Decomposition of complete graph $K_{n}$ into cycles when $n$=9 Clearly, in a complete graph $K_n$, number of edges $=\dfrac{n(n-1)}{2}$ 
so number of such cycle produced during decomposition= $\dfrac{n(n-1)}{2n}=\dfrac{n-1}{2}$
 clearly for $n=9$,
number of  decomposable  graph $=4$.
then why in graph theory by West has written that this construction fails for $n=9$
i am writing the solution from begining as
For $n = 5$ and $n = 7$, it sufﬁces to use cycles formed by traversing the
vertices with constant difference:
$(0,1,2,3,4)$ and $(0,2,4,1,3)$ for $n = 5$
$ (0, 1, 2, 3, 4, 5, 6)$, $(0, 2, 4, 6, 1, 3, 5)$, and $(0, 3, 6, 2, 5, 1, 4)$ for $n = 7$.
This construction fails for $n = 9$ since the edges with difference $3$ form
three $3$-cycles. The cyclically symmetric construction below treats the ver-
tex set as $\mathbb{Z}$8 together with one special vertex.
Please help...
 A: Suppose that you start at vertex $0$ and always go from vertex $k$ to vertex $(k+d)\bmod 9$ for some fixed distance $d$. When $d=1$ your path is
$$0\to 1\to 2\to 3\to 4\to 5\to 6\to 7\to 8\to 0\;,$$
represented by the cycle $(0,1,2,3,4,5,6,7,8)$. When $d=2$ it’s
$$0\to 2\to 4\to 6\to 8\to 1\to 3\to 5\to 7\to 0\;,$$
represented by the cycle $(0,2,4,6,8,1,3,5,7)$. (Note that $8+2=10$, and $10\bmod 9=1$.) So far we have two of the desired four $9$-cycles. When we try $d=3$, however, we don’t get a $9$-cycle: we get the path
$$0\to 3\to 6\to 0\;,$$
corresponding to the $3$-cycle $(0,3,6)$. The problem is that $9$, the desired cycle length, is a multiple of $d$ in this case. Whenever that happens, taking ‘steps’ of length $d$ will close the cycle too early. (In fact the cycle will close too early whenever $\gcd(n,d)>1$.)
We can go ahead and try $d=4$, and it works fine, because $\gcd(9,4)=1$; we get the path
$$0\to 4\to 8\to \to 3\to 7\to 2\to 6\to 1\to 5\to 0\;,$$
corresponding to the cycle $(0,4,8,3,7,2,6,1,5)$, and we now have three of the four desired $9$-cycles. What happens if we keep trying larger values of $d$?
When $d=5$ we get the path
$$0\to 5\to 1\to 6\to 2\to 7\to 3\to 8\to 4\to 0\;.$$
It’s the right length, but it’s not a new cycle: it’s simply the $d=4$ cycle in reverse, so it uses exactly the same edges of $K_9$. This is because $5\equiv-4\pmod 9$, so a step size of $5$ is the same as a step size of $-4$, which of course simply runs through the $d=4$ path in reverse. Similarly, $d=6$ produces the path
$$0\to 6\to 3\to 0\;,$$
the reversal of the $d=3$ path, since $6\equiv-3\pmod9$. I’ll leave it to you to check that $d=7$ produces the reversal of the $d=2$ path, and $d=8$ the reversal of the $d=1$ path. Of course $d=0$ goes nowhere. Values of $d$ that are congruent modulo $9$ produce exactly the same path, and every integer is congruent modulo $9$ to one of the integers $0,1,2,3,4,5,6,7$, and $8$, so we have found all of the paths that can be produced by taking ‘steps’ of a constant size $d$. There are five of them; three are $9$-cycles, one is a $3$-cycle, and one is a $1$-cycle. This approach therefore cannot produce a decomposition of $K_9$ into four $9$-cycles.
