MATH 118 PROBLEM SET 6

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1 MATH 118 PROBLEM SET 6 WASEEM LUTFI, GABRIEL MATSON, AND AMY PIRCHER Section 1 #16: Show tht if is qudrtic residue modulo m, nd b 1 (mod m, then b is lso qudrtic residue Then rove tht the roduct of the qudrtic residues modulo is congruent to +1 or 1 ccording s the rime is of the form 4k + or 4k + 1 Proof Since is qudrtic residue mod m, there exists n x Z such tht x (mod m Furthermore, since b 1 (mod m, we hve b 1 ( x 1 ( x 1 (mod m Hence, b is qudrtic residue mod m Now we need to show tht the roduct of the qudrtic residues modulo is congruent to +1 or 1 deending on whether is of the form 4k + or 4k + 1 We cn tke the roduct of the qudrtic residues in irs, by grouing x with x 1 The only clsses tht get ired with themselves re those for which x x 1 (mod, ie x 1 (mod, ie x ±1 (mod The roduct of the irs whose elements re distinct is just x x 1 1 (mod So the only contribution comes from the excetionl cses x ±1 (mod Now 1 is qudrtic residue if nd only if 1 (mod 4 Therefore, we see tht the roduct of the qudrtic residues is 1 if 1 (mod 4, nd 1 if (mod 4 #19: For ll rimes rove tht x 8 16 (mod is solvble Proof For = this is trivil, we cn tke x = 0 For odd, we know tht (16, = 1 nd we must show (by Theorem 7 tht 16 1 (8, 1 1 (mod Since 1 is even, we know tht (8, 1 cn only tke vlues, 4, nd 8 For (8, 1 = we get (mod by Fermt s Little Theorem For (8, 1 = 4 we get (mod, Dte: Mrch 1, 01 ALL REFERENCES ARE FROM CLASS NOTES 1

2 gin by Fermt s Little Theorem For (8, 1 = 8 we must use Legendre symbol identity, tht is ( ( (mod ( Since (8, 1 = 8, we hve 1 (mod 8, so solvble for ll rimes = 1 Thus x 8 16 (mod is #4: Suose tht m is n odd number Show tht if (, m = 1 then the number of solutions of the congruence x (mod m is ( ( m 1 + Show tht this holds for ll integers if m is n odd squre-free number Proof Let nd m be integers such tht m is odd nd (, m = 1 We wnt to show tht the number of solutions to the congruence x (mod m is ( ( m 1 + We strt by roving the following Lemm, Lemm 1 Let nd such tht is n odd rime nd (, = 1 Then, given nturl number α, the number of solutions to the congruence ( if = 1, x (mod α = ( 0 if = 1 First we note tht the right hnd side of the eqution is simly the number of solutions to x (mod Proof of Lemm 1 Cse 1: is not qudrtic residue mod We will show tht x (mod α hs no solutions Indeed, x 0 Z such tht x 0 (mod α = α x 0 = x 0 = x 0 mod This is contrdiction, ( since we ssumed ws qudrtic non-residue mod So we hve tht when = 1, (# of solutions to x (mod α = 0 ( Cse : is qudrtic residue mod, ie = 1 First we need to show there re t most solutions to x (mod α Suose, for some x, y 0 tht x

3 (mod α nd y (mod α = x y mod α = α x y = α (x y(x + y = (x y(x + y = (x y or (x + y since is rime Assume (x y Mini clim: (x + y Mini Proof: (x y = x y (mod Assume (x + y = x y (mod = y y (mod = y 0 (mod This is contrdiction since (, = 1 = 0 (mod = y 0 (mod since y (mod, thus roving this mini clim Now since (x + y, we hve ( α, x + y = 1 since the only rime divisor of α is Then by the Fundmentl Theorem of Arithmetic, α (x + y(x y imlies tht α x y, ie x y (mod α Similrly, if we hve (x + y, then we find tht x y, so ( α, x y = 1, so α x + y by FTA, nd hence x y (mod α We hve therefore shown tht there re t most two solutions to x (mod α, nmely, ny one solution nd its negtive ( Now we will show tht x (mod α hs t lest solutions if = 1 To do this we will show tht if is squre mod, then is squre mod α Consider (Z/ α Z We showed in clss tht this grou is cyclic, ie g (Z/ α Z : g = (Z/ α Z, where g = {g k : k Z} Also note tht the order of (Z/ α Z is even To see this simly recll tht # (Z/ α Z = φ ( α = α 1 ( 1 Since is n odd rime, we hve tht 1 is even, nd tht the order s whole is even Now consider for grou G, the subset of the squres, ie H = {h : h G} For cyclic grou G generted by n element g, the subset of squres is exctly g To see this, first consider n element m H G Then h G : m = h Since h G = g = k Z : h = g k = m = h = (g k = g k = (g k g Thus H g To show the oosite continment, consider n g Then k Z : n = (g k = g k = (g k H = g H = g = H, the subset of the squres Now lying this generl result to (Z/ α Z, generted by some element g (Z/ α Z, we cn see tht the subset of the squres, cll it A, is g, nd tht #(A = ord(g = ord(g (, ord(g = ord(g = #((Z/α Z

4 Hence, the index of g in (Z/ α Z, denoted [(Z/ α Z : A], is #((Z/α Z = #(A So we hve tht the subset A of the squres in (Z/ α Z is n index two subgou Also consider B = {b : b (Z/Z } (Z/Z, the set of squres in (Z/Z And just s in our clcultion for A, we similrly hve tht #(B = #((Z/Z Now, consider the reduction m, = φ( r : (Z/ α Z (Z/Z, given by r : (mod α (mod Clim: The m r is surjective homomorhism Proof: (Homomorhism Let x, y (mod α (Z/ α Z Consider r(xy (mod α = xy (mod (x (mod (y (mod = r(x (mod α r(y (mod α (Surjectivity Let x mod (Z/Z Then x mod α, stisfies r(x mod α = x (mod, showing the surjectivity of the reduction m, nd thus roving the clim We now wnt to show tht the inverse imge of B under r is A The fct tht A is contined in r 1 (B is cler, since the reduction of squre is squre, ie if something is squre modulo α then it is squre modulo So we hve tht A r 1 (B Since r is surjective, we know tht #(r 1 (B = #(B #(ker(r By the First Isomorhism Theorm, since r is surjective, #(ker(r = φ(α φ( Hence, #(r 1 (B = #(B #(ker(r = φ( φ(α φ( = φ(α = #(A Now since A r 1 (B nd the two sets hve the sme size, this imlies tht A = r 1 (B This mens( tht everything tht is squre modulo is lso squre modulo α, nd tht when = 1, then there is t lest one solution to x mod α Given one solution x, x is clerly nother solution nd x x (mod α since is odd nd (x, = 1 So there re t lest two solutions It ws shown bove tht ( there re t most two solutions to x mod α, nd thus we hve tht when = 1, there re exctly two solutions to x mod α, thereby roving the Lemm Now, we move bck to the originl roblem Let the rime fctoriztion of odd m = n i=1 α i i, where we re trying to find the number of solutions to x (mod m, with (, m = 1 4

5 By the Chinese Reminder Theorem, since (, m = 1 nd ll the α i i s re reltively rime to ech other, x (mod m x (mod α i i, i = 1,,, n We hve from this tht (# of solutions to x (mod m = n i=1 (# of solutions to x (mod α i i From the Lemm roved bove, we know tht (# of solutions to x (mod α i i = (# of solutions to x (mod i for ech i = 1,,, n, since (, i = 1 Also we know tht by the definition ( of the Legendre symbol, i = 1,,, n : (# of solutions to x (mod i = 1 + Hence we chieve, s desired, (# of solutions to x (mod m = i m i ( ( 1 + i For the second rt of the roblem, ssume tht m is squre free integer This imlies m hs rime fctoriztion of the form n i=1 i, with i j = i j (ie no rime tht divides m does so with mutiliity greter thn 1 By the Chinese Reminder Theorem, since ll i s re reltively rime to ech other, x (mod m x (mod i i = 1,,, n So, the number of solutions to the congruence x (mod m = s ws desired to show = n (#of solutions to x (mod i i=1 n i=1 ( ( 1 +, i Section #6: Decide whether x 150 (mod 1009 is solvble or not Proof We will show tht 150 is not qudrtic residue (mod 1009 There does not exist rime less thn 1009 such tht divides 1009, so 1009 is rime Furthermore, we notice tht 150 = = 5 Now we hve enough informtion to comute the Legendre Symbol, which will tell us whether 150 is qudrtic residue of 1009 Becuse the Legendre Symbol is multilictive function, it follows from this fctoriztion tht ( ( = Moreover, Proerty 1 nd Theorem 1 yield = 1 nd = since ny squre is immeditely qudrtic residue nd (mod 8 At this 5

6 ( oint, it suffices to show tht = 1 By lying Theorem 14 nd gin 1009 Proosition 1, we hve ( ( ( = ( = ( = since (mod 4 Consequently, ( 150 = (1 ( 1 (1 = 1, 1009 nd thus x 150 (mod 1009 is not solvble #10: Of which rimes is qudrtic residue? ( Proof Let be rime We will show is qudrtic residue mod (ie = 1 iff 1, (mod 8 It follows from Proerty 1, Theorem 1, nd the fct tht the Legendre Symbol is multilictive function tht ( 1 = = ( 1 1 ( ( 1 In rticulr, this shows tht = 1 if nd only if = 1 nd = 1, 1 or = 1 nd = 1 We el gin to Proosition 1 nd Theorem 1 to yield ( 1 1 = 1 nd ( = 1 when 1 (mod 4 nd 1, 7 (mod 8 resectively Since 1 (mod 4 nd 7 (mod ( 8 cnnot both ( be true, 1 it follows tht 1 (mod 8 must be true Similrly, = 1 nd = 1 when (mod 4 nd, 5 (mod 8 As before, if (mod 4 then 5 (mod 8 Thus, ( = 1 if nd only if 1, (mod 8 s desired 6

7 #: Suose (b, = 1 for rime > ( Show the # of solutions (x, y to the b congruence x + by 1 (mod is Proof Let E = (# of solutions to x + by 1 (mod, with (b, = 1 Now, E = (# solutions to x 1 by (mod = (# solutions to x 1 (1 by (mod So we hve tht ( 1 Since = Clim: 1 E = [ 1 + (, we get E = + ] 1 1 by ( ( 1 1 by 1 1 by b = Proof of Clim ( 1 1 by is equl to ( # of solutions to x 1 by (mod Now x 1 by (mod is the sme eqution s by 1 x (mod y b 1 (1 x (mod The number of solutions to this eqution is 1 ( ( b 1 1 x b 1 ( 1 x 1 + = + x=0 By Lemm from clss, we hve 1 1 x 1 = So we re left with x=0 1 1 by b =, thus roving the clim ( [ 1 1 by So reconsidering, E = + = + the results of the clim And we get wht ws desired, ( which is tht b E = (# of solutions to x + by 1 (mod = is n odd rime x=0 ( ] b, from, when (b, = 1 nd 7

8 Bullet #1: For which rimes, 5 does x 10 (mod hve solution? ( 10 5 Proof Consider = Clim: For, 5, ( { 5 1 if 1, 4 (mod 5 = 1 if, (mod 5 ( 5 ( Proof of clim = becuse 5 1 (mod 4 5 ( 1 4 =,,, or deending on whether 1,,, or 4 (mod resectively Note tht is not equivlent to 0 (mod 5 becuse is rime nd 5 Thus the clim is roven ( 1 = 1 5 ( = 1 since 5 5 (mod 8 5 ( 5 = = = 1 since (mod 8 5 ( 4 = = ( 1( 1 = Note tht ( { 1 if 1, 7 (mod 8 = 1 if, 5 (mod 8 ( 10 Now, x 10 (mod hs solution whenever = 1 5 Cse 1: = = 1 5 If = = 1 = 1, 7 (mod 8 nd 1, 4 (mod 5 Note tht (5, 8 = 1 By the Chinese Reminder Theorem = 1, , , or (mod 40 = 1, 1, 9, or 9 (mod 40 5 Cse : = = 1 5 If = = 1 =, 5 (mod 8 nd, (mod 5 8

9 By the Chinese Reminder Theorem , , , or (mod 40 Hence 7,, 7, 1 (mod 40 Thus, for, 5, the congruence x 10 (mod hs solution iff 1,, 9, 1, 7, 1, 7, or 9 (mod 40 9

10 Bullet #: Let m = 4 n + 1 for some integer n 1 Prove tht (m 1/ 1 mod m m is rime Proof = : Assume (m 1/ 1 (mod m ( (m 1/ ( 1 (mod m = m 1 = 1 (mod m = ord( (m 1, where ord( denotes the order of in (Z/mZ Now, we need to show tht (m 1 ord( Note tht m 1 = (4 n = 4 n = n Since ord( 4 n = n, ord( must be ower of Assume tht ord( < (m 1 = n This imlies tht ord( n 1 = m 1 since ord( is ower of less thn n Hence n 1 = (m 1/ 1 (mod m, which is contrdiction since (m 1/ 1 Therefore ord( (m 1 = ord( = (m 1 Now #(Z/mZ (m 1 since #Z/mZ = m, nd 0 is never unit Furthermore, this uer bound is chieved if nd only if m is rime Since m 1 1 (mod m, it follows tht is unit, nd ord( = m 1 imlies tht #(Z/mZ (m 1 Hence we must hve equlity here, nd m must be rime = : Assume m = 4 m + 1 is rime First we need to show (, m = 1 This is cler becuse m is rime nd m 5, so m ( By roerties of the Legendre symbol, we hve (m 1/ (mod m We ( m wnt to show tht = 1 Now (, m = 1 nd we know tht m 1 (mod 4 m ( ( m since m = 4 n + 1 Therefore, = Now, m = 4 n n + 1 (mod m since 4 1 (mod Hence ( m = So (m 1/ s desired ( = 1 (mod m, nd m ( = 1 (m 1/ 1 (mod m Det of Mthemtics, University of Cliforni, Snt Cruz, CA E-mil ddress: nd nd 10

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