Smallest Triplet of Consecutive Integers each Divisible by Fourth Power

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
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:

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

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:

Hence none of these numbers is in a triplet.

Case $3$: a number is divisible by $7^4$
There are $14$ multiples of $7^4$ less than $33 \, 616$, and:

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

Case $4$: the numbers are divisible by $2^4, 3^4, 5^4$ respectively
We utilise Chinese Remainder Theorem/General Result.

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:

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 $33 \, 616$.