Simple Arguments on Consecutive Power Residues

By some extremely simple arguments, we point out the following: (i) If n is the least positive kth power non-residue modulo a positive integer m, then the greatest number of consecutive kth power residues mod m is smaller than m/n. (ii) Let OK be the ring of algebraic integers in a quadratic field K = Q( √ d) with d ∈ {−1,−2,−3,−7,−11}. Then, for any irreducible π ∈ OK and positive integer k not relatively prime to ππ̄− 1, there exists a kth power non-residue ω ∈ OK modulo π such that |ω| < √ |π|+ 0.65. For an integer a relatively prime to a positive integer m, if the congruence x ≡ a (mod m) is solvable then a is said to be a kth power residue mod m, otherwise a is called a kth power non-residue mod m. The theory of power residues (cf. [L]) plays a central role in number theory. In this short note we aim to show that some classical topics on power residues can be handled just by some extremely simple observations. Our first observation concerns the least positive kth power non-residue modulo a positive integer. Theorem 1. (i) Suppose that n = nk(m) is the least positive kth power non-residue modulo a positive integer m. Then the greatest number R = Rk(m) of consecutive kth power residues mod m is smaller than m/n, consequently n < √ m + 1/2 if m is a prime. (ii) Let p be an odd prime, and let k be a positive integer with gcd(k, p− 1) > 1. Provided that −1 is a kth power residue mod p (i.e., (p− 1)/ gcd(k, p− 1) is even) and that nk(p) 6= 2, we have nk(p) < √ p/2 + 1/4. Let p be an odd prime. That n2(p) < √ p + 1 was first pointed out by Gauss. Using sophisticated analytic tools, A. Granville, R. A. Mollin and H. C. Williams [GMW] proved that if d > 3705 is a discriminant of a quadratic number field then 2000 Mathematics Subject Classification. Primary 11A15; Secondary 05A19, 11A07, 11R11. Partially supported by the National Science Fund for Distinguished Young Scholars (No. 10425103) and a Key Program of NSF (No. 10331020) in China. 1