30th Turkish Mathematical Olympiad

Answer: $-2<a<0$ and $0<a<2$.

Letting $c=x+y+z$, one sees that $x,y,z$ are the roots of the polynomial $P(t)=t^3-3t+a+3c$. Therefore, Vieta's theorem implies $x+y+z=0$, thus $c=0$ and $P(t)=t^3-3t+a$. Consequently the polynomial $P(t)=t^3-3t+a$ has three distinct real roots. Moreover, the denominators appearing in the problem statement must be nonzero, hence the roots of $P(t)$ are nonzero. On the other hand, the converse statement also holds: if the polynomial $P(t)=t^3-3t+a$ has three distinct nonzero real roots, the equalities

in the problem statement are satisfied by these roots $x,y,z$. Now for $a>2$, one would have $t^3+a>t^3+2\ge 3t$ for positive $t$, thus the polynomial $P(t)$ would have no positive root, so it could not have three real roots. Similarly for $a<-2$, there would be no negative root and there could not be three real roots. Moreover for $a=2$ and $a=-2$, the polynomial $P(t)$ would have a double root of $t=1$ and $t=-1$, respectively. Finally for $a=0$, one of the roots of $P(t)$ would be 0. Outside all these cases, one has $-2<a<0$ or $0<a<2$. Indeed for $0<a<2$, one has $P(0)=a>0$ and $P(1)=a-2<0$, hence there is a root in each of the intervals $(-\infty ,0)$, $(0,1)$ and $(1,\infty )$. Similarly for $-2<a<0$, there is a root in each of the intervals $(-\infty ,-1)$, $(-1,0)$ and $(0,\infty )$.