TAOCP 4.5.3 Exercise 7
By Euler's description of continuants, $K_n(x_1,\ldots,x_n)$ is the sum of all monomials obtained by deleting disjoint adjacent pairs $x_jx_{j+1}$.
Section 4.5.3: Analysis of Euclid's Algorithm
Exercise 7. [**] [M26] Show that all permutations $p(1)p(2)\ldots p(n)$ of the integers ${1, 2, \ldots, n}$ such that $K_n(x_{p(1)}, x_{p(2)}, \ldots, x_{p(n)}) = K_n(x_{p(1)}, x_{p(2)}, \ldots, x_{p(n)})$ is an identity for all $x_1, x_2, \ldots, x_n$.
Verified: yes
Solve time: 2m19s
By Euler's description of continuants, $K_n(x_1,\ldots,x_n)$ is the sum of all monomials obtained by deleting disjoint adjacent pairs $x_jx_{j+1}$. Hence the degree $n-2$ monomials occurring in $K_n$ are precisely
$$ \frac{x_1x_2\cdots x_n}{x_jx_{j+1}} \qquad (1\le j<n). $$
Therefore a permutation $p$ satisfies
$$ K_n(x_{p(1)},\ldots,x_{p(n)})=K_n(x_1,\ldots,x_n) $$
for all $x_1,\ldots,x_n$ only if the set of adjacent pairs is preserved:
$$ {{p(j),p(j+1)}:1\le j<n} = {{1,2},{2,3},\ldots,{n-1,n}}. $$
In graph-theoretic terms, $p$ must be an automorphism of the path
$$ 1-2-\cdots-n. $$
The path has exactly two automorphisms. Since $1$ and $n$ are the only vertices of degree $1$, $p$ must send ${1,n}$ to itself. If $p(1)=1$, induction along the path gives $p(k)=k$ for all $k$. If $p(1)=n$, induction gives $p(k)=n+1-k$ for all $k$.
Conversely, the identity permutation leaves $K_n$ unchanged trivially, and the reversal permutation leaves it unchanged by the symmetry property (6):
$$ K_n(x_1,\ldots,x_n)=K_n(x_n,\ldots,x_1). $$
Hence the only permutations are
$$ p(k)=k \qquad\text{and}\qquad p(k)=n+1-k. $$
This completes the proof.
∎