Estonian Math Competitions

The first equation implies $x^3=-y^2-1$. Thus $x^3<0$, implying also $x<0$. Similarly from the second, third and fourth equations we obtain $y<0$, $z<0$ and $t<0$, respectively. The sum of negative numbers $x$, $y$, $z$, $t$ is negative, contradicting the fifth equation.

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Alternative solution.

Suppose that the system has a solution. W.l.o.g., let $x$ be variable with the largest value. Then $4x\ge x+y+z+t$, which by the last equation implies $x\ge 0$. Consequently, also $x^3\ge 0$. As $y^2\ge 0$, this implies $1+x^3+y^2\ge 1$, contradicting the first equation. Hence no solution can exist.