{{begin-eqn}} {{eqn | l = 33 \, 614       | r = 14 \times 7^4       | c =  }} {{eqn | l = 33 \, 615       | r = 415 \times 3^4       | c =  }} {{eqn | l = 33 \, 616       | r = 2101 \times 2^4       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = 28 \, 559       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 28 \, 560       | r = 2^4 \times 3 \times 5 \times 7 \times 17 }} {{eqn | l = 28 \, 562       | r = 2 \times 14281 }} {{end-eqn}} Since neither $28 \, 559$ nor $28 \, 562$ are divisible by a fourth power of some prime number, $28 \, 561$ is not in a triplet. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = 14 \, 639       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 14 \, 640       | r = 2^4 \times 3 \times 5 \times 61 }} {{eqn | l = 14 \, 642       | r = 2 \times 7321 }} {{eqn | l = 29 \, 281       | r = 7 \times 47 \times 89 }} {{eqn | l = 29 \, 283       | r = 3 \times 43 \times 227 }} {{end-eqn}} Hence none of these numbers is in a triplet. {{qed|lemma}} === Case $3$: a number is divisible by $7^4$ === There are $14$ multiples of $7^4$ less than $33 \, 616$, and: {{begin-eqn}} {{eqn | l = 2399       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 2400       | r = 2^5 \times 3 \times 5^2 }} {{eqn | l = 2401       | r = 7^4 }} {{eqn | l = 2402       | r = 2 \times 1201 }} {{eqn | l = 4801       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 4802       | r = 2 \times 7^4 }} {{eqn | l = 4803       | r = 3 \times 1601 }} {{eqn | l = 7202       | r = 2 \times 13 \times 277 }} {{eqn | l = 7203       | r = 3 \times 7^4 }} {{eqn | l = 7204       | r = 2^2 \times 1801 }} {{eqn | l = 9603       | r = 3^2 \times 11 \times 97 }} {{eqn | l = 9604       | r = 4 \times 7^4 }} {{eqn | l = 9605       | r = 5 \times 17 \times 113 }} {{eqn | l = 12 \, 004       | r = 2^2 \times 3001 }} {{eqn | l = 12 \, 005       | r = 5 \times 7^4 }} {{eqn | l = 12 \, 006       | r = 2 \times 3^2 \times 23 \times 29 }} {{eqn | l = 14 \, 405       | r = 5 \times 43 \times 67 }} {{eqn | l = 14 \, 406       | r = 6 \times 7^4 }} {{eqn | l = 14 \, 407       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 16 \, 806       | r = 2 \times 3 \times 2801 }} {{eqn | l = 16 \, 807       | r = 7 \times 7^4 }} {{eqn | l = 16 \, 808       | r = 2^3 \times 11 \times 191 }} {{eqn | l = 19 \, 207       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 19 \, 208       | r = 8 \times 7^4 }} {{eqn | l = 19 \, 209       | r = 3 \times 19 \times 337 }} {{eqn | l = 21 \, 608       | r = 2^3 \times 37 \times 73 }} {{eqn | l = 21 \, 609       | r = 9 \times 7^4 }} {{eqn | l = 21 \, 610       | r = 2 \times 5 \times 2161 }} {{eqn | l = 24 \, 009       | r = 3 \times 53 \times 151 }} {{eqn | l = 24 \, 010       | r = 10 \times 7^4 }} {{eqn | l = 24 \, 011       | r = 13 \times 1847 }} {{eqn | l = 26 \, 410       | r = 2 \times 5 \times 19 \times 139 }} {{eqn | l = 26 \, 411       | r = 11 \times 7^4 }} {{eqn | l = 26 \, 412       | r = 2^2 \times 3 \times 31 \times 71 }} {{eqn | l = 28 \, 811       | r = 47 \times 613 }} {{eqn | l = 28 \, 812       | r = 12 \times 7^4 }} {{eqn | l = 28 \, 813       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 31 \, 212       | r = 2^2 \times 3^3 \times 17^2 }} {{eqn | l = 31 \, 213       | r = 13 \times 7^4 }} {{eqn | l = 31 \, 214       | r = 2 \times 15 \, 607 }} {{eqn | l = 33 \, 613       | o = \text {is}       | r = \text {prime} }} {{eqn | l = 33 \, 614       | r = 14 \times 7^4 }} {{eqn | l = 33 \, 615       | r = 415 \times 3^4 }} {{eqn | l = 33 \, 616       | r = 2101 \times 2^4 }} {{end-eqn}} Hence the smallest valid triplet is $\tuple {33 \, 614, 33 \, 615, 33 \, 616}$. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = x_0       | r = 3^4 5^4 b_1 + 2^4 5^4 \times 46 b_2 + 2^4 3^4 \times 231 b_3       | rr = \pmod {2^4 3^4 5^4} }} {{eqn | r = 50 \, 625 b_1 + 460 \, 000 b_2 + 299 \, 376 b_3       | rr = \pmod {810 \, 000} }} {{end-eqn}} 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 $33 \, 616$. {{qed}}