69th Belarusian Mathematical Olympiad

It is easy to see that the triples $(0;0;0)$ and $(-1;-1;-1)$ satisfy the system. We will show that there are no other solutions. From the equations $$(x+1)(x^2+1)=y^3+1,(1)$$ $$(y+1)(y^2+1)=z^3+1,(2)$$ $$(z+1)(z^2+1)=x^3+1(3)$$ it follows that if $x=-1$, then, from (1) $y=-1$, and from (2) $z=-1$. Similarly, if $x=0$, then $y=0$ and $z=0$. Further assume that none of $x$, $y$ and $z$ equals $-1$ or $0$. Factor the right-hand sides of (1), (2), (3) and multiply the results: $$(x+1)(x^2+1)(y+1)(y^2+1)(z+1)(z^2+1)=(y+1)(y^2-y+1)(z+1)(z^2-z+1)(x+1)(x^2-x+1).$$ Reducing by $(x+1)(y+1)(z+1)\ne 0$, we get $$(x^2+1)(y^2+1)(z^2+1)=(x^2-x+1)(y^2-y+1)(z^2-z+1).(\ast )$$ Note that if $x<-1$, then, according to (1), $y<-1$ and, according to (2), $z<-1$. In this case all factors in the right-hand side of () are greater than corresponding factors in the left-hand side — a contradiction. Now factor the left-hand sides of equations (1), (2) and (3), and subtracting 1 from both sides of these equations, we obtain the equations $$x(x^2+x+1)=y^3,(4)$$ $$y(y^2+y+1)=z^3,(5)$$ $$z(z^2+z+1)=x^3.(6)$$ Note that if $x>0$, then, according to (4), $y>0$ and, according to (5), $z>0$. In this case all factors in the right-hand side of () are less than corresponding factors in the left-hand side — a contradiction. It remains to consider the case, when $x,y$ and $z$ belong to the interval $(-1;0)$ and, in particular, are negative. Again, all factors in the right-hand side of (*) are greater than corresponding factors in the left-hand side. So, there are no other solutions.

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Alternative solution.

Consider the values of polynomials $x^3+x^2+x$ and $x^3$ on the intervals $(0;+\infty )$, $(-\infty ;-1)$ and $(-1;0)$. If $x>0$, then $x^3+x^2+x>x^3>0$. From the equations of the system successively obtain the inequalities $y>x>0$, $z>y>0$ and $x>z>0$, whence $x>z$. If $x<-1$, then $x^3+x^2<0$ and $x^2+x>0$, hence $-1>x^3+x^2+x>x^3$. From the equations of the system we obtain the inequalities $-1>y>x$, $-1>z>y$ and $-1>x>z$, and again $x>z$. If $x\in (-1;0)$, then $x^3+x^2>0$ and $x^2+x<0$, therefore $-1<x^3+x^2+x<x^3<0$. From the system we obtain the inequalities $-1>y>x$, $-1>z>y$ and $-1>x>z$, whence again $x>z$. In all three cases we got a contradiction and the two remaining variants $x=0$ and $x=1$ easily lead us to the triples from the answer.