Solve the following equation:

$$
(x-2)^{4}+(x-1)^{4}=\frac{17}{4}(x-2)^{2}(x-1)^{2}
$$
I. solution. The given equation can also be written as:

$$
\left[(x-2)^{2}+(x-1)^{2}\right]^{2}-2(x-2)^{2}(x-1)^{2}=\frac{17}{4}(x-2)^{2}(x-1)^{2}
$$

or

$$
\left[(x-2)^{2}+(x-1)^{2}\right]^{2}=\frac{25}{4}(x-2)^{2}(x-1)^{2}
$$

Taking the square root of both sides:

$$
(x-2)^{2}+(x-1)^{2}= \pm \frac{5}{2}(x-2)(x-1)
$$

from which

$$
x^{2}-3 x=0 \quad \text { and } \quad 9 x^{2}-27 x+20=0
$$

the roots of these equations are:

$$
x_{1}=0, x_{2}=3, x_{3}=\frac{5}{3}, x_{4}=\frac{4}{3}
$$

(Vilmos Erdös, Budapest.)

II. solution. Divide both sides of the equation by the product $(x-2)^{2}(x-1)^{2}$, then we get:

$$
\left(\frac{x-2}{x-1}\right)^{2}+\left(\frac{x-1}{x-2}\right)^{2}=\frac{17}{4}
$$

from which, if

$$
\left(\frac{x-2}{x-1}\right)=y
$$

we have:

$$
4 y^{4}-17 y^{2}+4=0
$$

the roots of this equation are: $2,-2, \frac{1}{2},-\frac{1}{2}$.

Setting the fraction $\frac{x-2}{x-1}$ equal to each of these values, we again get:

$$
x_{1}=0, \quad x_{2}=3, \quad x_{3}=\frac{5}{3}, \quad x_{4}=\frac{4}{3}
$$

(Arthur Paunz, Pécs.)

Number of solutions: 44.