(\ds 14 \times 7^4\)
\(\ds 33 \, 615\) \(=\) \(\ds 415 \times 3^4\)
\(\ds 33 \, 616\) \(=\) \(\ds 2101 \times 2^4\)


Each number in such triplets of consecutive integers is divisible by a fourth power of some prime number.

Only $2, 3, 5, 7, 11, 13$ are less than $\sqrt [4] {33 \, 616}$.


Case $1$: a number is divisible by $13^4$

The only multiple of $13^4$ less than $33 \, 616$ is $28 \, 561$, and:

\(\ds 28 \, 559\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 28 \, 560\) \(=\) \(\ds 2^4 \times 3 \times 5 \times 7 \times 17\)
\(\ds 28 \, 562\) \(=\) \(\ds 2 \times 14281\)

Since neither $28 \, 559$ nor $28 \, 562$ are divisible by a fourth power of some prime number, $28 \, 561$ is not in a triplet.

$\Box$


Case $2$: a number is divisible by $11^4$

The only multiples of $11^4$ less than $33 \, 616$ are $14 \, 641$ and $29 \, 282$, and:

\(\ds 14 \, 639\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 14 \, 640\) \(=\) \(\ds 2^4 \times 3 \times 5 \times 61\)
\(\ds 14 \, 642\) \(=\) \(\ds 2 \times 7321\)
\(\ds 29 \, 281\) \(=\) \(\ds 7 \times 47 \times 89\)
\(\ds 29 \, 283\) \(=\) \(\ds 3 \times 43 \times 227\)

Hence none of these numbers is in a triplet.

$\Box$


Case $3$: a number is divisible by $7^4$

There are $14$ multiples of $7^4$ less than $33 \, 616$, and:

\(\ds 2399\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 2400\) \(=\) \(\ds 2^5 \times 3 \times 5^2\)
\(\ds 2401\) \(=\) \(\ds 7^4\)
\(\ds 2402\) \(=\) \(\ds 2 \times 1201\)
\(\ds 4801\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 4802\) \(=\) \(\ds 2 \times 7^4\)
\(\ds 4803\) \(=\) \(\ds 3 \times 1601\)
\(\ds 7202\) \(=\) \(\ds 2 \times 13 \times 277\)
\(\ds 7203\) \(=\) \(\ds 3 \times 7^4\)
\(\ds 7204\) \(=\) \(\ds 2^2 \times 1801\)
\(\ds 9603\) \(=\) \(\ds 3^2 \times 11 \times 97\)
\(\ds 9604\) \(=\) \(\ds 4 \times 7^4\)
\(\ds 9605\) \(=\) \(\ds 5 \times 17 \times 113\)
\(\ds 12 \, 004\) \(=\) \(\ds 2^2 \times 3001\)
\(\ds 12 \, 005\) \(=\) \(\ds 5 \times 7^4\)
\(\ds 12 \, 006\) \(=\) \(\ds 2 \times 3^2 \times 23 \times 29\)
\(\ds 14 \, 405\) \(=\) \(\ds 5 \times 43 \times 67\)
\(\ds 14 \, 406\) \(=\) \(\ds 6 \times 7^4\)
\(\ds 14 \, 407\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 16 \, 806\) \(=\) \(\ds 2 \times 3 \times 2801\)
\(\ds 16 \, 807\) \(=\) \(\ds 7 \times 7^4\)
\(\ds 16 \, 808\) \(=\) \(\ds 2^3 \times 11 \times 191\)
\(\ds 19 \, 207\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 19 \, 208\) \(=\) \(\ds 8 \times 7^4\)
\(\ds 19 \, 209\) \(=\) \(\ds 3 \times 19 \times 337\)
\(\ds 21 \, 608\) \(=\) \(\ds 2^3 \times 37 \times 73\)
\(\ds 21 \, 609\) \(=\) \(\ds 9 \times 7^4\)
\(\ds 21 \, 610\) \(=\) \(\ds 2 \times 5 \times 2161\)
\(\ds 24 \, 009\) \(=\) \(\ds 3 \times 53 \times 151\)
\(\ds 24 \, 010\) \(=\) \(\ds 10 \times 7^4\)
\(\ds 24 \, 011\) \(=\) \(\ds 13 \times 1847\)
\(\ds 26 \, 410\) \(=\) \(\ds 2 \times 5 \times 19 \times 139\)
\(\ds 26 \, 411\) \(=\) \(\ds 11 \times 7^4\)
\(\ds 26 \, 412\) \(=\) \(\ds 2^2 \times 3 \times 31 \times 71\)
\(\ds 28 \, 811\) \(=\) \(\ds 47 \times 613\)
\(\ds 28 \, 812\) \(=\) \(\ds 12 \times 7^4\)
\(\ds 28 \, 813\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 31 \, 212\) \(=\) \(\ds 2^2 \times 3^3 \times 17^2\)
\(\ds 31 \, 213\) \(=\) \(\ds 13 \times 7^4\)
\(\ds 31 \, 214\) \(=\) \(\ds 2 \times 15 \, 607\)
\(\ds 33 \, 613\) \(\text {is}\) \(\ds \text {prime}\)
\(\ds 33 \, 614\) \(=\) \(\ds 14 \times 7^4\)
\(\ds 33 \, 615\) \(=\) \(\ds 415 \times 3^4\)
\(\ds 33 \, 616\) \(=\) \(\ds 2101 \times 2^4\)


Hence the smallest valid triplet is $\tuple {33 \, 614, 33 \, 615, 33 \, 616}$.

$\Box$


Case $4$: the numbers are divisible by $2^4, 3^4, 5^4$ respectively

We utilise Chinese Remainder Theorem.

We are to solve the system of linear congruences:

$x \equiv b_1 \pmod {2^4}$
$x \equiv b_2 \pmod {3^4}$
$x \equiv b_3 \pmod {5^4}$

where $\set {b_1, b_2, b_3} = \set {1, 0, -1}$.


First note the linear congruences:

$3^4 5^4 x \equiv 1 \pmod {2^4}$
$2^4 5^4 x \equiv 1 \pmod {3^4}$
$2^4 3^4 x \equiv 1 \pmod {5^4}$

have solutions $1, 46, 231$ respectively.


Thus our system of linear congruences has the solution:

\(\ds x_0\) \(=\) \(\ds 3^4 5^4 b_1 + 2^4 5^4 \times 46 b_2 + 2^4 3^4 \times 231 b_3\) \(\ds \pmod {2^4 3^4 5^4}\)
\(\ds \) \(=\) \(\ds 50 \, 625 b_1 + 460 \, 000 b_2 + 299 \, 376 b_3\) \(\ds \pmod {810 \, 000}\)


Now we assign $\set {b_1, b_2, b_3}$ to $\set {1, 0, -1}$.

The solutions are:

$\tuple {0, 1, -1}: 160 \, 624$
$\tuple {0, -1, 1}: -160 \, 624 \equiv 649 \, 376$
$\tuple {-1, 0, 1}: 248 \, 751$
$\tuple {1, 0, -1}: -248 \, 751 \equiv 561 \, 249$
$\tuple {-1, 1, 0}: 409 \, 375$
$\tuple {1, -1, 0}: -409 \, 375 \equiv 400 \, 625$

and none of these solutions are less than $33 \, 616$.

$\blacksquare$


Historical Note

This result is reported by David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$ as the work of Stephane Vandemergel, but details are lacking.


Sources

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