# 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.)

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You can't assume that the P_i are monic. Anyway, I think I did this problem some years ago and I remember there was a reasonably nice solution, so try not to jump into the algebra too quickly. – Qiaochu Yuan Aug 28 '10 at 3:44

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.

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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 $c_1=c_2=c_3=c_4$.

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Again, you can't assume that the P_i are monic. – Qiaochu Yuan Sep 2 '10 at 1:58
I must apologize, Qiaochu. The original problem had the condition that the leading coefficients are 1; I forgot to mention that. So perhaps there is a counterexample to the statement in the general case. – Weltschmerz Sep 2 '10 at 23:53