66th Belarusian Mathematical Olympiad

Let the trinomials $P(x)=x^2+ax+b$ and $Q(x)=x^2+cx+d$ with the roots $x_1<x_2$ and $x_3<x_4$, respectively, be friendly. By condition, $x_1+x_3$ and $x_2+x_4$ are the roots of the trinomial $x^2+(a+c)x+b+d$. By Vieta's theorem, we have $$b+d=(x_1+x_3)(x_2+x_4)=x_1{x}_2+x_1{x}_4+x_3{x}_2+x_3{x}_4.(1)$$ Since, by Vieta's theorem applied to the trinomials $P(x)$ and $Q(x)$ we have $x_1{x}_2=b$ and $x_3{x}_4=d$, from (1) it follows $$x_1{x}_4=-x_2{x}_3.(2)$$ If none of $x_1,x_2,x_3,x_4$ is zero, then $$\frac{x_1}{x_2}=-\frac{x_3}{x_4}.$$ In particular, the ratios of the roots of any trinomials from $M$ are of opposite signs, so $M$ cannot have more than two trinomials, a contradiction. Therefore, among the trinomials from $M$ there exists a trinomial with zero root. By condition, the other root of this trinomial differs from zero, so from (2) it follows that this trinomial cannot be friendly to the trinomial with two nonzero roots. Thus, zero is a root for all trinomials from $M$.