AustriaMO2013

We first note that $(0,0,0)$ is obviously a solution of the system of equations. We will now show that there are no others. Let $x=\frac{p}{q}$ with relatively prime integer values of $p$ and $q$ and $q>0$. We then have $$y={\left({\left(\frac{p}{q}\right)}^2+1\right)}^3-1=\frac{(p^2+q^2{)}^3-q^6}{q^6}=\frac{p^6+qQ}{q^6}=\frac{r}{q^6},$$ and this fraction cannot be simplified, since $p$ and $q$ are relatively prime. Further substitutions then yield $z=\frac{s}{q^{36}}$ and $x=\frac{t}{q^{216}}$, and since these fractions similarly cannot be simplified, $q^{216}=q=1$ follows. We see that $x$ (and also $y$ and $z$) must be integers. For integer values not equal to $0$, we have $(x^2+1{)}^3>x^2+1\ge x+1$, and since equality must hold if the three equations are multiplied, this yields a contradiction. We see that $(0,0,0)$ is indeed the only solution, as claimed. $\square$