Four polynomials with single-rooted sums 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.)
 A: You problem statement is too vague.  I guess you mean all four polynomials and their pairwise sums are non-degenerate, otherwise we can simply pick any four straight lines whose slopes are nonzero and of different magnitudes.  There should be other restrictions as well, or else {x^2, 2x^2, 3x^2, 4x^2} will also be a solution.
The following transformation of the problem may help.  Let P_i(x) = a_ix^2 + b_ix + c_i.  If P_1+P_2 is a polynomial with a single root (actually, do you mean a double root?), it must be equal to some k_1(x-x_1)^2 with k_1 \not = 0 and x_1 \in \mathbb{R}.  Thus a_1+a_2 = k_1 and vice versa for other pairwise sums.  So we get the following matrix equation:

[1 1 0 0] [a_1]   [k_1]
[1 0 1 0] [a_2]   [k_2]
[1 0 0 1] [a_3] = [k_3]
[0 1 1 0] [a_4]   [k_4]
[0 1 0 1]         [k_5]
[0 0 1 1]         [k_6].

Call the leftmost matrix M.  Hence the vector u = (k_1, ..., k_6)^T must lie inside the the column space of M.  By inspecting the coefficients b_i and c_i, we see that v = (k_1x_1, ..., k_6x_6)^T and w = (k_1x_1^2, ..., k_6x_6^2)^T also lie inside this column space.  Now, if you can prove that this is impossible, you are done.
A: So we have the equations
$(b_1+b_2)^2=8(c_1+c_2)$ (i)
$(b_1+b_3)^2=8(c_1+c_3)$ (ii)
$(b_1+b_4)^2=8(c_1+c_4)$ (iii)
$(b_2+b_3)^2=8(c_2+c_3)$ (iv)
$(b_2+b_4)^2=8(c_2+c_4)$ (v)
$(b_3+b_4)^2=8(c_3+c_4)$ (vi)
Substracting (ii) from (i) we get 
$(b_2-b_3)(b_2+b_3+2b_1)=8(c_2-c_3)$ (1)
and substracting (vi) from (v) we get
$(b_2-b_3)(b_2+b_3+2b_4)=8(c_2-c_3)$ (2)
Now, substracting (2) from (1) we get
$2(b_2-b_3)(b_1-b_4)=0$.
In a similar fashion we arrive at
$2(b_2-b_1)(b_3-b_4)=0$,
$2(b_1-b_3)(b_2-b_4)=0$.
The last three equations show that $b_1=b_2=b_3=b_4$. Using (i) through (vi), we see that $c_1=c_2=c_3=c_4$. 
