4-cycle decomposition of a complete graph Let $K_n$ be the complete graph with n vertices where the number of edges $\frac {n \times (n-1)} {2}$ is a multiple of 4. Can this graph be decomposed into 4-cycles? (i.e find a partition of the set of edges such that each subset of the partition contains a subset of the edges that form a 4-cycle).
I am particularly interested for the case where $n = 9$ but it would be interesting to know if there is any general theorem. Also, what happens in the case when instead of 1 edge between each pair of vertices we have $k$ edges instead?
 A: For the case of $K_9$, one such decomposition is drawn below: the cycles are
$$
   \{(0, 1, 2, 3), (0, 4, 1, 5), (0, 6, 1, 7), (0, 2, 4, 9), (1, 3, 5, 8), \\(2, 6, 3, 7), (2, 5, 6, 8), (3, 4, 7, 8), (4, 5, 7, 6)\}.
$$

Based on my comments to the other question, there is a general solution (found by a cooperative effort). For $K_{8k+1}$ we have a decomposition by taking the cycles $(i, i+4j, i+1, i+4j-2)$ for $i=0,1,\dots,8k$ and $j=1, \dots, k$, where all addition is done modulo $8k+1$.
The $n=8k+1$ case is the only case where a decomposition is possible. We need all vertex degrees to be even, since each cycle uses up $2$ edges out of its vertices, which means $n$ has to be odd. At the same time, the number of edges, which is $\frac{n(n-1)}{2}$, must be divisible by $4$, so $n(n-1)$ is divisible by $8$. Since $n$ is odd, this only happens if $n-1$ is divisible by $8$: if $n=8k+1$ for some $k$.
A: Draw a regular nonagon and note the the edges between vertices come in $4$ different lengths, $9$ of each length. So, if you can find a $4$ cycle that contains edges of all different lengths, then you can rotate that $4$ cycle by ninths of a full turn to get the rest of the cover.
Label the vertices $0$ through $8$ going around the nonagon. Find that $(0,1,5,2)$ meets our criteria. Now rotate by adding 1 to each vertex number mod $9$ to get:
$\{(0,1,5,2),(1,2,6,3),(2,3,7,4),(3,4,8,5),(4,5,0,6),(5,6,1,7),(6,7,2,8),(7,8,3,0),(8,0,4,1)\}$
