\section*{Problem 4 - 111244}

a) Determine all ordered triples \((x, y, z)\) of real numbers that satisfy the equation \(x^{3} z + x^{2} y + x z + y = x^{5} + x^{3}\).

b) Among the triples found in a), identify those in which exactly one of the three numbers \(x, y, z\) is positive, exactly one is negative, and exactly one is zero.
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a) The equation can be equivalently transformed to \(\left(x^{2}+1\right)\left(x z+y-x^{3}\right)=0\). The first factor is positive for all real numbers. Therefore, it suffices to consider the second factor. For \(x=0\), \(y=0\) and \(z \in \mathbb{R}\) can be arbitrary, so we have the triples \((0,0, z)\). For \(x \neq 0\), we can solve the equation for \(z\) to get \(z=x^{2}-\frac{y}{x}\) and obtain the solution triple \(\left(x, y, x^{2}-\frac{y}{x}\right)\) for any \(x, y \in \mathbb{R}\).

b) From \(x=0\) it follows that \(y=0\). From \(z=0\) it follows that \(y=x^{3}\). Therefore, the other two coordinates have the same sign. Hence, only \(y=0\) remains. In this case, \(z=x^{2}\). Only for \(x0\) and thus \(x<0\) follows.

Therefore, only triples \(\left(x, 0, x^{2}\right)\) with \(x<0\) can have the properties required in a) and b). A check shows that each such triple indeed has these properties.

Taken from \([2]\)