初等数论 习题解答

.ababbb1)b.n9qr(0r9)3当Z且rn27k3n9,；3nk)k)k)19kkk1)1若；3322nk)k)k)1kkk1)8若33232nnn32当Z326nnn322nnn323262nnn证=32是662nnnn(2nn1)(n1)n(2n1)322(nn(n2nn(nn(nn(n=.|n(nn(nn(n.由!k6|(nn(nn(nn(2n.,若(nn(2nn为1是3为3n为1(nn(2n是3.(nn(2n6(nn(2n22226(nn(2n当Z,+6若n．5nnnnn(mnl)!m!n!l!当N+(mnl(mnl)(mnl(nlnl)(nll1)l!=由km(mnl)(mnl(nl,!|(nl)(nllab|(ab)n；当Z且annab|(ab)当Z且annn．ab|(ab)5a≠nna=，当nababbbnnnnn1abnabb1nabbnn1n1当n5abaaaaa且2212345ii1a2kkZ,iiii5555a(2k4k(k52|k(k1)k(k,8|422iiiiiiiii1i1i1i15aqqZ2b(2k1)4k(k1)1bq1k,qZ，，，222**,而ii188cax2c0ppp,p,qa()bc02qqqbpqcq0即ap若pap若qbpqcq222bpq2cq2ap2。这123456789A2或4或8或3或?2和5A2余A5余4和A4余A余8和A8余余3和9A∴A3和9AAy5534．97被5能被3．1至的99?1234567891011121314212815161718192022232425262799599699799899910001001则9.9，．|aaaaaa的特征是：或或整除3210．证明：或或nn1|aaaaaa|nn13210aann1aaaa3210aaa(aaaaaa)．210nn13210aann1aaaaaaa713(aaaaaa)或∴3210nn13nn132101)3a，b，c2001200324|62742,,.．a和ab=?22Z6|．fxZ.fxcfffx2c363|62742,8|627423|),且8|有0,2,4,7,9,50;4;8;2;6.ab+a+b22．fffx,+,b.b.kbf;f=++a+b+,f19874520272461,5f(n)n4f(n)n5n9nf(n)3当n，n=44543212f(n)4f(n)n4n15n=n+n2422f(n)n4n44n(n2)(2n)(n2n2)(n2n2)422222221[(n1)1][(n1)1]f(n)1f(n)1nn22f(n)(n2nn)n6nn)(2n4n2n)(n2n1)2+543432322nnnnnf(n)2232f(n)(n3)(n15)3令n或n或4f(n)=22223nf(n)3∵n+n=n+n=+n+或n知424222222f(n)4f(n)41+nn42f(n)n1)(n1)f(n)当n5531；1.或2或(6n1)(6n1)36nn6n6n16(6nnnn)1，12121212126nnnnN12121kp,p,pN=6pp…pN12k12k1N1+1N,p为p,p,p12kp|设m|[(mm.mm1p是1p|且p|p|,,,aa则121pa小1小21小3pp2)Zn.+Z.+32求aa.p和q求.p和859433.qq.n2，n≥.10911,200345.58nn++n+++n+1.++Z,和+11.322+2∵a+是3a+4是3aa为3a,a,1<≤1或。.3得p33|3|33|.33.1第12p12q32p14p24p24p2设、k,,得k4kh,kqpq2qkp3kp4q5kp,h4q,只qhq2+11.1p,p,,pN=4pp…pNn12k12k1N1+1,p为p,p,,pp|12kn,设222）+2）+…+2+.nstsst-1st-2s+…当22.2n-1n6.或p,.若则3,,.73nnNnnnn1|当N(n1,n2|）即2|n,；==455767445311314445313143511131451122925112922732927344153342121011112343329742qP10Q441314576744535114453131432921314511273292251173(13145112)251113142511513142(445313143)513141744535(57674453)174453557671744532262742ab,求,.abab62742ab62742ab2ab83||3|+又8|abab8，15证∵，∴qq若lg2((p,q)p,qN,p2pp10225252,∵=∴5)=qpqqpqqpqpq25－∴.qpqqpq2p21515((p,q)p,qN,p.(p,q)115∵,∴pq∴2215.a+,,求,.abdtta,b)a,bdtt)。t,t:令(a,b)d,adt,bdt,则有12121212d是a+,,],1令,t+tttt,t为或。121212令,，t+ttt1212令,，t+ttt,11212∴,b为或，t,tx,使xtyt令(a,b)d,adt,bdt,则121212xtt)(yxttt,t)(tt,t)122122121tt,tt)1(,t)d.ta,b,即ab,a,b(a,b12121212122[a,,c2a,,c[a,b,cc,a](a,b,c)(c,a)2abca,,c222[a,,c21[a,b,c][,a](a,b),c)(,a)[,,c][,a]222a,,ca,,c22(a,b,c,a)[,](,),c,c)[,a](,a)(,,)(,a)A+xy是N是|)，y|。|；|；||||以；|。51|(136124－51|(13611－。bc，c35=c,.，，.或c.与，有1号2233对15512到除9，5的26112解：11211，,9为元.此=a77,得++77727k|25（n,即777mn+k为3则++k777+77b+aa与b77107是3为或或或7777ba。7107ba3)ab=］Z.nnn+42下.设ba+.A经B到ABCB6块765432倍abanbnnn，由定理1.14a,ba,b,a,b,a,bnnnnna,bna,bna,bna,bnababnn1，。,,a,ba,ba,ba,bab=）=annnn］）=ab，，abab=ab，nnnnnnnnnn∴］）=abab∴］=abnnnnnnnnn4500,1237549500,2750,1237524750,24750较小，2475027509.99=米.16.a+令(a,b)d,a,b,t,tx,y1,xtt)(yxt121212122tt,t)tt,t)tt,tt)tt,t)11221211212121(,t)da,b(ab,a,b1212()=(,b)=和米.2个.x岁，,,,,5.，，9×131×。3.332∴，．将85，，，，，，，，6．、或;和1;5款3437,由22222）=2[a,,c2a,,c）[a,b,cc,a](a,b,c)(c,a)rppNp|，+且|p为a的p)a的p)=.）r)=)=)。p,，p).p=。p([a,b][b,c][c,a])[a,b]b,c][c,a]lmllm[,,c2[a,b,c]2[a,,c]l,p∴m。2[a,b,c,a]p((a,b,cc,a))pa,bp,c)pc,a)mnnnm∵2a,b,c(a,bb,c)(c,a)[a,b,c]2[a,,c]2npm2∴。p654C3326若．解a=p，a15ppp1p(pp1)14p=，11a21，此时；k1kp112ia5若ip1i1i3ppp1p(pp2p,11，iiiiiiiiiia故5a642,a2或a4,a8a642。）aaaaa61264a42，即，aaa=a2⑴1p即ap3,或⑵为2app122若N,且a=c+nmb,c)1,bp,cq,pq(i,j)jiijiji1j1nmnmabc2pq,apiqj,N,Ni,jj故j22ii22ijiji1j1i1j12',2'N,'N(i,j得iijjij222'nnmmbpp,cqq,'2''jjiiiijji1i1j1j1（44个2与4个2个2与1个。a+b[=与=3,b32ab22233332222322231222222222bbc5abc52522225）））121315133276．3．229213151解：n222323。524).1，C）=3，求A，B，C.331耀12233nn41-4解答5abc9或B|,B,所以B是B是623237,A2323233,知由121212221若B242与3∴A237,B23,C23,,,而且与12112111有0;0,0;0,6组2111111知1-6部分习题解答x|x[sin],x|x[tan],k求2A11,(17)2设37371373771272,。372221232000求333312320001123200023333333312320008，2320007232000343。3则3333333320003413420003413333312320003333p2p(q1)p(p1)(q1),。qqq2kpkpZ1qkqqqkp(qk)pkpkppp1=qqqqp2p(q1)p1kp(qk)p(p1)(q1)q1∴。2qqqqq2k1log1log2log3log求2222log1log2log3log=222222448=222222224×2×23923243921102…23921101k7k7－100100777772323x5x0）xx13x5x0,3x35x故x50478x6.x.x,x385y503x3x8xxx，0xx35555y,5055..即y,yxx38883888x(xx)0xxxx0x22x2515151xxx。222y2x3设y+y3x25y2x33x253(x2)53(x2)53x1y，x+[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345.:，。。故Z1到这到这的3572=601212)2277(12)11(122214(12)22(12221(12+10×13×6+14×9×4+22×3×5+15×9×4+21×3×5+33×－300－42－求632?1635555234数345aaa求a12341到这343435453453003434354534534343545345a,而是4是3a6)2试讨论xy与xy的大小关系。若,yR,1求证：xyxy;1xyx＋xy＋yxyxy＋xy＋xy显然xy＋xy＋xy，证得。2，yZ时，xyxy0,有xyxy;＋而x1.2,y2.1,xy2.52,xy0.52,xy0.02,xyxy;又x1.9,y1.8,xy3.42,xy0.42,xy0.72,xyxy.1991199219996）979797515259622212515259697212229697975151952053853955893129797979797975785969797112311）20082221111111123200821223221111111112322120082232008222225720012001.个5775有235的盏1x1xx2x1x1121x1,2,21；xxxxx2x11x1,x2x,即x01x0,即22x0,2xZ,1x,xx1,x2x2∴1或≤或或19202191xxxx若10010010010077xyx8xx8x7.43.100x项m;8m取m取m取2(2m取8422m≡.bm≡dm)则m且mm是和m取bm≡dm)则a-且c-d由a≡bma≡bm；22≡1≡1323≡22222由a≡bma≡bm224.证明，xZ(15x＋24x＋32x－16x＋3－13)x＋3+3)5432≡5证:由15≡7，2432－16≡≡，－1330≡≡。－－=设md若abcdm)明c≡d证：abcdmm|m∴mc≡dm求3模求3≡)≡≡∴3模23))3),而3≡))≡)=≡=≡2223222222≡3≡2222223.25。n24kk,Z.+≡4≡4.设ma若aamaa2naamaa∴aaaa2323n设paab)ab)a)22p|a),即p|)+pp|或p|+22在≡≡4≡≡7≡1≡≡7+因7+≡1,。1若和d,与4.和d,d|－d|和d,d|－d|∴d|,或2n222证：由,n.设n8|n8222+则a+)≡y，y.22aa9999m,1,1rr,bm=m,bmmk取k,b设mx,xm12mmx+m;12mmmx+≡）122m3mx。rrm(m)m(m)xm;1222mm(m)m2mmmmm2m()m22222m2,m22221=m2=mm,m2222设ts>{x|=+mm1,≤st－st－mmtsu有mumv有m种st－st－tmm=m0到m1的mst－tss=sms)；()(m)：()；()；(2()=5323222(m)m(m)=kp1m的pi1iii12,1(m)2=,1mm=m=pp…mα>1p12ij∴p-pp-p-1-11k1若有无论p为奇为偶pppp1是偶数.则1iiiiiiiiii(m)m1pm2ik果mp(m)p1是偶数.iii1),22220129i>j222,，ijjjφ=109为m为φm最小简化剩余系φ41111111113392222943369777783355598696xxxm12k∴i≠jxxxi=iji∴xxx.12k又xxi≠jxiji∴xxxφm12k∴xxxm.12k7.∴799同.当m.===，====若m.)m=1或m当m.m.=14==p1-1ppk111=m.m=pp…-p-p=2-1kαk-1121km=m=2或m=m32n=.解===.若m.m=1.∵=φ.nn.+…+..m时，2=12.设q3p≡q.22=(3)=(3)=pq.∴p≡q.2222又∵qpp有p22q∴p≡q.∴p≡q.22222设p5p.4===ppp）≡.4422又p2p+8ppp.2244==∴p.41010101被6即故61010=6k46k66而66.104424；；；；∴88)×8×8）9488)×8≡×8≡22am.(77)1117160x的,与而66,≡,6≡,≡≡,23225≡)≡,≡≡)632252m+n(m)1(n)n(m)≡m(n)≡m(n)(m)≡.m(n)(m)≡1m(n)(m)≡.paa≡p-1aaa≡.p-1ap是质=1a为奇a≡.p-1∴a+≡.ap-1aa,a()≡p.n）p-1ap-1q是两相异的质aap-1q-1a≡.a≡∴a≡)≡a≡aa≡)≡a≡ppqqqpp=a≡.≡(m)1≡.设ABAB.当Z|x-|.2m∵()1()1≡a()b∵=a(m)=1mm.设B=<A≤=B<1+5=C≤4+6=又=1，=≡7≡7≡≡44∴B×××x-x-2|x或2|xx≡x-≡.又xx-x+∴当7|x或7|xx-x+≡.6666而x-x-x+x+∴当5|x或5|xxx+x+1≡.484484又x-xx+x+x+1当3|x或3|xxx+x+2284228x+≡.∴|x-4=x+x,232x=2|x-p|9999p2和5kp.p1p个9x-x-≡.又1pp-1x≡x-.2|x或2|xx-≡p-122当3|x或3|xx≡故x-≡6|x-.2p-1p10p1.=≡.p-1kp1k个9p|9999.∴）≡.∴p-1kp1p个9)4.故7)≡3257257)77()7()243..x717400130971218nqP1012773185627117491918311102838086213）.,5:5nqP101222337419748191知,和.故7)817）用同余变形法解：x8.32x）5）1143x1666）.1,∴,∴）6543276x177331）,.ax5(mod18),x5(mod18),x8(mod21).xa(mod35).xa(mod35),x8(mod21).即x=(tx3(mod7),x6(mod13),,x5(mod11).x7(mod24).5x7(mod11),2x4(mod8),6x90(mod19).15x5(mod35).≡，得≡3。MM=，12令M，得M，7MM，1122+≡，得≡7≡。MM=，12令M，得M7，11MM，22+≡≡..x1mod56,）x3mod9.≡y≡2得y≡1≡。x3mod11,x2mod12,）MMM，123x1mod13.令M，得M；111M，MM；222M，得M322+≡2,3,5,72222410mod3,41mod3,xx22x2mod9,4x20mod5,4x2mod5,x13mod25,2即2即x13mod49.24x30mod7.4x3mod7.2x2mod9,MM，x13mod2549.12令，得M；11，M22+≡7，。x12,x1mod10,x13,,:x15,x17.x1mod21.x41mod210,x41mod210,x210.x1mod11.x0mod11.x41mod210,x2mod11.4）））））x2mod3,x3mod5,MMM，123x1mod8.令M，得M1；11M，MM；222M，MM322≡+≡x2mod35,x2mod35,