Let's determine the value system of the coefficients $p, q, r$ (and $s$) under which the following products do not contain terms with odd powers of $x$ in their polynomial form:
a) $\left(x^{3}-x^{2}-6 x+2\right)\left(x^{3}+p x^{2}+q x+r\right)$
b) $\left(x^{4}+c x^{3}+d x^{2}+e x+f\right)\left(x^{4}+p x^{3}+q x^{2}+r x+s\right)$.

Write the polynomial form of the products for the found value systems.
For the polynomial forms, the terms containing $x$ with an odd exponent must have a coefficient of 0 in the product. This condition provides a system of equations for the sought coefficients. Writing this for the coefficients of the 5th, 3rd, and 1st degree terms:

$$
-1+p=0, \quad 2-6 p-q+r=0, \quad 2 q-6 r=0
$$

From here, $p=1$, then $q=-6$ and $r=-2$, and finally the second factor of (1) and the polynomial form of the product:

$$
x^{3}+x^{2}-6 x-2, \quad \text { or } \quad x^{6}-13 x^{4}+40 x^{2}-4
$$

In the case of (2), similarly, we need to solve the following system:

$$
\begin{gathered}
c+p=0 \\
e+d p+c q+r=0 \\
f p+e q+d r+c s=0 \\
f r+e s=0
\end{gathered}
$$

In any case, $p=-c$; eliminating $r$ from (3) for $q$ and $s$, the system

$$
\begin{aligned}
(e-c d) q+c s & =c f-c d^{2}+d e \\
-c f q+e s & =-c d f+e f
\end{aligned}
$$

must be satisfied. Subtracting $c$ times (5a) from $e$ times (4a):

$$
\left(c^{2} f-c d e+e^{2}\right) q=\left(c^{2} f-c d e+e^{2}\right) d
$$

First, consider the case where

$$
c^{2} f-c d e+e^{2}=K
$$

is not 0. In this case, the system (4a), (5a) has a unique solution: $q=d$, and furthermore $s=f$ (since at least one of the coefficients $c$ and $e$ is not 0, otherwise $K=0$ would hold). With these, from (3) $r=-e$, and thus the second factor of (2) and the polynomial form are

$$
\begin{gathered}
g(x)=x^{4}-c x^{3}+d x^{2}-e x+f \\
x^{8}+\left(2 d-c^{2}\right) x^{6}+\left(d^{2}-2 c e+2 f\right) x^{4}+\left(2 d f-e^{2}\right) x^{2}+f^{2}
\end{gathered}
$$

If $K=0$, then for $c \neq 0$, from (4a) and (3):

$$
\begin{gathered}
s=f-d^{2}+\frac{d e}{c}+\left(d-\frac{e}{c}\right) q=f+\left(d-\frac{e}{c}\right)(q-d) \\
r=-e-c(q-d)
\end{gathered}
$$

Using (8), the second factor of (2) can be written as:

$$
g_{1}(x)=g(x)+(q-d)\left(x^{2}-c x+d-\frac{e}{c}\right)
$$

for any $q$, and it is easily verifiable that the expansion of (2) in this case also contains only even powers of $x$.

If $K=0$ and $c=0$, then from these $e=0$ and $p=0$, so $s$ does not appear in the conditions (3) - (5), hence the value of $s$ is arbitrary, and from (3) $r=0$. For this case, (2) becomes:

$$
\left(x^{4}+d x^{2}+f\right)\left(x^{4}+q x^{2}+s\right)
$$

in this form, odd powers of $x$ do not appear either.

László Surányi (Budapest, Fazekas M. gyak. g. II. o. t.)

Remark. By separating the even and odd terms of the individual factors, it is easy to separate the even and odd terms of the product as well. In the first case,

$$
\begin{aligned}
& {\left[x\left(x^{2}-6\right)-\left(x^{2}-2\right)\right]\left[x\left(x^{2}+q\right)+\left(p x^{2}+r\right)\right]=x^{2}\left(x^{2}-6\right)\left(x^{2}+q\right)-} \\
& \quad-\left(x^{2}-2\right)\left(p x^{2}+r\right)+x\left\{\left(x^{2}-6\right)\left(p x^{2}+r\right)-\left(x^{2}-2\right)\left(x^{2}+q\right)\right\}
\end{aligned}
$$

Here, after simplifying the terms in the curly braces, all terms must cancel out, so the polynomial must vanish for all values of $x$. By substituting 0, 2, and 6 for $x^{2}$, $p, q, r$ can be easily calculated.

Similarly, from the second product, it follows that

$$
\left(c x^{2}+e\right)\left(x^{4}+q x^{2}+s\right)+\left(x^{4}+d x^{2}+f\right)\left(p x^{2}+r\right)
$$

must be identically 0. Here, it is natural to substitute $-e / c$ for $x^{2}$. Then the first factor of the second product is $e^{2} / c^{2}-d e / c+f=\left(e^{2}-c d e+c^{2} f\right) / c^{2}$, which sheds light on the role of the disappearance of this expression in solving the problem.