Belorusija 2012

Answer: $\sqrt{5}$. By condition, the trinomial $x^2+px+q$ has the roots, so its discriminant $D=p^2-4q\ge 0$. Since $p^2=4q+D$ and $p$ and $q$ are integer, we have $D\ne 2$ and $D\ne 3$ because the square of the integer number is congruent neither to 2 nor to 3 modulo 4. Moreover, the trinomial $x^2+px+q$ has the roots $${\alpha }_1=2^{-1}(-p-\sqrt{D})\text{и}{\alpha }_2=2^{-1}(-p+\sqrt{D}).(\ast )$$ If the trinomial is irrational, but ${\alpha }_1$ and ${\alpha }_2$ are irrational numbers if and only if $D$ is different from the square of some integer number. Thus, in particular, $D\ne 0,D\ne 1$ and $D\ne 4$. Therefore, $D\ge 5$. Since the roots ${\alpha }_1$ and ${\alpha }_2$ of the irrational trinomial are different from 0, there are only two possibilities:

1) ${\alpha }_1\cdot {\alpha }_2>0$

2) ${\alpha }_1\cdot {\alpha }_2<0$

For case 1) we have $q={\alpha }_1\cdot {\alpha }_2>0$ (the Vieta theorem). So $q\ge 1$. From () it follows that $|{\alpha }_1|+|{\alpha }_2|=|p|$. Since $D\ge 5$, we have $p^2=4q+D\ge 4\cdot 1+5=9$, i.e. $|p|\ge 3$. Thus, $|{\alpha }_1|+|{\alpha }_2|\ge 3$ for case 1).

For case 2) from () it follows that $|{\alpha }_1|+|{\alpha }_2|=\sqrt{D}\ge \sqrt{5}$. It suffices to show that the estimate $\sqrt{5}$ is admissible.

We find all irrational trinomials with the discriminants $D=5$. By Vieta's theorem, ${\alpha }_1\cdot {\alpha }_2<0$ if and only if $q={\alpha }_1\cdot {\alpha }_2<0$, i.e. case 2) holds if and only if $q\le -1$. If $q=-1$, then $D=5$ only if $p^2=1$, but if $q\le -2$, then $D=p^2-4q\ge -4q\ge -4\cdot (-2)=8$. Therefore, there exist exactly two irrational trinomials with $D=5$ and ${\alpha }_1\cdot {\alpha }_2<0$: $x^2-x-1$ and $x^2+x-1$.

Therefore, the smallest value of the sum of the modules of the roots of the irrational trinomial is equal to $\sqrt{5}$.