Qudrtic recirocity Frncisc Bozgn Los Angeles Mth Circle Octoer 8, 01 1 Qudrtic Recirocity nd Legendre Symol In the eginning of this lecture, we recll some sic knowledge out modulr rithmetic: Definition 1. Modulo Nottion Let, e integer numers nd n ositive integer numer, then we sy tht is congruent to modulo n nd we write mod if n divides the difference. Also recll the Fundmentl theorem of rithmetic: Every integer n cn e written uniquely s n ± 1 1... r r where r is ositive integer greter thn or equl to 1, 1,... r distinct rime numers nd 1,... r re ositive integers. The lst nd non-trivil result tht we hve to recll in order to roceed further is Fermt s Little Theorem which sys tht if is n integer nd is rime, then mod, or equivlently 1 1 mod if, 1. Now we cn strt tlking out the qudrtic recirocity method. This method ws develoed intuitively not ctully on this modern form tht you will see here y some gret numer theorists of the 17th nd 18th century like Fermt, Euler, Lgrnge nd Legendre. They egn y looking t the qudrtic olynomils modulo rime nd trying to solve it in the most generl forms. Nmely, they did not try to solve x + x + c 0 in the rel or comlex numers, which ws lredy known, ut rther to solve the eqution x + x + c 0mod where is rime numer. in the integer numers. We will cll such n eqution solvle, if there exists n integer x 0 such tht x 0 + x 0 + c 0 mod. Although with little success, these mthemticins stted some imortnt conjectures in this field tht would rodened the field of numer theory. A mjor rekthrough in this direction cme when Guss in 1798 roved wht 1
is now clled the Qudrtic Recirocity Lw, nmely, if, q re rime numers nd if q 1 mod 4, then x 0 mod q is solvle if nd only if x q 0 mod is solvle nd if q 1 mod 3, then x 0 mod q is solvle if nd only if x + q 0 mod is solvle. We will lern this theorem lter, ut in more modern formultion. Now let s discuss out the essentil results of this toic. Definition. Let m, n nd e integers, m 1, n 1 nd, m 1. We sy tht is residue of n-th degree modulo m if congruence x n mod m hs n integer solution; else is nonresidue of n-th degree. In rticulr, if n, we will cll qudrtic residue or qudrtic nonresidue. Now we stte our first theorem: Theorem 3. Given rime nd n integer, the eqution x hs zero, one, or two solutions modulo. Proof. Suose tht the considered congruence hs solution x 1. Then so clerly is x x 1. There re no other solutions modulo, ecuse x x 1 mod imlies x ±x 1. As n immedite corollry of the ove thereom, we hve tht for every odd rime, mong the numers 1,,..., 1 there re exctly 1 qudrtic residues nd s mny qudrtic nonresidues. Now we re redy to define the Legendre s symol: Definition 4. Given rime numer nd n integer, Legendres symol is defined s 1, if nd is qudrtic residue mod ; 1, if nd is qudrtic nonresidue mod ; 0, if. x As first result out the Legendre s symol, we see tht 1 for ech rime nd integer x, x. From now on, unless noted otherwise, is lwys n odd rime nd n integer. Clerly, is qudrtic residue modulo if nd only if so is + k for some integer k. Thus we my regrd Legendre s symol s function from the residue clsses modulo to the set { 1, 0, 1}. Fermts theorem sserts tht if, 1 then 1 1 mod, which imlies 1 ±1 mod. More recisely: Theorem 5. Euler s Criterion If is n integer nd rime, then 1 mod.
3 Proof. The sttement is trivil for divides. From now on we ssume tht. Let g e rimitive root modulo. Then the numers g i, i 0, 1,..., form reduced system of residues modulo. We oserve tht g i 1 g i 1 1mod if nd only if 1 divides i 1, or equivlently, divides i. On the other hnd, g i is qudrtic residue modulo if nd only if there exists j {0, 1,, } such tht g j g i mod, which is equivlent to j i mod 1. The lst congruence is solvle if nd only if divides i, tht is, exctly when g i 1 1 mod. Now, y Euler s Criterion we hve tht 1 1 1, for ll integers, nd rime numers, therefore we roved tht the Legendre s symol is multilictive. Theorem 6. for ll integers, nd rime numer. From Euler s Criterion, we lso get wht is clled the Theorem 7. First Sulement of the Qudrtic Recirocity Lw For every rime numer 3, 1 1. 1 Now we return to our initil rolem, the one tht strted ll this theory out qudrtic residues: Theorem 8. Let x, y e corime integers nd,, c e ritrry integers. If is n odd rime divisor of numer x +xy +cy which doesn t divide c, then D 4c is qudrtic residue modulo. In rticulr, if divides x Dy nd x, y 1, then D is qudrtic residue modulo. Proof. Denote N x + xy + cy. Since 4N x + y Dy, we hve x + y Dy mod. Furthermore, y is not divisile y ; otherwise so would e x+y nd therefore x itself, contrdicting the ssumtion. There is n integer y 1 such tht yy 1 1mod. Multilying the ove congruence y y 1 gives us xy 1 + yy 1 Dyy 1 Dmod, imlying the sttement. Let us get closer to the min theorem of this introduction to qudrtic recirocity, ut first we hve to stte nother intermedite theorem: Theorem 9. Second Sulement of the Qudrtic Recirocity Lw We hve 1 1 8. In other words, is qudrtic residue modulo rime 3 if nd only if ±1 mod 8.
4 The roof is retty comlicted nd uses the Guss Lemm, very eutiful nd helful trick. With ll this eing sid, we conclude with the most imortnt theorem of this rt, the Guss Lw of Qudrtic Recirocity: Theorem 10. For ny different odd rimes nd q, q 1 1 q 1. q The roof of this fct is lso retty tricky, nd it will e ommitted. Nevertheless, this won t sto us to use it frequently in deducing other roerties out Legendre s symol. You my sk why Guss did not stte his theorem in this more elegnt form. Well, you should know tht Legendre invented his symol long fter Guss roved the qudrtic recirocity lw, similrly s Guss invented the modulo sign long fter Fermt, Euler nd Lgrnge roved their theorems. Exercises 1. Wht is 1 if is rime numer of the form 19k + 3? Is 5 qudrtic residue modulo 79? Is 6 qudrtic residue modulo 79? Is 11718 qudrtic 11719 trust me, this numer is indeed rime?. Comute the following Legendre symols: 17 01 3 013 5 014 7 015 19 3. In fct, wht Guss roved ws 1 1 8, 13 59., 01 103, 1 [ +1 4 ]. Prove tht [ ] + 1 mod 4 for ny odd rime numer, so indeed Theorem 9 is exctly wht Guss roved. Note tht [x] is the gretest integer smller or equl to x 4. Is 3 squre mod 41? Is 15 squre mod 41? CHMMC, Winter 010 5. Comute the numer of rimes less thn 100 such tht divides n +n+1 for some integer n. CHMMC, Winter 010 6. Find ll the rimes with the roerty tht 7 + 3 4 is erfect squre. Junior Blkn MO 007Hint: Use Fermt s Little Theorem 7.
5 is qudrtic residue modulo if nd only if 1 or 3 mod 8; 3 is qudrtic residue modulo if nd only if 1 mod 6; c 3 is qudrtic residue modulo if nd only if ±1 mod 1; d 5 is qudrtic residue modulo if nd only if ±1 mod 10. 8. Show tht there exist infinitely mny rime numers of the form 10k + 9. 9.Prove tht for n N every rime divisor of numer n 4 n + 1 is of the form 1k + 1. 10. If is rime of the form 4k + 1, rove tht x 1! is solution of the congruence x + 1 0 mod. Here m! 1... m 11. There exists nturl numer < +1 tht is qudrtic nonresidue modulo [. Hint: Suose is qudrtic nonresidue. Wht cn you sy out ] + 1? 1. Chllenge rolem Prove tht n integer is qudrtic residue modulo every rime numer if nd only if is erfect squre. 13. Evlute [ ] [ ] [ ] 1 + + + + 003 003 003 [ 001 003 Note tht [x] is the gretest integer smller or equl to x 14. Prove Guss Lemm: Let e rime numer nd n integer tht is corime with. Consider the integers,, 3,..., 1 nd tke their lest ositive residue modulo. These residues re ll distinct, so there re 1 of them. Let m e the numer of qudrtic residues tht re greter thn. Show tht 1 m. Generliztion: The Jcoi Symol Definition 11. Let e n integer nd n odd numer, nd let α1 1 α1... αr r e the fctoriztion of onto rimes. Jcois symol is defined s roduct of Legendre s symols, nmely α1 1 α αr. r We see tht if is qudrtic residue modulo n, then clerly n 1. However the converse is not true nd we cn find counterexmle. For exmle, 15 3 5 1 1 1, ].
6 ut is not qudrtic residue modulo 15, s is not so modulo 3 nd 5. Nevertheless, if is qudrtic nonresidue modulo n, then we get n 1, which imlies tht there exists rime numer dividing n such tht 1 y the ove definition. All this discussion cn e summrrized in the following sttement: Theorem 1. Let e n integer nd ositive integer, nd let α1 1 α... αr r e the fctoriztion of onto rimes. Then is qudrtic residue modulo if nd only if is qudrtic residue modulo αi i for ech i 1,,..., r. We will not see the roof of this theorem here. One direction is trivil, nmely if there exists n x such tht x mod n, then clerly the sme x stisfies x mod αi i. The other direction should e n esy exercise for those of you who know the Chinese Reminder Theorem. The most eutiful thing out the Jcoi symol is tht it oeys most of the lws tht the Legendre s Symol is oeying. Theorem 13. For ll integers, nd odd numers c, d the following equlities hold: + c c c 1 c c c cd c d 3 We lso hve tht the Jcoi symol oeys the three reciroicity lws, nmely Theorem 14. For every odd integer, 1 1 1, 1 1 8 nd for ny two corime odd numers, it holds tht 1 1 1. 3 Exercises 1. If gcd, nm1, then nm n.. Let F x x 17x 19x 33. Prove tht for ech m ositive integer, the eqution F x 0 mod m
7 hs solution x in N. But F x 0 doesn t hve ny integer solutions not even rtionl solutions. 3. Let m, n 3 e ositive odd integers. Prove tht m 1 doesn t divide 3 n 1. 4. Show tht there re no ositive integers x, y, z, t such tht x + y + t 4xyz. Hint: Write the eqution s 4zt + 1 4zy 14zx 1; now look t the Jcoi symol. z 4yz 1 5. Chllenge Prolem The numer of qudrtic residues modulo n n 1 is equl to [ n 1 ] [ 1 n+1 ] 1 + for, nd + 1 for 3. 3 + 1 6. Chllenge Prolem Prove tht the eqution x y 3 5 hs no integer solutions x, y. 7. Chllenge Prolem Prove tht 4kxy 1 does not divide the numer x m + y n for ny ositive integers x, y, k, m, n.