Solving system of equations for rational numbers

Question

Question: How many rational number ordered triples $(x, y, z)$ satisfying the equation system $x + y + z = 0, xyz + z = 0$ and $xy + yz + zx + y = 0$?

My partial solution:

First simplify the $2$nd equation to be $xyz = -z$ and the $3$rd equation to be $xy + yz + zx = -y$.

We then notice something about the equations. If we let $x, y$, and $z$ be the roots of a cubic polynomial, then by Vieta's, the polynomial would be $x^3 + 0x^2 - yx + z$.

We can simplify this cubic polynomial to $x^3 - yx + z$.

Hence, we just need to solve the equation $x^3 - yx + z = 0$ for the rational numbers $x, y$, and $z$.

However, this is the part on where I am stuck with the problem. Maybe I could set values for $x$ and then solve for $y$ and $z$? I don't know.

Any help is appreciated. Thank you in advance.

+1 to your question, for good work shown. It is now moot, because of Dietrich Burde's answer, but if no one had answered, I would have suggested cardano's formula as the place to start. This article may not have resolved the problem, but absent the rational root test, which I overlooked, it would have been the place to start. — user2661923, Jul 01 '21 at 15:36

score 2 · Accepted Answer · answered Jul 01 '21 at 14:45

2

Hint: The only rational solutions are $$ (x,y,z)=(0,0,0),(-1,1,0). $$ To see this note that the only other case yields $x= - (z^2 + z + 1)$, $y=z^2 + 1$ with $$ z^3+z^2+2z+1=0. $$ By the rational root test, this cubic polynomial has no rational root.

answered Jul 01 '21 at 14:45

Dietrich Burde

130,978

Thanks for the hint :) – I am a person Jul 01 '21 at 14:49

Solving system of equations for rational numbers

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