Three rational numbers $x, y, z$ satisfy $x + y + z = 0$ and $xy + yz + zx = 0$. Show that all the rational numbers $x, y, z$ must be simultaneously zero.

Given, $x + y + z = 0$ and $xy + yz + zx = 0$

Since, $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.

Substituting the values, we have

$$0^2 = x^2 + y^2 + z^2 + 2(0)$$

$$\Rightarrow x^2 + y^2 + z^2 = 0.$$

Since squares of rational numbers are non-negative, and the sum of non-negative numbers is 0 only when each is 0,

$$x^2 = 0, y^2 = 0, z^2 = 0$$

$$\Rightarrow x = 0, y = 0, z = 0.$$

Hence, proved.