I wonder what is the motivation for this.

I cannot answer your question, but let me show you my own solution which I found quite naturally after playing around a little with identities in the spirit of IMO 2018, P2 (with all indices read cyclically): $$ \sum b_i^2= \sum (a_{i+1}-a_i)^2= 2\sum a_i^2-2 \sum a_ia_{i+1}=2 \sum a_i(a_i-a_{i+1})=-2 \sum a_ib_i=-2 \sum b_i=-2 \sum (a_{i+1}-a_i)=0.$$Hence all $b_i$ are zero which leads to the solutions $(a,0)$.

First we sum up the $a_i$, to get $\sum a_i= \sum a_i + \sum b_i$, or in other words, $\sum b_i=0$. Since this seemed like a successful idea, we do the same with the $b_i$, to get $\sum b_i= \sum a_ib_i$, which means $\sum a_i b_i=0$. Now we'd somehow like to use the $\sum a_ib_i=0$ identity, so it would be nice to find it somewhere else. This is why we square the first reccurence to get $a_{i+1}^2=a_i^2+b_i^2+2a_ib_i$. We sum up those equalities and end up with $\sum a_i^2=\sum a_i^2+\sum b_i^2 + 2 \sum a_ib_i$. This implies $\sum b_i^2=0$. Thus, all $b_i$ are zero.

Note that $1) b_0+...+b_{2021}=0$ and $*a_n^2-a_{n+1}a_n+b_{n+1}=0$ because $b_{n+1}$ is a root of $x^2-a_{n+1}x+b_{n+1}$. If we sum up all the * equations for $n=0,1,2,...,2021$, use the given condition with 1 we get $a_0^2+...+a_{2021}^2=a_0a_1+a_1a_2+...+a_{2020}a_{2021}+a_{2021}a_0$ which gives $a_i$ is constant. Thus $(a,0)$ is the only solution. Edit: I realized that this is the same as tintarn's solution.

a^2 - 2a - 2b works as well, since (a+b)^2 - 2(a+b) - 2ab = a^2 + b^2 -2a - 2b, hence we need b=0