Solve the equation $x^{3}+p x+q=0$ if the product of two roots is 1.
Let the roots be $a, b, c$. The left side of the equation is the product of the root factors:

$$
\begin{gathered}
(x-a)(x-b)(x-c) \equiv x^{3}+p x+q \\
x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c \equiv x^{3}+p x+q
\end{gathered}
$$

This identity means that the coefficients of equal powers of $x$ are equal, so:

$$
\begin{aligned}
& a+b+c=0 \ldots \\
& a b+a c+b c=p \ldots \\
& a b c=-q \ldots
\end{aligned}
$$

According to our assumption: $a b=1$, so from 3) $c=-q$ and from 1): $a+b=-c=q$.

From 2), $1+c(a+b)=p$ or $1-q^{2}=p \ldots$

This means that the coefficients $p$ and $q$ must satisfy the relationship 4) if the product of two roots is 1.

Now, if $a b=1$ and $a+b=q$, then $a$ and $b$ are the roots of the equation $u^{2}-q u+1=0$ and thus

$$
a=\frac{1}{2}\left(q \pm \sqrt{q^{2}-4}\right), \quad b=\frac{1}{2}\left(q \mp \sqrt{q^{2}-4}\right), \quad c=-q .
$$

Vilmos Csuri (State Kossuth Lajos g. VII. o. Pestszenterzsébet)