Let's determine $p$ and $q$ such that $x^{4}+p x^{2}+q$ is divisible by $x^{2}-6 x+5$.

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Determine $p$ and $q$ such that $x^{4}+p x^{2}+q$ is divisible by $x^{2}-6 x+5$.
I. Solution. If we actually perform the division, then the quotient is: $x^{2}+6 x+p+31$ and the remainder: $(6 p+156) x+q-5 p-155$. This remainder can only be 0 for all values of $x$ if

$$
6 p+156=0
$$

and

$$
q-5 p-155=0
$$

From these two equations, $p=-26, \quad q=25$.

(Lukhaub Gyula, Szeged.)

II. Solution. $x^{4}+p x^{2}+q$ is equal to the product of two quadratic expressions; the roots of these expressions are also roots of the trinomial $x^{4}+p x^{2}+q$. But the roots of the equation $x^{2}+6 x+5=0$ are $x_{1}=1, x_{2}=5$, so the roots of the equation $x^{4}+p x^{2}+q=0$ are $x_{2}= \pm 1, x_{2}= \pm 5$. Therefore, $x^{4}+p x^{2}+q$ is 0 when $x^{2}$ is equal to 1 and 25; since $p$ is equal to the sum of the roots with the opposite sign, and $q$ is the product of the roots, therefore $p=-26$ and $q=25$.

(Filkorn Jenö, Nyitra.)

III. Solution.

$$
x^{4}+p x^{2}+q=\left(x^{2}-6 x+5\right)\left(x^{2}+a x+b\right)
$$

Performing the multiplication:

$$
x^{4}+p x^{2}+q=x^{4}+(a-6) x^{3}+(b-6 a+5) x^{2}+(5 a-6 b) x+5 b
$$

This equation can only hold for all values of $x$ if the corresponding coefficients are equal. Therefore,

$$
a-6=0 \text { and } 5 a-6 b=0
$$

From these equations, $a=6$ and $b=5$. Thus,

$$
6-6 a+5=-26=p
$$

and

$$
5 b=25=q
$$

(Sasvári Géza, Pécs.)

The problem was also solved by: Appel S., Barcsay Á., Bayer B., Bender E., Benedek Zs., Burján K., Csete F.A., Czank K., Engel D., Faith F., Fekete S., Jankovich S., Kerekes T., Kiss A., Kornis F., Krausz B., Krisztián Gy., Lupsa Gy., Ovenden S., Perl Gy., Pollák L., Rozlosnik P., Stern D., Stromfeld F., Szabó J., Szőllősy J., Sztranyavszky S., Téglás G., Weisz J.