On The Least K-th Power Non-Residue
RICHARD I-I. HUDS0:N University of South Carolina, U.S.A.
Abstract
L e t Ck(p) d e n o t e t h e g r o u p o f t h e k - t h p o w e r s ( m o d p ) , p a p r i m e w i t h (k, p -- 1) > 1.
A n e w e l e m e n t a r y r e s u l t for t h e l e a s t k - t h p o w e r n o n - r e s i d u e is g i v e n a n d t h e r e s u l t is a p p l i e d t o f i n d i n g a n e w e l e m e n t a r y b o u n d for t h e m a x i m u m n u m b e r o f c o n s e c u t i v e i n t e g e r s i n a n y c o s e t o f Ck(p).
1. Introduction
Throughout this paper k will be an integer ~ 2 and /9 a prime with (k,p -- 1) > 1. Let Ck(p) denote the group of k-th powers (modp) and let S be the m a x i m u m number of consecutive integers in a n y eoset of Cl~(p). Finally, let g(p, k) denote the least positive k-th power non-residue.
Vinogradov [12], A. Brauer [2], Davenport and ErdSs [5], l~6dei [1% Burgess [4], and the author [6], [7], have given estimates for S. In particular, Vinogradov [12]
showed t h a t there exists at least one quadratic non-residue in each set of 3[p ~/2] -- 1 consecutive integers (mod p).
A. Brauer [2], using purely elementary arguments, showed t h a t for every p and k,
S < (2p) 1/2 + 2.
(I.i)
Using deep and non-elementary algebraic arguments, Burgess [4] showed t h a t S = O(p I/4 log p), an enormously stronger result t h a n has been obtained using elementary methods alone.
Tile major contribution of this paper is to give a small improvement of the elementary upper bounds of Brauer and Reynolds [3], Skolem [ill, N~gell [9], and Rddei [10], for g(p, k). Namely, we show t h a t
g(p, k) < (p13) ~/2 d- 2 (1.2)
2 ~ 8 I:r Yl:. : K U D S O N
for all p except p = 23 and p = 71. While much stronger results are known using analytic methods, the proof of (1.2) for each prime p for which -- 1 is a k-th power residue is not only elementary, but synthetic, and possibly even simple enough to be included in a textbook on elementary number theory.
We also note t h a t (1.2) can be used to improve (1.1), for the author has given a purely elementary argument in [7] t h a t
s < m a x {p~/z + (3/4)2~/:p ~/~ + 2,
(1/2)(g(p, ~) + (g(p, ~)~ +4p)~/2)}. (1.3)
From (1.2) and (1.3) it follows t h a tS < ( ( % / ~ ~- 1)/2 %/'
3--)p :/2 + (p/3) '/4
-~ 2 ~1.3295 pl]2 _~ (p/3)1/4 _~_
2. (1.4) Results such as (1.2) and (1.4) are of interest almost solely for the method by which t h e y are obtained. However, t h e y have additional interest for small primes where results such as Burgess's m a y be un-informative.2. The proof of (1.2) and (1.4)
THEOREM 1. For each k _> 2 and p :/: 23 or 71,
g(p,
k) < (p/3) 1/2 -~ 2.Proof.
Henceforth, for simplicity, we denoteg(p, k)
by g. Let p be a prime for which -- 1 is a k-th power residue and assume t h a t g >(p/3) 1/2
~- 2 so t h a t g > 3. Then each integerI,
such t h a t( p - - g ) / 3 < I < ( p ~ - g ) / 3 ,
is a k-th power residue. Further, each odd integer J satisfying either of the following in- equalities is a k-th power residue.(p -- 2g)/3
< J < (p -- g)/3. (2.1)
(p + g)/3 < J -< (p ~-
2g)/3.(2.2)
For if, (2.1) holds, t h e n the positive integer (p -- 3J)/2 is less t h a n g, and since -- 1, 2, and 3 are k-th power residues, it follows t h a t J is a k-th power residue.
Likewise, if (2.2) holds, J is a k-th power residue since, then, ( 3 J -
p)/2
is a positive integer less t h a n g.Now the number of integers in the closed integer interval A --~ [(p -- 2g)/3, (p -~ 2g)/3] exceeds g and, consequently, A must contain a multiple of g, say
ag.
Note t h a t a > 0 for obviously
g < p/2.
Nowa < ( p / 3 ) 1/2-~ 1 < g - - 1. (2.3) For if
a > ( p / 3 ) ~/2+
1, thenag > ((p/3) lr2 + 1)((p/3) 1~2 + 2) = p/3 + (3p) ll2 + 2 > (p + 2g)/3
(2.4)
O:N" T]-I~ L E A S T K - T H :FOVC':ER : N O ~ - I ~ E S I D ~ 219 since it is well k n o w n t h a t g < p~/2. (See, for e x a m p l e , W e s t e r n a n d Miller [13, p p . x i - - x i i ] ) . T h u s a a n d a § 1 are ]c-th p o w e r residues. N o w ag =/= (p - - bg)/3, b = • 1, ~= 2, for ag = ( p - - bg)/3 ~ p = (3a § b)g, b u t p is p r i m e . H e n c e , ag is a n e v e n i n t e g e r l y i n g e i t h e r in t h e i n t e r v a l (2.1) or in t h e i n t e r v a l (2.2). F o r ag is a k - t h p o w e r n o n - r e s i d u e since a < g - - 1 and, c o n s e q u e n t l y , c a n n o t be an i n t e g e r in t h e i n t e r v a l [(p - - g)/3, (p § g)/3], n o r c a n it be a n o d d i n t e g e r in e i t h e r o f t h e i n t e r v a l s (2.1) or (2.2). I f ag lies in t h e i n t e r v a l (2.1) n o t e t h a t (a § 1)g is a n o d d n o n - r e s i d u e in t h e i n t e r v a l (2.2), a c o n t r a d i c t i o n . I f ag lies in t h e i n t e r v a l (2.2), t h e n (a - - 1)9 is a n o d d n o n - r e s i d u e in t h e i n t e r v a l (2.1), again, a c o n t r a d i c t i o n . W e conclude t h a t g < (p/3) ~/2 § 2 if - - 1 is a /c-th p o w e r residue.
I f - - 1 is a k - t h p o w e r n o n - r e s i d u e it follows f r o m [1] t h a t
g(p, k) < (2p) 2/~ § 3(2p) ~/5 § 1. (2.5) I t is e a s y to see t h a t (2/)) : / 5 § 3 (2p) ~/~ § 1 ~ (p/3) ' / 2 § 2 for p ~ 10 ~ a n d it c a n be v e r i f i e d f r o m e x i s t i n g tables, (see, for e x a m p l e , D. H . L e h m e r , E m m a L e h m e r , a n d D a n i e l S h a n k s [8]), t h a t g ~ (p/3) ~/~§ 2 for 7 1 ~ p ~ 10 ~.
THeOreM
2. S < + 1)/2 + (p/3) + 2.Proof. T h e r e s u l t is easily c h e c k e d if p ~ 71.
I f p > 71 we see f r o m T h e o r e m 1 t h a t
(1/2)(g -~ (g2 § 4p)~/2) < (1/2)((p/3)1/2 § 2 § (13/)/3 § 4(p/3) ~l: § 4) ~/2)
< (1/2)((p/3) + 2 § (13p/3) §
2(p/3) § 2) (2.6)
= ( ( ~ § 1)/2 ~ 3 ) p l / 2 § (p/3)1/4 § 2 ~ 1.3295 V p § (p/3) 1/r § 2, a n d t h e r e s u l t follows f r o m (1.3).
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Received M a y 26, 1972 Richard ~ . I-Iudson
University of South Carolina U.S.A.
