Bulgarian Mathematical Competitions

Solution:

a.
Using Vieta's formulas we get $x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} = (x_{1} + x_{2})^{2} - x_{1} x_{2} = 4$. Hence $|x_{1}^{3} - x_{2}^{3}| \leq 4 \Longleftrightarrow |x_{1} - x_{2}| \leq 1$. Let $D = 16 - 3 a^{2}$ be the discriminant of $f(x)$. Then $D \geq 0$ and $|x_{1} - x_{2}| = \sqrt{D}$. It follows that $0 \leq 16 - 3 a^{2} \leq 1$, and therefore $a \in \left[ -\frac{4 \sqrt{3}}{3}, -\sqrt{5} \right] \cup \left[ \sqrt{5}, \frac{4 \sqrt{3}}{3} \right]$.

b.
If $D = 16 - 3 a^{2} \leq 0$, i.e. $a \in \left( -\infty, -\frac{4 \sqrt{3}}{3} \right] \cup \left[ \frac{4 \sqrt{3}}{3}, \infty \right)$, then $f(x) \geq 0$ for every $x$. If $D > 0$, then $|x_{1} - x_{2}| \leq 1$ (since otherwise there is an integer $x \in (x_{1}, x_{2})$, i.e., $f(x) < 0$). Hence $a \in \left[ -\frac{4 \sqrt{3}}{3}, -\sqrt{5} \right] \cup \left[ \sqrt{5}, \frac{4 \sqrt{3}}{3} \right]$ and we have to find all $a$ from these intervals such that $f(x) \geq 0$ for every integer $x$.
If $f(x) < 0$ for some integer $x$, then the distance between $x$ and $\frac{a}{2}$ is at most $\frac{1}{2}$. Since $-\frac{3}{2} < -\frac{2 \sqrt{3}}{3} \leq \frac{a}{2} \leq -\frac{\sqrt{5}}{2} < -1$ or $\frac{3}{2} > \frac{2 \sqrt{3}}{3} \geq \frac{a}{2} \geq \frac{\sqrt{5}}{2} > 1$, we conclude that $x = \pm 1$ and by the inequalities $f(-1) \geq 0$ and $f(1) \geq 0$ we get $a \in \left[ -\frac{4 \sqrt{3}}{3}, \frac{-1 - \sqrt{13}}{2} \right] \cup \left[ \frac{1 + \sqrt{13}}{2}, \frac{4 \sqrt{3}}{3} \right]$. In conclusion, the desired values of $a$ are
$$a \in \left( -\infty, \frac{-1 - \sqrt{13}}{2} \right] \cup \left[ \frac{1 + \sqrt{13}}{2}, +\infty \right)$$