Theorem in the Additive Number Theory

THEOREM. Each set of 2n-1 integers contains some subset of n elements the sum of which is a multiple of n. PROOF. Assume first n = p (p prime). Our theorem is trivial for p = 2, thus henceforth p > 2. We need the following LEMMA. Let p > 2 be a prime and A = {a,, a 2 ,. . ., a,} 2 5 s < p a s tegers each prime to p satisfying ca, a 2 (mod p). Then the set a i a + , s =10 or 1 contains at least s + 1 distinct congruence classes. +=1 We use induction. If s = 2, a, i a 2 , a, + a 2 are all incongruent (since a, * a2, a, * 0, a 2 * 0). Thus the lemma holds for s = 2. Assume that it holds for s-1, we shall prove it for s. Let b,, b 2 ,. . ., bk be all the congruence classes of the form a+ai. By assumption +=1 k z s. If k >-s + 1 there is nothing to prove. Thus we can : assume k. = s < p. But then since a, * 0 (mod p) it is easy to see (see e .g. [1]) that at least one of the integers b i + a " 1 5 i-<-k is incongruent to all the b's. Thus the number of integers of the s form a + a+, a+ = 0 or 1 is at least s + 1, which proves the Lemma. t=1 Let there be given 2p-1 residues (mod p). Arrange them according to size OSa,_5a 2 5. . .5a 2p-1 <p. ++p-1 We can assume a + 96 ai+p-, (for otherwise a; = pa i-0 (mod p)) and that j =i Y a t-c * 0 (mod p). Put b + = ap+i-ai+1, 1 < i 5 p-1. Clearly-c +=1 P-1 at b +, ai = 0 or 1 is solvable. If the b's are not all congruent this follows from +=1 our Lemma and if the b's are all congruent the statement is evident. Clearly set of s in-p p-1 a+ + ai bi-0 (mod p) i=1 +=1 is the sum of p a's. Thus our Theorem is proved for n = p. Now we prove that if our Theorem is true for n = u and …