Problem 10.1. Let $f(x)=x^{2}-a x+a^{2}-4$, where $a$ is a real number. Find all $a$, for which:

a) the equation $f(x)=0$ has two real roots $x_{1}$ and $x_{2}$ such that $\left|x_{1}^{3}-x_{2}^{3}\right| \leq$ 4 ;

b) the inequality $f(x) \geq 0$ holds for all integers $x$.

Peter Boyvalenkov
10.1. a) Using Vieta's formulas we get $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=$ 4. Hence $\left|x_{1}^{3}-x_{2}^{3}\right| \leq 4 \Longleftrightarrow\left|x_{1}-x_{2}\right| \leq 1$. Let $D=16-3 a^{2}$ be the the discriminant of $f(x)$. Then $D \geq 0$ and $\left|x_{1}-x_{2}\right|=\sqrt{D}$.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 $\left|x_{1}-x_{2}\right| \leq 1$ (since otherwise there is an
integer $x \in\left(x_{1}, x_{2}\right)$, i.e., $\left.f(x)\frac{2 \sqrt{3}}{3} \geq \frac{a}{2} \geq \frac{\sqrt{5}}{2}>1$, 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

$$
a \in\left(-\infty, \frac{-1-\sqrt{13}}{2}\right] \cup\left[\frac{1+\sqrt{13}}{2},+\infty\right)
$$