CONSECUTIVE NUMBERS WITHTHESAMELEGENDRESYMBOL

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 130, Number 9, Pages S (0) Article electronically ublished on Aril 17, 00 CONSECUTIVE NUMBERS WITHTHESAMELEGENDRESYMBOL ZHI-HONG SUN (Communicated by David E. Rohrlich) Abstract. Let be an odd rime, R be a comlete set of residues (mod ). The goal of the aer is to determine all the values of n (n R ) such that ( ) ( n n+1 ) ( or n 1 ) ( n ) ( n+1 ) (,where ) is the Legendre symbol. 1. Introduction Let be an odd rime, ( ) be the Legendre symbol, let R be a comlete set of residues modulo. It is well known that (see [D]) n ( n) (n +1) } (1) 3 1,n R [ ] () n ( n ) (n +1) } 1 1, n R [ ], where [ ] is the greatest integer function. In this aer we construct two or three consecutive numbers with the same value of Legendre symbols by roving the following two theorems. Theorem 1. Let be an odd rime, R be a comlete set of residues (mod ), let g be a rimitive root of. Then n ( n) (n +1) } 1, n R x k xk (gk 1) g k (mod ), x k R, k 1,,...,[ 3 ] } n ( n) (n +1) } 1, n R y k yk (gk 1 1) g k 1 (mod ), y k R, k 1,,...,[ 1 } ]. Received by the editors February 7, Mathematics Subject Classification. Primary 11A15; Secondary 11A07. Key words hrases. Prime, Legendre symbol. 503 c 00 American Mathematical Society

2 50 ZHI-HONG SUN Theorem. Let be an odd rime, F Z/Z be the residue class ring modulo, letf F be the field with elements. If g is a generator of the cyclic subgrou of F ( F 0}) of order ( 1 ),then n ( n 1) (n) (n +1) }, n F ± ( 1 g ) +s g s 1 s 1,,...,[ 3 ] }. We remark that if 1 (mod ), then g is a rimitive root (mod ), if 3(mod)wemaytakeF a + bi a, b F } write g a + bi with a, b F a + b 1. In the aer we also establish the following result. Theorem 3. Let be an odd rime, n Z with n 0, ±1 (mod). Then (n 1 ) (n ) (n +1) (x +1) n x 3 x (mod ) for some x Z. Throughout this aer, we denote the set of integers by Z as usual, identify ( a+z )with( a )fora Z. For later convenience, we will also denote the Legendre symbol ( ) n by χ(n).. Proof of Theorem 1 For k 1,,...,( 3)/letm k R be given by m k (g k 1) /(g k )(mod). Then m k +1 (g k +1) /(g k ) (mod ). So χ(m k ) χ(m k +1) ( 1) k. If s, t 1,,..., 3 } with s t, then gs+t 1(mod) sog s g t (g s g t )/g s+t (mod ). This imlies that g s + g s g t + g t (mod ) so m s m t (mod ). Since n } [ 3 ] [ 1] 3 χ(n) χ(n +1),n R + by (1) (), we obtain } } n χ(n) χ(n +1),n R m1,m,...,m 3. This together with the fact that χ(m k )( 1) k yields the result. Remark 1. Let >3bearime,n Z with n(n + 1). It follows from Theorem 1 that (x 1) χ(n) χ(n +1) n (mod ) for some x Z. x UsingTheorem1onecanalsoderivethat χ(n)χ(n+1)1 n R χ(n)χ(n+1) 1 n R n n 3+χ( 1) 3 χ( 1) (mod ) (mod ).

3 CONSECUTIVE NUMBERS WITH THE SAME LEGENDRE SYMBOL Proof of Theorem For s 1,,...,[ 3 ]} let n s g ( χ( 1))/+s /(g s 1). Then n s F since g s 1. We claim that n s F. If 1 (mod ), then g 1 1sog g. Hence g F therefore n s F. If 3 (mod ), then g +1 g χ( 1) 1 hence g 1 g.sowehave g +1 +s +1 ( + g +s) g +1 +s +(g +1 +s ) tr(g +1 +s ) F, where tr( ) is the trace function. Now, using the above the fact that g (+1)/ 1 weseethat n s g +1 +s + g F +1. ( +s) So the assertion holds. Since g ( χ( 1))/ 1 it is easily seen that From this one can check that n s ± 1 n s 1 n s 1 ( g ( χ( 1))/+s +1 ) / (g s 1), n s ( 1+g ( χ( 1))/) g s / (g s 1), n s +1 ( g ( χ( 1))/ + g s) / (g s 1). (g ) s + g s + g χ( 1) s χ( 1) ( + g s). If 3 (mod ), then g k + g k g k + g k tr(g k ) F. If 1(mod), then g F so g k + g k F. Thus, by the above we see that n s +1n s x n s 1n s y for some x, y F. Observe that n s (n s 1)(n s +1) 0since 1 s ( 3)/. Then we have χ(n s 1) χ(n s )χ(n s + 1) hence χ( n s 1) χ( n s )χ( n s +1). If s, t 1,,...,[ 3 ]} with s t, then clearly gs t,g s+t ±1 so g s+t (g t ± g s ) g s ± g t. This imlies g s (g t 1) ±g t (g s 1) hence g s 1 ± g t 1. Thus n s ±n t. According to [BEW] or [D], if b, c Z with b c 0(mod), then g s gt (3) Set () 1 χ(n + bn + c) 1. n0 R n χ(n 1) χ(n) χ(n +1)1,n F } (5) N n χ(n 1) χ(n) χ(n +1) 1, n F }.

4 506 ZHI-HONG SUN Then we see that (6) ( )( )( ) 1+χ(n 1) 1+χ(n) 1+χ(n +1) R n ( )( )( ) 1 χ(n 1) 1 χ(n) 1 χ(n +1) N. n So, by (3) we have ( )( )( ) (R + N) 1+χ(n 1) 1+χ(n) 1+χ(n +1) n + ( 1 χ(n 1) )( 1 χ(n) )( 1 χ(n +1) )} 1+χ(n n)+χ(n + n)+χ(n 1) } n ( 3) + 1 ( χ(n n)+χ(n + n)+χ(n 1) ) χ() χ( 1)} n0 ( 3) 6 χ() χ( 1) 16[ 3 ]. That is, R + N [ 3 ]. Now, combining the above we rove the theorem. Remark. Let 1(mod)bearime, a + b (a, b Z), a 1(mod), g be a rimitive root of. IfR N are defined by () (5) resectively, using (3), (6) the fact that 1 ( n ) 1 ( n0 0 n ) ( 3 n n0 ) a (cf. [J], [BE, Theorem.]) we see that R 1 1 ( )( )( ) 1 1+χ(n 1) 1+χ(n) 1+χ(n +1) 1 χ() 1 n0 + 1 ( χ(n n)+χ(n + n)+χ(n 1) + χ(n 3 n) )} 1 1 χ() n0 1 ( 3 χ()a) 1 1 χ() 17 a 1 therefore if 1(mod), 5 + a 1 if 5(mod) N [ 3 ] R 1 + a 1 if 1(mod), 5 a 1 if 5(mod).

5 CONSECUTIVE NUMBERS WITH THE SAME LEGENDRE SYMBOL 507. Proof of Theorem 3 Let Q 0 () be defined as in [S]. From [S, Theorem.] [S, Corollary 3.] we see that (n 1) (n) (n +1) n k +1(mod) for some k Q 0 () n ( x 6x +1) +1 (mod ) for some x Z x 3 x n (x +1) (x 3 (mod ) for some x Z x) n (x +1) x 3 (mod ) for some x Z. x So the theorem is roved. References [BE] B.C. Berndt R.J. Evans, Sums of Gauss, Jacobi, Jacobsthal, J. Number Theory 11 (1979), MR 1j:1005 [BEW] B.C. Berndt, R.J. Evans K.S. Williams, Gauss Jacobi Sums, John Wiley & Sons, Inc., New York, Chichester, 199,. 5. MR 99d:1109 [D] H. Davenort, The Higher Arithmetic, 5th edition, Cambridge University Press, London, New York, 19, MR a:10001 [J] E. Jacobsthal, Über die Darstellung der Primzahlen der Form n +1 als Summe zweier Quadrate, J. ReineAngew. Math. 13 (1907), 3-5. [S] Zhi-Hong Sun, Sulements to the theory of quartic residues, Acta Arith. 97 (001), MR 00c:11007 Deartment of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 3001, Peole s Reublic of China address: hyzhsun@ublic.hy.js.cn