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:
 \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'

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.