QUADRATIC RESIDUES MATH 372. FALL INSTRUCTOR: PROFESSOR AITKEN

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QUADRATIC RESIDUES MATH 37 FALL 005 INSTRUCTOR: PROFESSOR AITKEN When is n integer sure modulo? When does udrtic eution hve roots modulo? These re the uestions tht will concern us in this hndout 1 The Legendre Symol Definition 1 Let F where is n odd rime We cll sure if there is n element F such tht = Non-zero sures re lso clled udrtic residues The set of udrtic residues is written U or Q We will see lter tht U is closed under multiliction in other words, it is sugrou of U Remrk Oserve tht is udrtic residue if nd only if there is non-zero such tht = One direction is esy: if is udrtic residue, then y definition it is non-zero sure So there is such tht = This cnnot e zero since is not zero The other direction is not too d: if = where is not zero, then is sure Now is non-zero: otherwise would e zero divisor, ut we know tht the field F hs no zero divisors So is udrtic residue Definition Let Z, nd let e n odd rime Then the Legendre symol is defined to e 0, +1, or 1 The Legendre symol is defined to e 0 when = 0 in F In other words, it is 0 if nd only if The Legendre symol is defined to e +1 when is udrtic residue In other words, it is +1 if nd only if U The Legendre symol is defined to e 1 in ny other cse In other words, it is 1 if nd only if is in U ut not in U Exercise 1 Clculte 11 for ll 0 < 11 directly from the definition without using the roerties elow Lemm 1 Let e n odd rime If = +1 then 1/ = 1 Proof The hyothesis imlies tht = for some U Then 1/ = 1/ 1 = = 1 y Fermt s Little Theorem Dte: Fll 005 Version of Novemer 9, 005 1

The contrositive gives the following: Corollry 1 Let e n odd rime If 1/ 1 then U Lemm Let e n odd rime Let g e rimitive element of U Then g 1/ = 1 So y the ove corollry, g is not udrtic residue Proof Recll tht g hs order 1 since it is genertor Let = g 1/ So = g 1/ = g 1 = 1 Since = 1, the element is root of x 1 From n erlier result, this imlies tht is 1 or 1 However, = g 1/ is not 1 since the order of g is 1 which is greter thn 1/ Remrk Recll tht every element of U is ower of rimitive element g In fct, U = { g 0, g 1,, g } Thus hlf of the elements of U cn e written s g k with 0 k even, nd the other hlf cn e written s g k with 0 k odd Lemm 3 Let e n odd rime, nd let g U e rimitive element If = g k with k even, then = +1 If = g k with k odd, then 1/ = 1 nd = 1 Proof If = g k with k even, then k = l for some l Thus = g l So is sure It is non-zero since it is unit owers of g re units Thus = +1 If = g k with k odd then 1/ = g k 1/ = g 1/ k = 1 k = 1 using the fct tht k is odd together with Lemm Finlly, y Corollry 1 we know tht the unit is not udrtic residue, so = 1 Corollry Of the 1 elements of U, there re 1/ udrtic residues nd there re 1/ tht re not udrtic residues Proof Recll, U = {g 0, g 1,, g } In the rnge 0 k there re 1/ even vlues of k nd 1/ odd vlues of k Theorem 1 If is n odd rime nd is n integer, then = 1/ Remrk In the ove theorem we re considering s tking vlues 0, 1, 1 U insted of 0, 1, 1 Z So, techniclly we should ut ig r over Proof There re three cses to consider First suose tht = 0 By definition, = 0 Thus, 1/ = 0 1/ = 0, nd the result follows Next suose tht = +1 Then 1/ = 1 y Lemm 1

Finlly, suose tht = 1 Let g e rimitive element of U Since g genertes U, there is k such tht g k = By Lemm 3, this k cnnot e even So k is odd The result follows from Lemm 3: 1/ = 1 Exercise Clculte 11 for ll 0 < 11 using Theorem 1 Bsic roerties of the Legendre Symol Here re some very useful roerties to know in order to clculte Throughout this section, let e n odd rime Proerty 1 If 0 mod then = 0 In rticulr, = 0 Proof This follows stright from the definition 1 Proerty If 0 mod nd Z is sure, then = 1 In rticulr, = 1 Proof If is sure, then is sure modulo So = 1 since 0 1 Proerty 3 = 1 1/ In rticulr: 1 If 1 mod 4, then = 1 If 3 mod 4, then = 1 Proof The first eution follows from Theorem 1 If 1 mod 4, then 1 = 4k for some k Thus 1/ = k In this cse 1 1/ = 1 k = 1 If 3 mod 4, then 3 = 4k for some k Thus 1 = 4k +, nd 1/ = k + 1 In this cse 1 1/ = 1 k+1 = 1 Proerty 4 For, Z we hve = Proof This follows from Theorem 1: = 1/ = 1/ 1/ = 1 r Proerty 5 If r mod then = Proof If r mod then = r By Definition 1, = r clerly imlies = r Exercise 3 Use Proerty 4 to show tht the roduct of two udrtic residues is udrtic residue Thus the set U of udrtic residues is closed under multiliction In fct, it is sugrou of U Exercise 4 Use Proerty 4 to show tht if, U re units such tht one of them is udrtic residue ut the other is not, then is not udrtic residue 3

Exercise 5 Use Proerty 4 to show tht if, U re units tht re oth non-udrtic residues, then is udrtic residue Remrk For those of you who hve tken strct lger, oserve tht Proerty 4 tells us tht the m is grou homomorhism U {±1} The kernel of this homomorhism is the sugrou U of udrtic residues The udrtic residues form sugrou, ut the non-udrtic residues only form coset Exercise 6 Give multiliction tle for U 11 Hint: it should hve 5 rows nd columns 3 Advnced roerties of the Legendre Symol The roofs of the roerties of this section will e ostoned Proerty 6 Let e n odd rime, then is determined y wht is modulo 8 If 1 or 7 mod 8, then If 3 or 5 mod 8, then The following is celerted theorem of Guss = 1 = 1 Proerty 7 Qudrtic Recirocity Let nd e distinct odd rimes Then = 1 1 1 1 Remrk As we discussed ove, is even if 1 mod 4, ut is odd if 3 mod 4 Similrly, for So 1 1 is even if either or is congruent to 1 modulo 4, ut is odd if oth re congruent to 3 So If 1 or 1 mod 4, then = If 3 nd 3 mod 4, then 4 Sure roots If = in F then is clled sure root of = Lemm 4 Let e n odd rime If is not zero, then Proof Suose otherwise, tht = = 1 Since is unit, it hs multilictive inverse Multily oth sides of = 1 y 1 This gives 1 = 1 So 1 1 mod This mens tht divides 1 1 = However, >, contrdiction Proosition 1 Let e n odd rime If hs sure root, then is lso sure root Furthermore, ± re the only sure roots of 4

Proof Since = 1 =, if = then = So the first sttement follows Now we must show tht ± re the only sure roots of First ssume 0 Then y Lemm 4, ± re two distinct solutions to x = However, the olynomil x hs t most two roots y Lgrnge s theorem Thus x = hs no other solutions In other words, there re no other sure roots Finlly, consider the cse where = 0, so = 0 nd = 0 s well Now if c is non-zero sure root of = 0 then it is zero divisor Zero divisors do not exist in F since it is field So = 0 is the only sure root Proosition Let e n odd rime Then the numer of sure roots of in F is given y the formul + 1 Proof There re three cses Cse = 0 By definition, = 0, which hs 0 for sure root By Proosition 1 the sure roots re ±0 So 0 is the uniue sure root: there is exctly one sure root Oserve tht + 1 = 0 + 1 = 1 gives the correct nswer in this cse Cse = 1 By definition, is non-zero sure, so it hs sure root in F Clerly is non-zero otherwise would e 0, ut is non-zero By Proosition 1 nd Lemm 4 there is exctly one other sure root, nmely So there re two sure roots Oserve tht + 1 = 1 + 1 = gives the correct nswer in this cse Cse = 1 By definition, is not sure in F So there re no roots Oserve tht + 1 = 1 + 1 = 0 gives the correct nswer in this cse Exercise 7 Find ll the sure roots of ll the elements of F 11 For more rctice try F 7 or F 5 Exercise 8 For which rimes is it true tht 1 hs sure root? Find the first eight rimes with this roerty For few of these, find sure roots of 1 5 Qudrtic eutions modulo odd rimes The revious section considered the roots of x = 0 which re clled sure roots In this section we consider the generl udrtic eution x + x + c = 0 in F with n odd rime Lemm 5 Comleting the sure Let e n odd rime, nd consider the udrtic olynomil x + x + c where 0 Then r is root of this olynomil if nd only if r + is sure root of 4c Proof Oserve tht r + = 4 r + 4r + = 4 r + 4r + 4c 4c + = 4r + r + c + 4c So if r + r + c 0 mod, then r + 4c mod Conversely, suose r + 4c mod So 4r + r + c = r + 4c 0 mod 5

But is unit modulo y ssumtion, nd 4 so 4 is lso unit modulo Thus we cn cncel the 4 fctor in the ove eution leving us with r + r + c 0 mod Remrk We cll 4c the discriminnt of x + x + c Corollry 3 Let e n odd rime, nd consider the olynomil x + x + c where 0 If this olynomil hs root in F then the discriminnt hs sure root in F Remrk You might hve seen something like the ove lemm in the context of deriving the clssicl udrtic formul for F = R or F = C In fct, the ove lemm is vlid in ny field F such tht 1 + 1 0 However, it it fils in F = F Theorem Let e n odd rime, nd consider the olynomil x + x + c where 0 If this olynomil hs t lest one root in F nd if δ F is sure root of the discriminnt 4c which exists y the revious corollry, then the roots re given y the formul ± δ 1 This formul is trditionlly written s ± 4c Finlly, if the discriminnt is sure in F then the olynomil hs t lest one root Proof According to Lemm 5, if r is root of x + x + c, then r + is sure root of the discriminnt By Proosition 1 the only sure roots of the discriminnt re δ nd δ So either r + = δ or r + = δ Now solve for r Now suose the discriminnt is sure with sure root δ Let r e + δ 1 This imlies tht r + = δ So r is root y Lemm 5 Proosition 3 Let e n odd rime, nd consider the olynomil x + x + c where 0 Then the numer of roots in F is given y the following Legendre Symol sed formul: 4c + 1 Proof There re three cses Cse 4c = 0 In other words, discriminnt is 0, which is oviously sure So y Theorem, the olynomil hs t lest one root Oserve tht δ = 0 is sure root of the discriminnt in this cse So y Theorem, the roots re ± δ 1 Since δ = 0, oth ossiilities give the sme nswer: there is exctly one root nd it is 1 Cse 4c = 1 In other words, the discriminnt is non-zero sure So y Theorem, the olynomil hs t lest one root Let δ e sure root of the discriminnt Since the discriminnt is non-zero, δ 0 So δ nd δ re distinct y Lemm 4 By Theorem, the roots re ±δ 1 Clim: these roots re distinct To see this suose +δ 1 = δ 1 From this eution it is esy to derive δ = δ, contrdiction Thus there re exctly two roots 6

Cse 4c = 1 In other words, the discriminnt does not hve sure root in F So y Corollry 3 contrositive, there re zero roots 6 Additionl Prctice Prolems Exercise 9 Comute 5 71 using the ove roerties Likewise, comute 3 71 Exercise 10 Use the Legendre symol to decide if 14 is sure in F 101 Exercise 11 How mny roots does x + 3x + 4 hve in F 39? Exercise 1 When is 5 sure modulo where is n odd rime? List the first eight rimes where this hens Check few of these to see if you cn find sure roots of 5 Hint: the nswer deends on wht is modulo 5 Exercise 13 When is 7 sure modulo where is n odd rime? List the first eight rimes where this hens Check few of these to see if you cn find sure roots of 7 Hint: the nswer deends on wht is modulo 8 Divide into two cses: 1 mod 4 nd 3 mod 4 Use the Chinese Reminder Theorem Exercise 14 Show tht 3 i = 3, ii 1 mod 4, nd iii 3 mod 4 with 3 = 3 for ll odd rimes Hint: divide into three cses Exercise 15 For wht odd rimes re there elements nd + 1 tht re multilictive inverses to ech other? List the first eight rimes where this hens Check few of these to see if you cn find Hint: show this hens if nd only if x + x 1 = 0 hs roots Exercise 16 For wht odd rimes re there elements nd in F tht re oth dditive nd multilictive inverses to ech other? List the first eight rimes where this hens Check few of these to see if you cn find nd Hint: show this hens if nd only if x = 1 hs solutions Exercise 17 For wht odd rimes re there elements nd in F tht dd to 3 ut multily to? Exercise 18 For wht odd rimes re there elements nd in F tht dd to ut multily to 3? List the first eight rimes where this hens Check few of these to see if you cn find nd Hint: the nswer deends on whether is sure modulo Comute the Legendre symol for ech ossile vlue of modulo 8 Oserve tht knowing modulo 8 gives you knowledge of modulo 4 Exercise 19 For wht odd rimes is there non-zero element in F whose cue is eul to 3 times itself? List the first eight rimes where this hens Check few of these rimes to see if you cn find the desired element in F Hint: show this hens if nd only if x = 3 hs solution Slit into three cses: = 3 nd 1 mod 4 nd 3 mod 4 Dr Wyne Aitken, Cl Stte, Sn Mrcos, CA 9096, USA E-mil ddress: witken@csusmedu 7