1.8 Let $a<b<c<d$ and $(x, y, z, t)$ be any permutation of $(a, b, c, d)$. How many different values can the expression
$$
n=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-x)^{2}
$$
take?
$$
\begin{aligned}
& \text{[Solution 1]} \text{ Since the expression for } n \text{ is cyclically symmetric with respect to } (x, y, z, t), \text{ we can set } x=a. \\
& \text{Also, since the expression is symmetric with respect to } y \text{ and } t, \text{ for the 6 different permutations of } (b, c, d), n \text{ can take at most 3 different values.} \\
& \text{On the other hand, since } a>0, \\
& n(a, c, b, d)-n(a, b, c, d) \\
= & 2(a b+c d-a c-b d)=2(c-b)(d-a)>0,
\end{aligned}
$$

This implies
$$
n(a, c, b, d)>n(a, b, c, d)>n(a, b, d, c),
$$

which means that $n$ can take 3 different values.
［Solution 2］Note that the expression
$$
\begin{aligned}
& n(x, y, z, t)+(x-z)^{2}+(y-t)^{2} \\
= & n(x, y, z, t)+\left(x^{2}+y^{2}+z^{2}+t^{2}\right)-2(x z+y t)
\end{aligned}
$$

is symmetric with respect to $(x, y, z, t)$, so its value is independent of the permutation of $(x, y, z, t)$ being $(a, b, c, d)$. Since the second term on the right side of the above equation is also symmetric with respect to $(x, y, z, t)$, the number of different values that $n(x, y, z, t)$ can take is equal to the number of different values that $x z+y t$ can take.

The value of the expression $v=x z+y t$ depends on how $(a, b, c, d)$ is divided into two pairs, and there are exactly 3 such divisions: $\{a, b ; c, d\},\{a, c ; b, d\},\{a, d ; b, c\}$. Therefore, $v$ has 3 values:
$$
v_{1}=a b+c d, v_{2}=a c+b d, v_{3}=a d+b c .
$$

It is easy to verify that $v_{1}>v_{2}>v_{3}$, hence $n$ can take 3 different values.