On quadruples of consecutive 𝑘th power residues

Abstract

In [2] it was shown that A(k, 4)= oo for k? 1048909 and it was conjectured that A(k, 4) = cfor all k. In this paper we establish this conjecture with the following THEOREM. A(k, 4)-= co. PROOF. It suffices to prove the theorem for values of k which are prime. The proof makes use of the following proposition which is a special case of a result of Kummer [1 ] (see also [3]). PROPOSITION. Let k be a prime and let y', * * *, ye be an arbitrary sequence of kth roots of unity. Then there exist infinitely many primes p with corresponding kth power character X modulo p such that x (Pi) = -Y , 1< i _ n, where pi denotes the ith prime. Thus, for any n and prime k, there exists a prime p with corresponding kth power character X modulo p such that X(2) 5 1, x(pi) = 1, 2 < i? n. Now consider any four consecutive positive integers all less than pn. It is clear that exactly one of these integers must equal 2(2d+1) for some integer d. But we have %(2(2d + 1)) = X(2)X(2d + 1) = x(2) * 1 = 1 since 2d +1 is the product of odd primes less than pn. Therefore