From a 2005 Russian olympiad. Prove that there do not exist four (pairwise) different quadratic polynomials, with leading coefficient 1, such that the sum of any two of them has a single root.
(Optionally, this is what I've done so far:
If $P_i(x)=x^2+b_ix+c_i$, then $P_i+P_j$ is single-rooted iff $(b_i+b_j)^2=8(c_i+c_j)$, with root $-(b_i+b_j)/2$. If two of the six roots are equal, we're done, so suppose they're different, and that $P_1+P_2$ is single-rooted, that is, $(b_1+b_2)^2=8(c_1+c_2)$. If we want $P_1+P_3$ and $P_2+P_3$ to be single-rooted as well, we need $(b_1+b_3)^2=8(c_1+c_3)$ and $(b_2+b_3)^2=8(c_2+c_3)$, from which, taking $b_1$ and $c_1$ as known, $b_2, c_2, b_3, c_3$ could be found.
This way looks cumbersome.)