We now show that the necessary conditions are also sufficient. Since Pk=Sk for k=1,2, the result holds for k=1,2 by Proposition 1. So it remains to consider the case k≥3. The proof is divided into cases n≥2k, n=k and k<n<2k, which are treated in Lemmas 4, 5, and 6, respectively.

Proof.
Suppose that n=k+r. Then, 0<r<k from the assumption k<n<2k. Let A1={a0,a1,…,ak-1}, B1={b0,b1,…,bk-1}, A2={ak,ak+1,…,ak+r-1}, and B2={bk,bk+1,…,bk+r-1}. Let Gi=Kn,n[Ai∪B1] for i=1,2 and H=Kn,n[A∪B2]. Then, Kn,n=G1+G2+H. Note that G1 is isomorphic to Kk,k, H is isomorphic to Kk+r,r, and G2 is isomorphic to Kr,k, which is Sk-decomposable by Proposition 1. Hence, it is sufficient to show that G1+H is (Pk,Sk)-multidecomposable.

Let t=r2/k. Since k∣n2, we have k∣r2, which implies that t is a positive integer Let C=(b0,a0,b1,a1,…,bk-2,ak-2,bk-1,ak-1). Then, C is a 2k-cycle in G1. Let p=⌊t/2⌋. For odd t, define a k-path P in G1 as follows:
(3)P={b2pa0b2p+1a1⋯b2p-1+k/2ak/2-1b2p+k/2if k is even,b2pa0b2p+1a1⋯b2p+(k-1)/2a(k-1)/2if k is odd,
where the subscripts of b are taken modulo k. Since t=r2/k and r<k, we have
(4)t≤r-1≤k-2.
Thus, 2p+1=t≤k-2, which implies the labels of the edges in P are 2p and 2p+1. Note that for i=0,1,…,p-1, C+2i is a 2k-cycle which consists of all of the edges with labels 2i and 2i+1 in G1. Thus, C,C+2,…,C+2(p-1) and P are edge-disjoint in G1.

Define a subgraph W of G1 as follows:
(5)W={∑i=0p-1C+2ifor even t,∑i=0p-1C+2i+Pfor odd t≥3,Pfor t=1.
Since C2k can be decomposed into 2 copies of Pk and 2p=t for even t as well as 2p+1=t for odd t, W can be decomposed into t copies of Pk. Let δ=1 for even k and δ=0 for odd k. Note that for even t, degG1-E(W)ai=k-2p=k-t, and for odd t,
(6)degG1-E(W)ai ={k-t-1if i=0,1,…,⌈k2⌉-2,k-t-δif i= ⌈k2⌉-1,k-t+1if i=⌈k2⌉,⌈k2⌉+1,…,k-1.
Let Xi=(G1-E(W))[{ai}∪B1] for i=0,1,…,k-1. Then for even t, Xi=Sk-t, and for odd t,
(7)Xi={Sk-t-1if i=0,1,…,⌈k2⌉-2,Sk-t-δif i=⌈k2⌉-1,Sk-t+1if i=⌈k2⌉,⌈k2⌉+1,…,k-1
with the center at ai. In the following, we will show that H can be decomposed into r copies of Sk with centers in B2, and into k copies of St with centers in A1 for even t, and into ⌈k/2⌉-1 copies of St+1 with centers in {a0,a1,…,a⌈k/2⌉-2}, an St+δ with the center at a⌈k/2⌉-1, and k-⌈k/2⌉ copies of St-1 with centers in {a⌈k/2⌉,a⌈k/2⌉+1,…,ak-1} for odd t.

We show the required star decomposition of H by orienting the edges of H. For any vertex x of H, the outdegree degH+x (indegree degH-x, resp.) of x in an orientation of H is the number of arcs incident from (to, resp.) x. It is sufficient to show that there exists an orientation of H such that
(8)degH+bj=k,
where j=k,k+1,…,k+r-1, and for even t(9)degH+ai=t,
where i=0,1,…,k-1, and for odd t(10)degH+ai={t+1if i=0,1,…,⌈k2⌉ -2,t+δif i=⌈k2⌉-1,t-1if i=⌈k2⌉,⌈k2⌉+1,…,k-1.

We first consider the edges oriented outward from A1 according to the parity of t. Let β=k+(t+1)(⌈k/2⌉-1) and γ=β+t+δ. If t is even, then the edges aibk+ti,aibk+ti+1,…,aibk+ti+t-1 are all oriented outward from ai, where i=0,1,…,k-1. If t is odd, then the edges aibk+(t+1)i,aibk+(t+1)i+1,…,aibk+(t+1)i+t for i=0,1,…,⌈k/2⌉-2, and a⌈k/2⌉-1bβ,a⌈k/2⌉-1bβ+1,…, a⌈k/2⌉-1bβ+t+δ-1, as well as aib(t-1)(i-⌈k/2⌉)+γ,aib(t-1)(i-⌈k/2⌉)+γ+1,…,aib(t-1)(i-⌈k/2⌉)+γ+t-2 for i=⌈k/2⌉,⌈k/2⌉+1,…,k-1 are all oriented outward from ai. In both cases, the subscripts of b are taken modulo r in the set of numbers {k,k+1,…,k+r-1}. Note that for even t we orient t edges from each ai and for odd t we orient at most t+1 edges from ai. By inequality (4), we have t+1≤r, which assures us that there are enough edges for the above orientation.

Finally, the edges which are not oriented yet are all oriented from B2 to A. From the construction of the orientation, it is easy to see that (9) and (10) are satisfied, and for all bw,bw′∈B2, we have
(11)|degH-bw-degH-bw′|≤1.
So, we only need to check (8).

Since degH+bw+ degH-bw=k+r for bw∈B2, it follows from (11) that
(12)|degH+bw-degH+bw′|≤1
for bw,bw′∈B2. Note that for even t, ∑i=0k-1degH+ai=tk, and for odd t,
(13)∑i=0k-1degH+ai=(t+1)(⌈k2⌉-1)+t+δ+(t-1)(k-⌈k2⌉)=(t-1)k+2(⌈k2⌉)-1+δ={(t-1)k+2(k2)-1+1if k is even,(t-1)k+2(k+1)2 -1if k is odd.=tk.
Thus,
(14)∑w=kk+r-1degH+bw=|E(Kk+r,r)|-∑i=0k-1degH+ai=(k+r)r-tk=kr+r2-r2=kr.
Therefore from (12), we have degH+bw=k for bw∈B2. This establishes (8). Hence, there exists the required decomposition 𝒟 of H. Let Xi′ be the star with center at ai in 𝒟 for i=0,1,…,k-1. Then, Xi+Xi′ is a k-star. This completes the proof.

Now, we are ready for the main result. It is obtained form the arguments above, Lemma 4 and Lemmas 3, 4, 5, and 6.