Let's determine the coefficients $C$ and $D$ in the equation

$$
x^{4}+4 x^{3}-34 x^{2}+C x+D=0
$$

such that the roots of the equation form an arithmetic sequence.
I. solution: Denoting the difference of the sequence by $2 d$, the roots of the equation can be written in the following form:

$$
x_{1}=a-3 d, \quad x_{2}=a-d, \quad x_{3}=a+d, \quad x_{4}=a+3 d
$$

Based on the relationship between the roots and the coefficients, the coefficient of the cubic term is

$$
-\left(x_{1}-x_{2}+x_{3}+x_{4}\right)=-4 a=4
$$

the coefficient of the quadratic term is

$$
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=6 a^{2}-10 d^{2}=-34
$$

From (1) and (2), $a=-1, d= \pm 2$, and thus the roots are: $-7,-3,1,5$.

Again, based on the relationship between the roots and the coefficients,

$$
C=-\left(x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}\right)=-(21+105-35-15)=-76
$$

and

$$
D=x_{1} x_{2} x_{3} x_{4}=105
$$

László Csernyák (Kaposvár, Táncsics g. IV. o. t.)

II. solution: By substituting $x=y+k$, we eliminate the cubic term. In the expression $(y+k)^{4}+4(y+k)^{3}+\ldots$, the coefficient of $y^{3}$ is $4(k+1)$, which is 0 if $k=-1$.

By substituting $x=y-1$ in our equation, we get

$$
\begin{gathered}
(y-1)^{4}+4(y-1)^{3}-34(y-1)^{2}+C(y-1)+D=y^{4}-40 y^{2}+(76-C) y+ \\
+(D-C-37)=0
\end{gathered}
$$

The roots of this equation also form an arithmetic sequence, but since the coefficient of the cubic term is 0, the sum of the four roots must necessarily be 0, meaning the four roots are symmetrically placed around the 0 point.

Thus,

$$
y_{1}=-3 u, \quad y_{2}=-u, \quad y_{3}=u, \quad y_{4}=3 u
$$

The factorization by roots is

$$
\left(y^{2}-9 u^{2}\right)\left(y^{2}-u^{2}\right)=y^{4}-10 u^{2} y^{2}+9 u^{4}=0
$$

Thus, from comparing (1) and (2),

1.) $-10 u^{2}=-40$, so $u^{2}=4$,

2.) $76+C=0$, hence $C=-76$

3.) $D-C-37=9 u^{4}$, from which

$$
D=9 u^{4}+C+37=9 \cdot 4^{2}-76+37=105
$$

József Vigassy (Bp. I., Petőfi g. IV. o. t.)