6.79. Let e0,…,en−1 be the vectors of the sides of a regular n-gon. It is sufficient to prove that by reordering these vectors, we can obtain a set of vectors {a1,…,an} such that ∑k=1nkak=0. A number n, which is not a power of a prime number, can be represented as n=pq, where p and q are coprime numbers. We will now prove that the set {e0,ep,…,e(q−1)p;eq,eq+p,…,eq+(q−1)p,…,;e(p−1)q+p,…,e(p−1)q+(q−1)p} is the desired one. First, note that if x1q+y1p≡x2q+y2p(modpq), then x1≡x2(modp) and y1≡y2(modq), so in the considered set, each of the vectors e0,…,en−1 appears exactly once.
The endpoints of the vectors eq,eq+p,…,eq+(q−1)p with a common origin form a regular q-gon, so their sum is zero. Moreover, the vectors e0,ep,…,e(q−1)p transform into eq,eq+p,…,eq+(p−1)q upon rotation by an angle φ=2π/p. Therefore, if e0+2ep+…+qe(q−1)p=b, then (q+1)eq+(q+2)eq+n+…+2qeq+(q−1)p=q(eq+…+eq+(q−1)p)+eq+2eq+p+…+qeq+(q−1)p=Rφb, where Rφb is the vector obtained from b by rotation by an angle φ=2π/p. Similar reasoning shows that for the considered set of vectors, ∑k=1nkak=b+Rφb+…+R(p−1)φb=0.