Belarus2022

Let us prove the following Statement: $n=k^2-k+1$ for some positive integer $k$. Proof: Let $d$ be the largest of the greatest common divisors found. Then the equality $\gcd (a_i,a_{i+\ell })=d$ is equivalent to the fact that each of the numbers $a_i$ and $a_{i+\ell }$ is a multiple of $d$. Let $(i_1,i_2,\dots ,i_k)$ be all indexes of numbers in the given sequence that are multiples of $d$, written in ascending order. Note that each pair $i_j<i_s$ of indices yields $d$ twice: for $\ell =s-j$ and for $\ell =j+n-s$. Therefore $d$ will occur as the greatest common divisor exactly $2\left(\frac{k(k-1)}{2}\right)$ times and by assumption this number is $n-1$. Thus $n=1+2\left(\frac{k(k-1)}{2}\right)=k^2-k+1$. The assertion is proved.

b) The equation $k^2-k+1=2021$ is equivalent to $k(k-1)=2020$. The left side of this equation increases for $k>0.5$ while $90\cdot 89=8010$ and $91\cdot 90=8190$, therefore this equation doesn't have a positive integer solutions. According to the proven statement, the answer at this item is: "no".

a) Note that $21=5^2-5+1$, i.e. $n=21$ satisfies the statement for $k=5$. It is easy to construct the sequence $(i_1,i_2,i_3,i_4,i_5)$ from the proof of this statement, for example $(1,11,13,18,19)$. Let us set $a_1=a_2=a_3=\cdots =a_{21}=1$ then choose $20$ different primes $p_1>p_2>\cdots >p_{20}$ and for each $i$ from $1$ to $20$ multiply by $p_i$ the numbers $a_i,a_{i+10},a_{i+12},a_{i+17},a_{i+18}$. It is easy to see that for each $\ell$ from $1$ to $20$ the set of greatest common divisors from the problem statement coincides with the set $\{p_1,p_2,\dots ,p_{20}\}$.