5・17 Let $a<b<c<d$. If the variables $x, y, z, t$ are some permutation of the numbers $a, b, c, d$, how many different values can the expression
$$
n=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-x)^{2}
$$

take?
【Sol】Let $n(x, y, z, t)=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-$ $x)^{2}$. Since the expression
$$
\begin{aligned}
& n(x, y, z, t)+(x-z)^{2}+(y-z)^{2} \\
= & n(x, y, z, t)+\left(x^{2}+y^{2}+z^{2}+t^{2}\right)-2(x z+y t)
\end{aligned}
$$

is independent of the permutation of the numbers $a, b, c, d$ chosen as the values of the variables, and $x^{2}+y^{2}+z^{2}+t^{2}$ also takes the same value for any permutation of the values $a, b, c, d$, therefore the value of $n$ depends only on the value of $V=x z+y t$, and reaches its maximum and minimum values together with it.
$V$ can take three different values:
$$
V_{1}=a b+c d, V_{2}=a c+b d, V_{3}=a d+b c .
$$

and satisfies the inequalities $V_{1}>V_{2}>V_{3}$, because
$$
\begin{array}{l}
V_{1}-V_{2}=(d-a)(c-b)>0, \\
V_{2}-V_{3}=(b-a)(d-c)>0 .
\end{array}
$$

Therefore, the values taken by $n$ can be arranged in size as follows:
$$
n(a, c, b, d)>n(a, b, c, d)>n(a, b, d, c) .
$$