Does there exist positive integers $n_1, n_2, \dots, n_{2022}$ such that the number $$ \left( n_1^{2020} + n_2^{2019} \right)\left( n_2^{2020} + n_3^{2019} \right) \cdots \left( n_{2021}^{2020} + n_{2022}^{2019} \right)\left( n_{2022}^{2020} + n_1^{2019} \right) $$is a power of $11$?
Problem
Source: 2022 Pan-African Mathematics Olympiad Problem 6
Tags: number theory
IAmTheHazard
26.06.2022 15:30
At a glance it seems like a slightly modified solution to India TST 2019/1/2would work
rafaello
26.06.2022 16:49
Firstly deduce that $\gcd(n_i,11)=1$ for any $i$.
Take mod $11$, thus $n_i\equiv -1\pmod{11}$. Take $n_i=11x_i-1$ and mod $11^2$. Get something like $x_i\equiv \left(\frac{6}{7}\right)^{\text{blah}\not\equiv 0\pmod{10}} x_i\pmod{11}$, thus $x\equiv 0\pmod{11}$. Go even further $11^i$ and get that all $n_i\rightarrow \infty$, hence answer is no.
China_BW
09.07.2022 16:45
IAmTheHazard wrote: At a glance it seems like a slightly modified solution to India TST 2019/1/2would work But in my opinion, it's pretty hard to move the solution to this question due to '2019"vs"1"
China_BW
09.07.2022 16:46
China_BW wrote: IAmTheHazard wrote: At a glance it seems like a slightly modified solution to India TST 2019/1/2would work But in my opinion, it's pretty hard to move the solution to this question due to '2019"vs"1" And I'm think it may be possible to find another solution or improve this solution into a stronger one.
China_BW
09.07.2022 17:06
China_BW wrote: China_BW wrote: IAmTheHazard wrote: At a glance it seems like a slightly modified solution to India TST 2019/1/2would work But in my opinion, it's pretty hard to move the solution to this question due to '2019"vs"1" And I'm think it may be possible to find another solution or improve this solution into a stronger one. finally it works and it's similar