Let $a, b, c$ be numbers, none of which are equal, and furthermore,

$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=p
$$

Determine the possible values of $p$!
From the condition $a+\frac{1}{b}=b+\frac{1}{c}$, after rearrangement,

$$
b c(a-b)=b-c
$$

is obtained. Similarly, we get

$$
\begin{aligned}
a c(b-c) & =c-a \\
a b(c-a) & =a-b
\end{aligned}
$$

Multiplying these equalities,

$$
(a b c)^{2}(a-b)(b-c)(c-a)=(b-c)(c-a)(a-b)
$$

From this, using the fact that $a, b$, and $c$ are pairwise distinct, $a^{2} b^{2} c^{2}=1$ and thus $|a b c|=1$.

Since $a+\frac{1}{b}=p$, after multiplying by $b$, we get $a b+1=b \cdot p$, and similarly $b c+1=c \cdot p$.

Subtracting these,

$$
b(a-c)=p(b-c)
$$

Multiplying this by (2),

$$
a b c(a-c)(b-c)=p(b-c)(c-a)
$$

According to the condition,

$$
p=-a b c
$$

Thus, the possible values of $p$ are 1 and -1.

It remains to show that these cases can indeed occur, that is, we can choose the numbers $a, b, c$ such that $p=1$, and also such that $p=-1$. For the first case, an example is $a=-1, b=1 / 2, c=2$; for the second case, $a=1, b=-1 / 2, c=-2$.