\documentclass{article}
\def\Sum{\displaystyle\sum}
\def\Prod{\displaystyle\prod}
\begin{document}
The $n$-tuple $(x_1, x_2, \dots, x_n)$ is a \emph{Markov n-tuple} if $\Sum_{i=1..n}
x_i^2 = n\Prod_{i=1..n} x_i$. Notice that the $n$-tuple $(1,1, \dots,
1)$ satisfies this relation. Notice also that if we replace $x_k$ by
$x_k' = \displaystyle\frac{1}{x_k}\cdot \Sum_{i\not=k} x_i^2 = \frac{1}{x_k}
\cdot (n(\Prod_{i=1..n} x_i) -x_k^2) = n\Prod_{i\not=k}x_i - x_k$, then the
new $n$-tuple also satisfies the relation:
\begin{eqnarray*}
\sum_{i \not= k} x_i^2 + x_k'^2
& = & \sum_{i \not= k} x_i^2 + (n\prod_{i \not= k}x_i - x_k)^2
\\
& = & \sum_{i \not= k} x_i^2 + x_k^2 - 2n\prod_{i=1..n}x_i
+n^2(\prod_{i \not= k}x_i)^2 \\
& = & \sum_{i=1..n} x_i^2- 2n\prod_{i=1..n}x_i +n^2(\prod_{i
\not= k}x_i)^2 \\
& = & n\prod_{i=1..n}x_i - 2n\prod_{i=1..n}x_i +n^2(\prod_{i
\not= k}x_i)^2 \\
& = & n^2(\prod_{i \not= k}x_i)^2- n\prod_{i=1..n}x_i \\
& = & n\prod_{i\not=k}x_i \cdot (n\Prod_{i\not=k}x_i - x_k)\\
& = & n\prod_{i\not=k}x_i \cdot x_k'
\end{eqnarray*}
Claim: Each $n$-tuple of integers that satisfy the above relation can be
obtained from the $n$-tuple $(1, 1, \dots , 1)$ by a series of switches.
Proof: Let $(x_1, x_2, \dots , x_n)$ be any $n$-tuple that satisfies the
relation. We may assume that $x_1$ is the largest among them.
Following the example of the Markov triples, we can think of a set of Markov
$n$-tuples sitting at the vertices of an infinite graph in which each vertex
has degree $n$. Each of the edges coming from a vertex corresponds
to exactly one of the numbers $x_i$ and moving across this edge to another
vertex corresponds to changing $x_i$ to $x_i'$.
This graph is exactly the Cayley graph of the group $G =(\bf{Z} /2)^{*n}=\bf{Z}
/2*\bf{Z} /2* \cdots *\bf{Z} /2$, the free product of $\bf{Z} /2$, the cyclic
group of order 2, with itself $n$ times.
\end{document}