PRIMES AND QUADRATIC RECIPROCITY ANGELICA WONG Abstrct We discuss number theory with the ultimte gol of understnding udrtic recirocity We begin by discussing Fermt s Little Theorem, the Chinese Reminder Theorem, nd Crmichel numbers Then we define the Legendre symbol nd rove Guss s Lemm Finlly, using Guss s Lemm we rove the Lw of Qudrtic Recirocity 1 Introduction Prime numbers re esecilly imortnt for rndom number genertors, mking them useful in mny lgorithms The Fermt Test uses Fermt s Little Theorem to test for rimlity Although the test is not gurnteed to work, it is still useful strting oint becuse of its simlicity nd efficiency An integer is clled udrtic residue modulo if it is congruent to erfect sure modulo The Legendre symbol, or udrtic chrcter, tells us whether n integer is udrtic residue or not modulo rime The Legendre symbol hs useful roerties, such s multilictivity, which cn shorten mny clcultions The Lw of Qudrtic Recirocity tells us tht for rimes nd, the udrtic chrcter of modulo is the sme s the udrtic chrcter of modulo unless both nd re of the form 4k + 3 In Section 2, we discuss interesting fcts bout rimes nd fke rimes seudorimes nd Crmichel numbers) First, we rove Fermt s Little Theorem, then show tht there re infinitely mny rimes nd infinitely mny rimes congruent to 1 modulo 4 We lso resent the Chinese Reminder Theorem nd using both it nd Fermt s Little Theorem, we give necessry nd sufficient condition for number to be Crmichel number In Section 3, we define udrtic residues nd the Legendre symbol, then exmine the udrtic chrcter of modulo which deends only on whether is 1 or 3 modulo 4) We show the multilictive roerty of the Legendre symbol nd rove Guss s Lemm Then, using the roerties of the Legendre symbol nd Guss s Lemm, we give net roof of the Lw of Qudrtic Recirocity 2 Generl Fcts About Primes nd Prime-Look-Alikes Proosition 21 fx) x is the identity utomorhism of Z/Z Proof First, we wnt to show tht f is field utomorhism of Z/Z; tht is, tht f reserves multiliction nd ddition Tht fxy) xy) x y fx)fy) follows from the commuttivity of multiliction We wnt to show tht x + y) x + y mod ) The binomil theorem gives x + y) i0 i) x i y i It suffices to show tht ech term of the exnsion Dte: August 11, 2008 1
2 ANGELICA WONG excet the first nd lst is zero modulo ; this follows from the fct tht i) if nd only if 1 i 1 This fct is true becuse! but i! i)! for such i,! so i! i)! i) Therefore, x + y) x + y mod ) Now to show f is the identity, let be rime By induction on x, we show tht x x mod ) This is clerly true for x 0 mod ) Assume the clim for x Alying the conclusion tht x + y) x + y mod ) to the x + 1 cse, we hve x + 1) x + 1 But this is just x + 1 by the induction hyothesis So x x mod ) holds for ll x Corollry 22 Fermt s Little Theorem) If is rime, then for ll x such tht x 0 mod ), x 1 mod ) Proof For x 0 mod ), we my multily both sides of the eulity x x mod ) by x, yielding x x x 1 mod ) Theorem 23 There exist infinitely mny rimes Proof Suose for contrdiction tht there re only finitely mny rimes; cll them 1,, k Consider the number x 1 k + 1 If x is not rime, there exists some rime i such tht i divides x This imlies x 0 mod i ), but this is imossible since x 1 mod i ) from the definition of x Therefore x must be rime Since x is greter thn ech rime i, this is contrdiction Thus, there must exist infinitely mny rimes Theorem 24 There re infintely mny rimes congruent to 1 modulo 4 Proof Suose for contrdiction tht there re only finitely mny rimes congruent to 1 modulo 4; cll them 1,, k Let us consider the numbers x 2 1 k nd N x 2 + 1 Becuse x is divisible by 2, x 2 0 mod 4), so N 1 mod 4) Note tht N 1 mod i ) for ech i Assume N is not rime; then there exists rime such tht divides N Since x 2 + 1), x 2 mod ), so x 4 1 mod ) Since 4 is the smllest such exonent, Theorem 22 imlies tht 4 divides, so 1 mod 4) Thus, i for some i, but this is contrdiction: N 0 mod ) while N 1 mod i ) for ll i It follows tht N is rime But since N is greter thn ech i nd N 1 mod 4), this is gin contrdiction Since either ssuming N is rime or ssuming N is comosite leds to contrdiction, there must be infinitely mny rimes congruent to 1 modulo 4 Fermt s Little Theorem cn be used to test for rimlity If there exists such tht n 1 mod n), then n is not rime Choosing rndomly nd testing this eulity yields n efficient rndomized rimlity test known s the Fermt Test Unfortuntely, there exist non-rime numbers tht lwys ss the Fermt Test see Definition 26) Definition 25 An odd comosite number n such tht b n 1 mod n) is clled seudorime to the bse b Note tht seudorime to the bse b my ss the Fermt Test when the rndomly chosen number is eul to b The smllest seudorime is 341, which is seudorime to the bse 2 [1] Definition 26 An odd comosite number n is clled Crmichel if n 1 mod n) for ll Z/nZ)
PRIMES AND QUADRATIC RECIPROCITY 3 The difference between seudorime nd Crmichel number is tht seudorime sses the Fermt Test for some numbers, but Crmichel numbers ss the Fermt Test for ll numbers In other words, Crmichel number is seudorime to every bse The smllest Crmichel number is 561 Theorem 27 Chinese Reminder Theorem) If gcd, b) 1 then for ll α, β Z there exists uniue γ mod b) such tht γ α mod ) γ β mod b) Proof We know x α mod ) if nd only if x α + k for some k We wnt to show tht there exists n integer k such tht α + k β mod b) Since gcd, b) 1, the Eucliden lgorithm gives integers r, s such tht 1 r+sb Multilying both sides of this eution by β α), we get β α) β α)r β α)sb Tking k rβ α), we get k β α) mod b), nd thus α + k β mod b) In order to rove uniueness modulo b, we first rove tht if y 0 mod ) nd y 0 mod b), then y 0 mod b) Since y nd b y, then lcm, b) y In ddition, since gcd, b) 1, this imlies tht lcm, b) b Therefore, y 0 mod b) Now, let x α mod ), x β mod b), nd x α mod ), x β mod b) Subtrcting x from x, we get x x 0 mod ) nd x x 0 mod b) so by bove we hve x x 0 mod b), s desired Proof tht 561 is Crmichel number We need to show tht for ll, 560 1 mod 561) We know 561 fctors s 561 3 11 17 Therefore, by the Chinese Reminder Theorem it suffices to show 560 1 mod 3) 560 1 mod 11) 560 1 mod 17) By Fermt s Little Theorem, since 3 is rime, we know 2 1 mod 3) So, 560 2 ) 280 1 280 1 mod 3) Similrly, since 10 1 mod 11), 560 10 ) 56 1 mod 11), nd since 16 1 mod 17), 560 16 ) 35 1 mod 17) Therefore, 561 is Crmichel number Proosition 28 If x 1 2 k where i re distinct rimes nd i ) x) for ll i, then x is Crmichel number Proof Exmining the roof tht 561 is Crmichel, we see tht the sme roof works for ny x stisfying the hyotheses of the roosition Consider the Crmichel number 561, which is eul to 3 11 17 The key to our roof ws using Fermt s Little Theorem for the rimes 3, 11, nd 17 to show tht 2 1 mod 3), 10 1 mod 11), nd 16 1 mod 17) We were then ble to show tht 560 1 mod 3) becuse 3 1) 560 nd 560 3)280 1 mod 3) We used the sme roch for 11 nd 17 The converse of the bove roosition is true s well [3]
4 ANGELICA WONG 3 Qudrtic Recirocity Definition 31 An integer k is udrtic residue modulo n if it is congruent to erfect sure modulo n; tht is, there exists n x Z such tht x 2 k mod n) Definition 32 The Legendre symbol for n integer nd n odd rime is defined s +1 if x) x 2 mod ) 0 if 0 mod ) otherwise For nonzero, the Legendre symbol euls 1 when is udrtic residue modulo nd when is not udrtic residue modulo The Legendre symbol is lso known s the udrtic chrcter of modulo Lemm 33 If is n odd rime nd P 1 2 1), then P mod ) Proof Cse 1 Sy 0 Then P 0 0 Cse 2 ) Sy is nonzero udrtic residue Then it suffices to show tht P 1 Let b 2 for some b Then P b 2P b But we know tht b 1 mod ) by Fermt s Little Theorem Thus, P Cse 3 ) Sy is not udrtic residue Then it suffices to show tht P Consider P ) 2 2P Substituting for P we hve 2P, b but by Fermt s Little Theorem this is congruent to 1 Thus P is sure root of 1 modulo, so P must be congruent to 1 or Consider the olynomil P 1 0 This is degree P olynomil so it cn hve t most P roots By Cse 2, for ny udrtic residue, P 1, so ech udrtic residue is root of this olynomil Since the function x x 2 is two-to-one on Z/Z), exctly hlf of the nonzero elements modulo re udrtic residues Thus, the P udrtic residues re exctly the P roots of the olynomil x P 1, so if is not udrtic residue, P b b Proosition 34 The Legendre symbol is multilictive: Proof Write s P b nd s b P b Then, P b P b) P Lemm 35 Let be rime Then the udrtic chrcter of modulo deends only on whether is 1 or 3 modulo 4, tht is, { 1 if 1 mod 4) if 3 mod 4) Proof Cse 1 1 mod 4) If 1) then there exists n x such tht x 1 mod ), but x b 1 mod ) for ny 0 < b < see [2] for roof of this fct) Since 4 1), there exists n x such tht x 4 1 mod ), but x 2 1 mod ) We know x 2 ) 2 1; therefore, x 2 is either 1 or We hve lredy ruled out
PRIMES AND QUADRATIC RECIPROCITY 5 x 2 1 mod ), so it must be the cse tht x 2 mod ) Thus, is udrtic residue modulo Cse 2 3 mod 4) Suose for contrdiction tht, tht is, there exists n x such tht x 2 mod ) Then, tking the sure of both sides of the eution, we hve x 4 1 By Fermt s Little Theorem, we know tht x 1 mod ) Substituting 1 4k + 2, we hve x 4k+2 1 We cn rewrite this s x 4k+2 x 4k x 2 x 4 ) k x 2 1 k x 2 But this is contrdiction since x 4k+2 1 Thus, is not udrtic residue modulo Lemm 36 Guss s Lemm) Let 0 mod ) where is rime greter thn 2 Let P 1 2 1) Form the numbers, 2, 3,, P nd reduce ech of these numbers to fll within the intervl 2, ) 2 by tking them modulo ) Let ν be the number of negtive numbers in the resulting set Then ) ν In other words, is udrtic residue if ν is even, nd non-residue if ν is odd Proof Exress the numbers, 2,, P s congruent to ±1, ±2,, ±P No number in the set 1, 2,, P will occur more thn once, whether ositive or negtive: if number occured twice with the sme sign, it would men tht two of the numbers in, 2,, P were congruent to one nother modulo, which cnnot hen since multiliction by is injective If the sme number occured twice with oosite signs, it would men tht the sum of two numbers in, 2,, P ws congruent to zero modulo, which lso cnnot hen Therefore, we hve tht {, 2,, P } {±1, ±2,, ±P } with definite sign for ech number We get )2) P ) ±1)±2) ±P ) mod ) Cnceling P! we get P ±1)±1) ) ±1) ) ν where ν is the number of negtive signs bove Therefore, P ) ν by Lemm 33 Theorem 37 Lw of Qudrtic Recirocity) Let nd be two different odd rimes The udrtic chrcter of mod ) is the sme s the udrtic chrcter of mod ) unless both nd re of the form 4k+3, in which cse the chrcters re oosite Another wy of sying this is tht ) if) nd re rime numbers, then the roduct of their Legendre symbols is ) 2 2 This roduct deends only on the rity of the exonent, nd is only when both nd re of the form 4k + 3 There re two cses in roving this theorem: one where mod 4) nd one where mod 4) We will need one lemm for ech cse, but their roofs re very similr Lemm 38 Let be ny ) nturl number, nd nd be odd rimes If mod 4), then Proof Using Guss s Lemm, we know ) is determined by ν, the number of the integers, 2,, P where P 2 ) tht lie between 1 2 nd, or between 3 2 nd 2, etc These intervls corresond to vlues between 2 nd 0 when reduced modulo s in Guss s Lemm) Since P is the lrgest multile of tht is less thn 1 2, the lst intervl tht we hve to consider is b 1 2 ), b), where b is 1 2 or 1 2 1) whichever is n integer) Thus ν is the number of multiles of tht
6 ANGELICA WONG lie in the intervls 1 2, ), 3 2, 2),, b 1 2, b) Dividing through by, we get the new intervls 2, 3 ), 2, 2 2b) ),, 2, b ), nd ν is now the number of integers in the union of these intervls Now write s 4k + r nd substitute for yielding 2k + r 2, 4k + r 3r 2r 2b)4k+r) ), 6k + 2, 8k + ),, 2, 4bk + br ) We cn disregrd the 2k nd 4k, etc nd consider r 2, r 3r ), 2, 2r 2b)r ),, 2, br ) becuse chnging the endoints of n intervl by n even integer does not chnge the rity of the number of integers in the intervl, nd it is only the rity of ν tht determines It is now cler from the form of these intervls tht the rity of ν deends only on r, nd not on) the secific rime which leves the reminder r when divided by 4; thus, deends only on the reminder of modulo 4 Lemm 39 Let be ny ) nturl number, nd be n odd rime If mod 4), then Proof Write s 4k r So, 4k 1) + 4 r) Alying the sme method from Lemm ) 38, we know tht the reminder of modulo 4 comletely determines In order to find the number of integers in the intervls in the cse, we need to relce r by 4 r After simlifiction, nd gin removing even integers from both ends of the interls, we hve the intervls r 2, r ), 3r 2, 2r 2b)r ),, 2, br ) Since these intervls re exctly the negtives of the intervls r 2, r 3r ), 2, 2r 2b)r ),, 2, br ) the number of integers in the union of the ) intervls is the sme in both cses Thus, we hve the sme result, Proof of Qudrtic Recirocity Suose nd re odd rimes Cse 1 Let mod 4); then 4 for some nd we hve the following eulities: ) ) 4 4+ ) 4 4 ) 4 4 ) ) In the bove eulitites we cn disregrd the 4 since multilying by udrtic residue does not chnge the udrtic chrcter of or of Tking the roduct of the Legendre symbols, we hve By Lemm 38, since mod 4), the exression on the right simlifies to 2 Then, by Lemm 35, ) which is exctly wht we needed ) { 1 if 1 mod 4) if 3 mod 4) )
PRIMES AND QUADRATIC RECIPROCITY 7 Cse 2 Let mod 4); then one of or is congruent to 1 mod 4) nd the other congruent to 3 mod 4) Thus mod 4), so + 4k for some k, nd we hve Similrly, 4k 4k 4 k k ) k k k Since mod 4k), by Lemm 39 There- ) 2 k k k 1, s we wnted fore ) References [1] Bch, Eric, nd Jeffrey Schllit Algorithmic Number Theory: Efficient Algorithms United Sttes of Americ: Integre Technicl Publishing Co, Inc, 1996 [2] Dvenort, H The Higher Arithmetic: An Introduction to the Theory of Numbers Cmbridge: Cmbridge University Press, 1982 [3] Korselt Probléme Chinois L intermédiire des mthémticiens, 6, 142-143, 1899 [4] Joyner, D, Kreminski, R, Turisco, J Alied Abstrct Algebr: Rough Drft htt://web ewusnedu/~wdj/book/node37html Johns Hokins University Press, 2002