$$
x^{2}-m x-1=0
$$

The roots of the equation are $a$ and $b$;

$$
x^{2}+m x-1=0
$$

The roots of the equation are $c$ and $d$; arrange these roots in order of magnitude without solving the equations, if $m>0$.
$$
x^{2}-m x-1=0
$$

$$
x^{2}+m x-1=0
$$

Since the constant term is negative in both equations, each equation has real roots of different signs; $a$ and $c$ are positive, $b$ and $d$ are negative. Since the coefficient of $x$ in (1) is negative, the positive root is greater in magnitude than the absolute value of the negative root, that is,

$$
|a|>|b|
$$

However, for two quadratic equations that differ only in the sign of the first-degree term, their roots also differ only in sign, and thus

$$
|a|=|d|
$$

and

$$
|b|=|c|
$$

Substituting (4) and (5) into (3):

$$
|d|>|b|
$$

$$
|a|>|c|
$$

Since both $b$ and $d$ are negative, it follows from (6) that

$$
|d|>|b|
$$

From (8) and (9), it follows that

$$
|a|>|c|>|b|>|d| \text {. }
$$

(Bernát Friedmann).

The problem was also solved by: Brandt D., Dénes A., Erdős A., Goldziher K., and Kármán T., Manheim E., Prode Gy., Spitzer Ö., Szabó I., Szabó K.