Solve the following equation:

$$
(3+x)^{4}+(2+x)^{4}=337
$$
I. Solution. Let

$$
(3+x)=y, (2+x)=z
$$

then

$$
y^{4}+z^{4}=337
$$

and

$$
y-z=1
$$

Raising (2) to the fourth power and subtracting from (1) yields:

$$
4 y^{3} z-6 y^{2} z^{2}+4 y z^{3}=336
$$

or

$$
y z\left(2 y^{2}-3 y z+2 z^{2}\right)=168
$$

but

$$
2(y-z)^{2}=2 y^{2}-3 y z+2 z^{2}-y z
$$

so

$$
2 y^{2}-3 y z+2 z^{2}=2+y z
$$

thus

$$
y z(2+y z)=168
$$

from which

$$
(y z)_{1}=12, (y z)_{2}=-14
$$

that is

$$
\begin{gathered}
(x+3)(x+2)=12 \\
(x+3)(x+2)=-14
\end{gathered}
$$

from which

$$
x_{1}=1, x_{2}=-6, x_{3}=-\frac{1}{2}(5+i \sqrt{55}), x_{4}=-\frac{1}{2}(5-i \sqrt{55})
$$

(Miklós Kamenitzky, Eperjes.)

II. Solution. Let $x=y-\frac{5}{2}$; then the equation takes the following form:

$$
\left(y+\frac{1}{2}\right)^{4}+\left(y-\frac{1}{2}\right)^{4}=337
$$

Performing the exponentiation, we get

$$
2 y^{4}+3 y^{2}-336 \frac{7}{8}=0
$$

from which

$$
y_{1}=\frac{7}{2}, y_{2}=-\frac{7}{2}, y_{3}=\frac{1}{2} i \sqrt{55}, y_{4}=-\frac{1}{2} i \sqrt{55}
$$

and

$$
x_{1}=1, x_{2}=-6, x_{3}=-\frac{1}{2}(5-i \sqrt{55}), x_{4}=-\frac{1}{2}(5+i \sqrt{55})
$$

(Géza Messik, Budapest.)

Number of solutions: 41.

General solution of the problem:
I.

$$
\begin{gathered}
(x+a)^{4}+(x+b)^{4}=c \\
x+a=y, x+b=z
\end{gathered}
$$

then

$$
\begin{gathered}
y^{4}+z^{4}=c \\
y-z=a-b
\end{gathered}
$$

from these two equations we find the value of $(y z)$ and from the value of $(y z)$ we obtain the four roots of $x$.

II. Let $x=y-\frac{a+b}{2}$, then

$$
\begin{gathered}
x+a=y+\frac{a-b}{2}, x+b=y-\frac{a-b}{2} \\
\left(y+\frac{a-b}{2}\right)^{4}+\left(y-\frac{a-b}{2}\right)^{4}=c
\end{gathered}
$$

performing the exponentiation, we get a quartic equation which can be reduced to a quadratic equation.