Some Results on Cubic Residues

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1 Interntionl Journl of Algebr, Vol. 9, 015, no. 5, HIKARI Ltd, htt://dx.doi.org/ /ij Some Results on Cubic Residues Dilek Nmlı Blıkesir Üniversiresi Fen-Edebiyt Fkültesi Mtemtik Bölümü Çğış Kmüsü, Blikesir, Turkey Coyright c 015 Dilek Nmlı. This rticle is distributed under the Cretive Commons Attribution License, which ermits unrestricted use, distribution, nd reroduction in ny medium, rovided the originl work is roerly cited. Abstrct In this er, we exmine the solubility of the cubic congruence x where is rtionl rime nd nd x re integers. Here, we give some results nd exmles relted with the cubic residues. Mthemtics Subject Clssifiction: 11A41, 11A15 Keywords: Rtionl rime, cubic residue, rimitive root 1. Introduction Let be rtionl rime nd be n integer. If there is n integer x such tht x then is sid to be cubic residue in mod. Z[ω] = { + bω, b Z} where ω = 1+. For Z[ω], the norm of is given by N =. = b + b where is the comlex conjugte of. If θ Z[ω], then the cubic residue chrcter θ { 0 if θ of θ in modulo is defined by ω i if θ N 1/ ω i where i {0, 1, }. Cubic residues hve been studied by severl uthors in [1],[],[],[4] nd [5]. In this er, we obtine some results nd exmles relted with the cubic residues.

2 46 Dilek Nmlı. Min Results Theorem 1. Let be rtionl rime for which 1. Then the equivlence x is solvble if nd only if 1/ 1. Proof. This theorem is the secil cse k = of the Euler s Criterion. Theorem. If is rtionl rime nd Z, then = 1. Proof. We know tht = find =. As = 1. is equivlent to ω or to ω, we Exmle 1. Let us consider whether 9 is cubic residue in mod 7 or not. Since , 7 7 ω 4 7. Thus 9 7 ω. Therefore 9 is not cubic residue in mod 7. Exmle. We consider the equivlence x Then, 15 is cubic residue in mod 7. In other words, the equivlence x 15 7 is solvble. In fct, x = 1, x = ω nd x = ω re the roots of the equivlence x Since ω = 1+, we get the roots of this equivlence s x 17, x 47 nd x 7. Exmle. Is the equivlence x 41 7 solvble? Since , 41 we obtin ω 8 7 nd 7 41 = ω. Therefore x 41 7 is unsolvble. Theorem. If is rtionl rime nd is ositive integer such tht, = 1, then in mod is cubic residue. Proof. Let be rime nd let be ositive integer such tht, = 1.Since, we cn write = k +, k Z. In this cse, nd N =. = k + k + = 9k + 1k + 4 N 1 = k + 4k + 1.

3 Some results on cubic residues 47 From, = 1 nd the Fermt s little theorem, we hve Thus 1 = k+1 1. N 1/ = k +4k+1 = k+1k+1 k+1 k+1 1 k+1 1. Corollry 4. If is rtionl rime, then there re exct cubic residues in mod different from ech other. In other words, ll elements of Z re cubic residues. Proof. Let be rtionl rime nd let g be rimitive root. Also let us choose {1,,..., 1} nd k {0, 1,..., } roviding the equivlence g k. Since, 1 = 1, there re integers x nd y such tht x + 1y = 1. If we tke x = x k nd y = y k, then we cn write s x + 1y = k. Since g 1 1, we find g k = g x+ 1y = g x g 1 y g x. Tht is, is cube in mod. Since 0 0, there re exct different cubes in mod. Exmle 4. Let = 11. Since , , 7 11, 9 11, , 5 11, , , 8 11, , nd , ll numbers in Z 11 re cubic residues. Theorem 5. If 1 is rtionl rime, then the number of different cubic residues in mod is +. Proof. Let 1 be rtionl rime. For every k element in {, 6, 9,..., 1}, g k = g t = g t is cube where g is rimitive root nd t Z. Here ll these g k s re different. Then, there re t lest 1 nonzero cubes in mod. On the other hnd, ech cube is the form b. From 1 nd the Fermt s little theorem, 1/ b 1 1. By the Lgrnge s Theorem for olynomils, there is the most 1 root of the equivlence 1/ b 1 1, tht is, 1 is n uer bound for the totl number of cubes in mod. Then there re exct 1 non-zero cubes. When counting the zero, then there re = + cubes in mod.

4 48 Dilek Nmlı Theorem 6. If is n odd rime number then.if nd only if is 0. Proof. Since, = 1, we get 0 0. Corollry 7. If is n odd rime number then cubic residues in Z re , 1,,,,,,, 1 1. Corollry 8. Let be n odd rime number. An integer is cubic residue in Z if nd only if is cubic residue in Z. Proof. If is solution of x k then k. Since is lso solution of x k. = k k, Theorem 9. Let be rime such tht. The sum of cubic residues roviding the equivlence x k is equivlent to zero in mod. Proof. Let be rime. From the Corollry.4, ll cubic residues re different nd these re 0, 1,,..., 1. Their sum is =. As is rime nd, 1 is n even number, tht is, 1 Z. Then, we find = Theorem 10. Let 1 be rime. The sum of the cubic residues roviding the equivlence of x is equivlent to zero in mod. Proof. Let 1 be rime. From the Theorem.5, there re + different cubic residues. If 1 then we cn write = k + 1, k Z. As is rime, k is n even number. As + + = k + 1, is n odd number. One of the cubic residues is zero. Let 0 = 0. Then, there re + 1 = 1 different cubic residues. Also, from the Corollry.8, if is cubic residue in Z, then is cubic residue in Z too. In this cse, the sum of the cubic residues is =

5 Some results on cubic residues 49 Theorem 11. If one of solutions of the equivlence x m is x, then the others re xω nd xω. Proof. If = 1 then, we know tht the solutions of x 1 m re x = 1, xω = 1.ω nd xω = 1.ω. If 1 nd if one of the solutions of x m is x, then xω = x ω x m, nd xω = x ω 6 x m. Corollry 1. The sum of solutions of the equivlence x m is equivlent to zero in mod m. Proof. From the Theorem.10, we know tht the solutions of the equivlence x m re x, xω nd xω. Then we hve x + xω + xω = x + xω + x 1 ω = 0. References [1] D. S. Xing, Z. F. Co, X. L. Dong, Identity bsed signture scheme bsed on cubic residues, Sci. Chin Inf. Sci , no. 10, htt://dx.doi.org/ /s [] K. Irelnd, M. A. Rosen, A clssicl introduction to modern number theory, Second edition. Grdute Texts in Mthemtics, 84. Sringer-Verlg, New York, xiv+89. ISBN: X. htt://dx.doi.org/ / [] D. Nmli, Cubic residue chrcters, Int. Mth. Forum 8 01, no. 1-4, [4] Z. Sun, On the theory of cubic residues nd nonresidues, Act Arith , no. 4, [5] Z. Sun, Cubic residues nd binry qudrtic forms, J. Number Theory , no. 1, htt://dx.doi.org/ /j.jnt Received: My 7, 015; Published: June, 015