Similar question in old russian olympiad )

I think I got that all $a_i$ are equal, otherwise we can get that there exists $i$, such that $a_i.Q(x)+b_i=a_{2022}.Q(x)+b_{2022}$ has a root since $Q$ has odd degree, and then we get that $P(x)$ can't have the same sign for all $x$, so it has a root. The harder case seems to be when all $a_i$ are equal, and no $b_i$ is equal to $b_{2022}$. How to proceed then? @below Oh yes, indeed, you're right, comparing leading coefficients and taking large x seems to suffice, thanks.

@above when all $a_i$ are equal the left hand side will have a much larger leading coefficient and so the equality will fail for sufficiently large $x$, I think?