Show an infinite sequence $a_1, a_2, \ldots$ of integers with both of the following properties: • $a_i \neq 0$ for every positive integer $i$, that is, no term in the sequence is equal to zero; • for all positive integer $n$, $a_n + a_{2n} + \ldots + a_{2023n} = 0$.
Problem
Source: Brazilian Mathematical Olympiad 2023, Level 3, Problem 1
Tags: number theory, Sequence
VicKmath7
21.10.2023 22:13
Basically like USAMO 2012/3 but much easier.
Set $a_1=1$, $a_p=1$ for all primes $p \neq 2017$, $a_{2017}=-2022$ and $a_{rs}=a_r\cdot a_s$ for all positive integers $r, s$. Notice that $a_n+a_{2n}+\ldots+a_{2023n}=a_n(a_1+a_2+\ldots+a_{2023})=0$ for all positive integers $n$ ($2017$ can be changed to any prime greater than $1011$).
levifb
21.10.2023 23:14
You can also just put $a_i = 1 ~ \forall i$ such that $2023 \nmid i$ and let the guys with multiples of $2023$ at index constructed in a way such that the sums in which they are the last terms be zero. Note that this guys can't be zero since by induction all terms of the sequence are $\equiv 1$ (mod $ 2023$), and we're done.
oiiooi
26.10.2023 05:12
Another solution: Let $a_i = (-2022)^{v_{2017}(i)}$, where $v_{2017}(n)$ is the greatest integer $\alpha$ such that $2017^{\alpha} \mid n$. Clearly, all terms are nonzero and, for every $n \geq 1$ and each $1 \leq i \leq 2023$, $a_{in} = (-2022)^{v_{2017}(n) + 1}$ if $i = 2017$ and $a_{in} = (-2022)^{v_{2017}(n)}$ otherwise $\Rightarrow \Sigma^{2023}_{i=1} a_{in} = 2022 \cdot (-2022)^{v_{2017}(n)} + (-2022)^{v_{2017}(n) + 1} =$ $2022 \cdot (-2022)^{v_{2017}(n)} + (-2022) \cdot (-2022)^{v_{2017}(n)} = 0$.