Let's determine the parameters $p$ and $q$ such that the roots of equation (1) are real (not necessarily distinct) and can be arranged in such a way that (2) is satisfied.

$$
x^{4} + p x^{2} + q = 0
$$

$$
\frac{x_{2}}{x_{1}} = \frac{x_{3}}{x_{2}} = \frac{x_{4}}{x_{3}}
$$
Let $c$ denote the common value of the quotients in (2), then

$$
x_{2}=c x_{1}, \quad x_{3}=c^{2} x_{1}, \quad x_{4}=c^{3} x_{1}
$$

Since in (1) $x$ only appears with even exponents, $x_{1}$ and $(-x_{1})$ together satisfy the equation. Therefore, one of $c x_{1}, c^{2} x_{1}$, and $c^{3} x_{1}$ is equal to $(-x_{1})$. Since $x_{1} \neq 0$—otherwise the first quotient in (2) would be undefined—some power of $c$ is $(-1)$, which implies $c=-1$, because no other real number has $(-1)$ as a power. Thus, for the roots of the equation, we have

$$
x_{1}=-x_{2}=x_{3}=-x_{4}
$$

Therefore, the quadratic equation in terms of the squares of the roots

$$
y^{2}+p y+q=0
$$

has $x_{1}^{2}$ as a double root, which is true if and only if the discriminant of (3) is 0, i.e.,

$$
p^{2}=4 q
$$

From this, $x_{1}^{2}=-p / 2$, and since $x_{1}$ is a real number,

$$
-\frac{p}{2}>0, \quad \text { i.e., } \quad p<0
$$

follows. If these conditions are satisfied, the original equation is

$$
x^{4}+p x^{2}+\frac{p^{2}}{4}=0, \quad \text { i.e., } \quad\left(x^{2}+\frac{p}{2}\right)^{2}=0
$$

Since $p$ is negative, the roots of the equation are, in the desired order,

$$
x_{1}=\sqrt{-\frac{p}{2}}, \quad x_{2}=-\sqrt{-\frac{p}{2}}, \quad x_{3}=\sqrt{-\frac{p}{2}}, \quad x_{4}=-\sqrt{-\frac{p}{2}}
$$

thus (2) is indeed satisfied.

Remark. The generalization of the problem: let the equation under consideration be

$$
x^{2 n}+p_{1} x^{2 n-2}+\ldots+p_{n-1} x^{2}+p_{n}=0
$$

and we want the roots to satisfy:

$$
\frac{x_{2}}{x_{1}}=\frac{x_{3}}{x_{2}}=\ldots=\frac{x_{2 n}}{x_{2 n-1}}
$$

From this, the roots are

$$
x_{1}, \quad c x_{1}, \quad c^{2} x_{1}, \quad \ldots, \quad c^{2 n-1} x_{1}
$$

where $x_{1} \neq 0$ and $c \neq 0$. Similarly to the above, it can be shown that $c=-1$, and thus the equation has $x_{1}$ and $(-x_{1})$ as $n$-fold roots. This means that on the left side of the equation,

$$
\left(x^{2}-x_{1}^{2}\right)^{n}
$$

stands, from which the coefficients can be easily determined.