Belorusija 2012

First, for $R(x)=x^3+px+q$ to have three distinct real roots it is necessary that $p<0$ (because the derivative $R^{\prime }(x)=3x^2+p$ cannot be nonnegative). Let now $p<0$, then the equation $R^{\prime }(x)=0$ has two real roots $x_1=-\sqrt{-p/3}$, $x_2=\sqrt{-p/3}$. Now, the condition that $R(x)$ has three distinct real roots is equivalent to the inequality $R(x_1)\cdot R(x_2)<0$, which can be written as $$4p^3+27q^2<0.(\ast )$$ Let () be valid; denote by ${\alpha }_1<{\alpha }_2<{\alpha }_3$ the roots of $R(x)$. Since $R(x)$ is irrational, it has no zero roots. By Vieta's formula ${\alpha }_1+{\alpha }_2+{\alpha }_3=0$. Hence we have two possibilities:

i) ${\alpha }_1<{\alpha }_2<0<{\alpha }_3$; ii) ${\alpha }_1<0<{\alpha }_2<{\alpha }_3$.

Without loss of generality we can assume that case i) holds (if $x^3+px+q$ satisfies ii), then $x^3+px-q$ satisfies i)).

So, let ${\alpha }_1<{\alpha }_2<0<{\alpha }_3$, then $q=-{\alpha }_1{\alpha }_2{\alpha }_3<0$. Further, $$|{\alpha }_1|+|{\alpha }_2|+|{\alpha }_3|=|{\alpha }_1+{\alpha }_2|+{\alpha }_3=2{\alpha }_3.$$ Hence we need to find $R(x)$ satisfying (), $p,q\in \mathbb{Z}$, $p<0$, $q<0$, and for which $2{\alpha }_3$ is the smallest possible.

First, note that due to $p<0$, $q<0$ we have $R(1)=1+p+q<0$ which implies ${\alpha }_3>1$. Further, $R(2)=8+2p+q$. If $R(2)\le 0$, then ${\alpha }_3\ge 2$. Hence, if ${\alpha }_3<2$, then $$8+2p+q>0.(\ast \ast )$$ Now, it is easy to see that all $R(x)$ satisfying (), ($\ast \ast$) with $p<0$, $q<0$ are the following: $x^3-2x-1$ and $x^3-3x-1$. The first of them is not irrational since it has $-1$ as a root. The second trinomial satisfies the condition.

Remark.* One can verify that the value of $|{\alpha }_1|+|{\alpha }_2|+|{\alpha }_3|$ is equal to $4\cos 20^{\circ }$ for the founded polynomials.