vuihoc24h.vn-chuyendesohocvmf split 5 (1)

Chng

5Phng trnh ng d

5.1 Phng trnh ng d tuyn tnh 895.2 Phng trnh ng d bc cao 905.3 H phng trnh ng d bc nht

mt n 905.4 Bc ca phng trnh ng d 955.5 Bi tp 95

5.6 ng dng nh l Euler giiphng trnh ng d 96

5.7 Bi tp 101

Trn Trung Kin (Ispectorgadget)Nguyn nh Tng (tungc3sp)

5.1 Phng trnh ng d tuyn tnh

nh ngha 5.1 Phng trnh ng d dng ax b (mod m) c gil phng trnh ng d tuyn tnh vi a, b,m l cc s bit.x0 l mt nghim ca phng trnh khi v ch khi ax0 b (mod m).Nu x0 l mt nghim ca phng trnh th cc phn t thuc lp x0cng l nghim. 4V d 5.1. Gii phng trnh ng d sau: 12x 7 (mod 23)Li gii. Do (12; 23) = 1 nn phng trnh lun c nghim duy nht.Ta tm mt s nguyn sao cho 7 + 23k chia ht cho 12. Chn k = 7suy ra 12x 7.24 (mod 23) x 14 (mod 23)

89

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90 5.2. Phng trnh ng d bc cao

V d 5.2. Gii phng trnh 5x 2 (mod 7) 4Li gii. V (5; 2) = 1 nn tn ti s k = 4 sao cho 2 + 7k chia ht cho

5. Khi y 5x 2 + 6.7 (mod 7) ta c nghim x 305 6 (mod 7)

hay x = 6 + 7k

V d 5.3. Gii phng trnh: 5x 4 (mod 11) 4Li gii. Ta c: {

5x 4 (mod 11)4 4 (mod 11)

p dng tnh cht bc cu ta c: 5x 4 (mod 11) 5x = 11t+ 4Ta c th ly t = 1;x = 3. T phng trnh c nghim duy nht lx 3 (mod 11) Nhn xt. Cch xc nh nghim ny l n gin nhng ch dng ctrong trng hp a l mt s nh hoc d thy ngay s k.

5.2 Phng trnh ng d bc cao

V d 5.4. Gii phng trnh 2x3 + 4 0 (mod 5) 4Li gii. Ta thy x = 2 suy ra 2x3 4 (mod 5).Nn x = 2 l nghim duy nht ca phng trnh cho.

5.3 H phng trnh ng d bc nht mt n

nh ngha 5.2 H phng trnh c dng sau c gi l h phngtrnh ng d bc nht mt n

x b1 (mod m1)x b2 (mod m2)....x bk (mod mk)

Vi m1;m2; ...mk l nhng s nguyn ln hn 1 v b1; b2; ...; bk l nhngs nguyn ty . 4

Din n Ton hc Chuyn S hc

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5.3. H phng trnh ng d bc nht mt n 91

Nhn xt. Trong trng hp tng qut, chng ta c th chngminh c rng: iu kin cn v h phng trnh (5.2) cnghim l UCLN(mi;mj) chia ht bi bj vi i 6= j(1 i, j k). Gi s m = p11 pa22 ...pkk l phn tch tiu chun ca m. Khi

y phng trnh ng d f(x) 0 (mod m) tng ng vi hphng trnh ng d f(x) 0 (mod p1i ), i = 1, 2, ..., k. T suy ra rng nu x b1 (mod p11 ) l mt nghim ca phngtrnh f(x) 0 (mod pi), i = 1, 2, ..., k th nghim ca h phngtrnh ca h phng trnh ng d

x b1 (modp11 )x b2 (modp22 )

...x bk

(modpkk

)cho ta nghim ca phng trnh f(x) 0(modm).Vy trong Trng hp tng qut gii mt phng trnh ngd dn n gii h trn. Vi cc module m1,m2, ...,mk i mtnguyn t cng nhau.

Phng php chung gii:

Trng hp 1: h 2 phng trnh{x b1 (mod m1)x b2 (mod m2)

Vi gi thit d = (m1,m2) chia ht cho b1b2. Trc tin ta nhnxt rng, mi s x = b1 +m1t, t Z l nghim ca phng trnhth nht. Sau ta tm cch xc nh t sao cho x nghim ngphng trnh th hai, ngha l h hai phng trnh trn tngng vi h phng trnh{

x = b1 +m1tb1 +m1t b2 (mod m2)

Chuyn S hc Din n Ton hc

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92 5.3. H phng trnh ng d bc nht mt n

V gi thit d = (m1,m2) l c b1 b2 nn phng trnh: b1 +m1t b2 (mod m2) tng ng vi phng trnh:

m1dt b2 b1

d(mod

m2d

)

Nhng (m1d,m2d

) = 1 nn phng trnh ng d ny cho ta

nghim t t0 (mod m2d

), l tp hp tt c cc s nguyn

t = t0 +m2du, u Z

Thay biu thc ca t vo biu thc tnh x ta c tp hp ccgi tr ca x nghim ng c hai phng trnh ng d ang xtl:x = b1 +m1(t0 +

m2du) = b1 +m1t0 +

m1m2d u, hay x = x0 +mu

vi x0 = b1 +m1t0,m = BCNN(m1,m2).

Vy x x0 (mod m) l nghim ca h hai phng trnh ng dang xt.

Trng hp 2: H gm n phng trnh. u tin gii h haiphng trnh no ca h cho, ri thay trong h hai phngtrnh gii bng nghim tm thy, ta s c mt h gm n 1phng trnh tng ng vi vi h cho. Tip tc nh vysau n 1 bc ta s c nghim cn tm.

V d 5.5. Gii h phng trnh:

x 26 (mod 36)x 62 (mod 60)x 92 (mod 150)x 11 (mod 231)

4

Li gii. H hai phng trnh:{x 26 (mod 36)x 62 (mod 60)

{x = 26 + 36t26 + 36t 62 , t Z.

26 + 36t 62 (mod 60) 36t 36 (mod 60) t 1 (mod 5)


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5.3. H phng trnh ng d bc nht mt n 93

Vy nghim ca h l: x 26 + 36.1 (mod 180) hay x 62 (mod 180)Do h phng trnh cho tng ng vi h:

x 62 (mod 180)x 92 (mod 150)x 11 (mod 231)

V d 5.6. Gii h phng trnh{x 62 (mod 180)x 92 (mod 150)

{x = 62 + 180t

62 + 180t 92 (mod 150) , t Z.

Li gii. Ta c:

62 + 180t 92 (mod 1)50)180t 30 (mod 150)6t 1 (mod 5) t 1 (mod 5)

Vy nghim ca h l:

x 62 + 180.(1+) (mod 900) x 242 (mod 900)H cho tng ng vi:{

x 242 (mod900)x 11 (mod231)

H ny c nghim x 242 (mod 69300) , v y cng l nghim cah cho cn tm.

V d 5.7. Tm s nguyn dng nh nht tha tnh cht: chia 7 d 5,chia 11 d 7 v chia 13 d 3. 4Li gii. Ta c: n1 = 7;N1 = 11.13 = 143;n2 = 11;N2 = 7.13 =91;n3 = 13;N3 = 7.11 = 77.Ta c N1b1 3b1 1 (mod 7) b1 = 2. Tng t b2 = 4; b3 = 1Vy a = 143(2)5 + (91)(4)(7) + (77)(1)(3) = 1430 + 2548 231 =887 vy cc s cn tm c dng b = 877 + 1001k.Vy 877 l s cn tm.


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94 5.3. H phng trnh ng d bc nht mt n

V d 5.8 (Chn i tuyn KHTN). Xt h ng d gm 3 phngtrnh:

xy 1 (mod z) (5.1)yz 1 (mod x) (5.2)xz 1 (mod y) (5.3)

Hy tm s b (x, y, z)nguyn dng phn bit vi1 trong 3 s l 19.4Li gii. T ba phng trnh, theo tnh cht ng d ta ln lt c

xy + 1...z v yz 1...x v zx 1...y

Suy ra

(xy + 1)(yz 1)(zx 1)...xyzx2y2z2 x2yz xy2z + xyz2 + xy yz zx+ 1...xyzxy yz zx+ 1...xyz

Nhn thy do x, y, z nguyn dng cho nn xyz 1. Suy ra xy yz zx+ 1 2xyzMt khc yz + zx xy 1 2xyz (yz + zx xy 1) 2xyzDo ta c bt phng trnh kp 2xyz xy yz zx+ 1 2xyzM xyyzzx+1...xyz xyyzzx+1 = 2xyz, 1xyz, 0,1xyz,2xyz Trng hp 1: xy yz zx+ 1 = 2xyz xy 1 (mod z), yz 1(mod x), zx 1 (mod y)Cho nn ta ch cn tm nghim ca xy yz zx+ 1 = 2xyz l xong.V x, y, z c mt s bng 19 nn ta thay ln lt vo.Nu x = 19 19y yz 19z + 1 = 38yz 39yz 19y + 19z = 1 (39y + 19)(39z 19) = 322 Vi y = 19 hoc z = 19 th tng t. Trng hp 2,3,4,5: xyyzzx+1 = 1xyz, 0,1xyz,2xyz lm honton tng t, ta y c v phng trnh c dng au+bv = ab+uv+xvi x l hng s.a v (a v)(b u) = x v gii kiu phng trnh c s. Bi tonhon tt.


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5.4. Bc ca phng trnh ng d 95

Nhn xt. Bi ton ny m khng cho iu kin mt s bng 19 thkhng a c dng au+ bv = ab+ uv + x (a v)(b u) = x lc suy ra v hn nghim.

5.4 Bc ca phng trnh ng d

nh ngha 5.3 Xt phng trnh ng d f(x) = 0 (mod m) vif(x) = a0x

n + a1xn1 + ...+ an, ai N, i = 0, 1, ..., n

Nu a0 khng ng d 0 (mod m) th ta ni n l bc ca phng trnhng d. 4

V d 5.9. Xc nh bc ca phng trnh 15x68x4 +x2 +6x+8 0(mod 3) 4Li gii. Ta thy 15 0 (mod 3) nn bc ca phng trnh khngphi l bc 6. Phng trnh trn tng ng vi 8x4 + x2 + 2 0(mod 3)

V 8 6 0 (mod 3) nn bc phng trnh l n = 4.

5.5 Bi tp

Bi 1. Gii cc phng trnh sau: a) 7x 6 (mod 13) b) (a + b)x a2 + b2 (mod ab) vi (a, b) = 1 c) 17x 13 (mod 11) d) x2 +x 2 1 (mod 3)

Bi 2. Gii cc h phng trnh: a)

x 1 (mod 3)x 4 (mod 4)x 2 (mod 7)x 9 (mod 11)

b)

5x 1 (mod 12)5x 2 (mod 8)7x 3 (mod 11)

Bi 3. Tm a nguyn h phng trnh sau c nghim


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96 5.6. ng dng nh l Euler gii phng trnh ng d

a)

x 3 (mod 3)x 1 (mod 4)x 11 (mod 7)x a (mod 11)

b){

2x a (mod 3)3x 4 (mod 10)

Bi 4. Mt lp gm 40 hc sinh ng thnh vng trn v quay mtv trong vng trn chi bng. Mi hc sinh nhn c bngphi nm qua mt 6 bn bn tay tri mnh. Chng minh rngtt c hc sinh trong lp u nhn c bng nm ti mnhsau 40 ln nm bng lin tip.

5.6 ng dng nh l Euler gii phng trnhng d

Qua bi vit ny ti xin gii thiu mt phng php gii phngtrnh ng d bng cch khai thc nh l Euler

Trc ht, xin nhc li vi kin thc quen thuc.

nh ngha 5.4 Hm Euler (m) vi s nguyn dng m l cc s tnhin nh hn m l cc s nguyn t vi m. 4

5.6.1 nh l Euler.

nh l 5.1 (Euler) Cho m l s nguyn dng v (a,m) = 1 tha(m) 1 (mod m)

Hm c tnh cht sau:

(mn) = (m)(n) vi (m;n) = 1

Nu p nguyn t (p) = p 1;(pn) = pn pn1(n > 1)


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5.6. ng dng nh l Euler gii phng trnh ng d 97

Nu m = p11 p22 ...pk

k , pi l cc s nguyn t th

(m) = m

(1 1

p1

)(1 1

p2

)...

(1 1

pk

)

By gi ta xt m = a.b trong (a; b) = 1 th c cc kt qu sau

nh l 5.2

a(b) + b(a) 1 (mod ab) (5.4)Chng minh. Theo nh l Euler ta c: a(b) 1 (mod b) m b(a) 0(mod b)Nn a(b) + b(a) 1 (mod b).Tng t ta c:a(b) + b(a) 1 (mod a)Theo tnh cht ng d th : a(b) + b(a) 1 (mod ab) nh l 5.3 Gi s c k(k 2) s nguyn dng m1;m2; . . .mk vchng nguyn t vi nhau tng i mt. t M = m1.m2. . .mk = mitivi i = 1, 2, 3. . . , k ta c

t(m1)1 + t

(m2)2 + ...+ t

(mk)k 1 (mod M) (5.5)

Chng minh. T gi thit ta c (mi, ti) = 1 vi mi i = 1, 2, . . . , k nntheo nh l Euler th

t(m1)1 1 (mod mi) (5.6)

Mt khc vi i; j thuc tp 1;2;. . . ;k v i 6= j th tj chia ht cho mjnn (tj ;mi) = mi hay

t(mi)j 0 (mod mi) (5.7)

t S = t(m1)1 + t(m2)2 + ...+ t

(mk)k

T (5.6) v (5.7) c S tmii 1 (mod mi)V m1;m2; . . .mk nguyn t vi nhau tng i mt, nn theo tnh chtng d thc cS 1 0 (mod m1.m2...mk) S 1 (mod M), tc l c (5.5).


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Khi m rng (5.4) theo hng nng ln ly tha cc s hng ta c ktqu sau.

nh l 5.4 Vi (a, b) = 1 v n, v l hai s nguyn dng no th

an(b) + bv(a) 1 (mod ab) (5.8)Chng minh. tin lp lun t x = a(b).Theo nh l Euler th x = a(b) 1 (mod b) x 1 0 (mod b)ng thi x = a(b) 0 (mod a).T c x(x1) 0 (mod a) v x(x1) 0 (mod b) nn x(x1) 0(mod ab)T x3 x2.x x.x x2 x (mod ab) v c lp lun nh th cxn x (mod ab) hay an(b) a(b) (mod ab)Tng t ta c: bv(a) b(a) (mod ab) nn theo (5.4) c an(b) +bv(a) b(a) + a(b) 1 (mod ab).(5.8) c chng minh.

H qu 5.1 Vi (a; b) = 1 th an(b) + bn(a) 1 (mod ab) H qu ny c th chng minh trc tip khi nng hai v ca h thc(5.4) ln ly tha bc n (s dng khi trin nh thc Newton) v ch rng ab 0 (mod ab). Nn lu rng trong ng d thc th a 6 0(mod ab)!

Vi k hiu nh nh l 5.3 ta c ti.tj 0 (mod M) vi i khc j v mii; j thuc tp 1,2,...,k (nhng t 6 0 (mod M) vi mi i = 1, 2, 3, ...k)

T khi nng hai v ca (5.5) ln ly tha bc n ta c kt qu sau.

nh l 5.5 Vi cc gi thit nh nh l 5.3 ta c:

tn(m1)1 + t

n(m2)2 + ...+ t

n(mk)k 1 (mod M) (5.9)

Vi cc k hiu nh trn ta t a = mi v b = ti th theo (5.4) c

mn(ti)i + t

n(mi)i 1 (mod M) (5.10)

Cng tng v ca k ng thc dng (5.10) v s dng (5.5) ta ckt qu sau:


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5.6. ng dng nh l Euler gii phng trnh ng d 99

nh l 5.6 Vi cc gi thit nh l 5.3 ta c:

m(t1)1 +m

(t2)2 + ...+m

n(tk)k k 1 (mod M) (5.11)

Khi nhn 2 v ca (??) vi mi ta c

m1+(ti)1 +mi.t

(mi)i + mi (mod M) (5.12)

Do mi.t(mi)i = mi.ti.t

(mi)1i = M.t

(mi)1i nn

m1+(t1)i mi (mod M), i = 1, k (5.13)

Cng tng v k ng thc dng (5.13) ta c kt qu sau:


m1+(t1)1 +m

2+(t2)2 + ...+m

1+(tk)k m1 +m2 + ...+mk (mod M)

(5.14)

Khi nhn 2 v ca (5.10) vi ti ta c

m1+(t1)1 +m

2+(t2)2 + ...+m

1+(tk)k m1 +m2 + ...+mk (mod M)

(5.15) t1+(mi)i ti (mod M), i = 1, k (5.16)

Cng tng v ca k ng d dng (5.16) ta c kt qu sau


t1+(m1)1 +t

1+(m2)2 + ...+t

1+(mk)k t1+t2+ ...+tk (mod M) (5.17)

Ch rng ti.tj 0 (mod M) nn khi nng ln ly tha bc n catng t1 + t2 + ...+ tk ta c kt qu sau.


tn1 + tn2 + ...+ t

nk (t1 + t2 + ...+ tk)n (mod M) (5.18)


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Kh nng tm ra cc h thc ng d mi cha phi ht mi bnc nghin cu thm. nm r c nhng phn trn ta tm hiuqua mt s v d sau y.

V d 5.10. Tm t nht bn nghim ca phng trnh ng d:

x3 + y7 1 (mod 30) (5.19)Li gii. Do 30 = 5.6 v (6; 5) = 1 nn theo (5.4) c 5(6) + 6(5) 1(mod 30)v (6) = (2).(3) = 2 v (5) = 4; 62 6 (mod 30).Tng t ta c: 257 25 (mod 30) v 63 6 (mod 30) nn 63 +257 26 + 6 1 (mod 30)Nu phn tch 30 = 3.10 vi (3; 10) = 1 th theo (5.4) c 3(10)+10(3) 1 (mod 30). Tnh ton tng t nh trn ta c 34 + 102 1 (mod 30).V 34 = 81 21 (mod 30) v 102 10 (mod 30) nn theo (5.8) c(34)3 + (102)7 1 (mod 30) v (34)7 + (102)3 1 (mod 30)Suy ra phng trnh trn c t nht bn nghim (x; y) l (25; 6); (6; 25);(21; 10); (10; 21).

V d 5.11. Chng minh rng phng trnh ng d sau c nghim(x; y; z; t) khc (0; 0; 0; 0):

x4 + y4 + z4 + t4 t3 (mod 60).Li gii. 60 = 3.4.5 v (5; 3) = 1; (5; 4) = 1; (3; 4) = 1 nn t m1 =3;m2 = 4;m3 = 5; t1 = 15; t2 = 1; t3 = 20 theo (5.18)

154 + 124 + 204 (15 + 20 + 12)4 1 (mod 60)

V d 5.12. Tm t nht mt nghim ca phng trnh ng d x17 +y19 1 (mod 35) 4Li gii. Ta c: 35 = 5.7 m (5; 7) = 1 nn theo (5.4): 5

7+ 7

5 1(mod 35))V (5) = 4;(7) = 6 nn 54 + 76 1 (mod 35)Theo (5.8): 1417 + 3019 14 + 30 1 (mod 35)Vy phng trnh ng d c t nht mt nghim (x; y) = (14; 30)


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5.7. Bi tp 101

5.7 Bi tp

Bi 1. Chng minh rng phng trnh ng d sau c nghim (x; y; z; t)khc (0; 0; 0; 0):

a) x3 + y3 + z3 t3 (mod 210)b) x5 + y5 + z5 t5 (mod 1155)

Bi 2. Tm t nht mt nghim ca phng trnh ng d sau:

x11 + y13 1 (mod 45)

Bi 3. Chng t rng mi phng trnh sau c nghim nguyn dng.a) 2x + 3y + 5z + 7t 3 (mod 210)b) 3x + 5y + 7z 2 (mod 105)


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Li gii thiuc v Bic s, c s chung, c s chung ln nhtBi s, bi s chung, bi s chung nh nhtBi tp ngh

S Nguyn TMt s kin thc c bn v s nguyn tMt s bi ton c bn v s nguyn tBi tpPh lc: Bn nn bit

Bi ton chia htL thuyt c bnPhng php gii cc bi ton chia ht

Phng trnh nghim nguynXt tnh chia htS dng bt ng thcNguyn tc cc hn, li v hn

Phng trnh ng dPhng trnh ng d tuyn tnhPhng trnh ng d bc caoH phng trnh ng d bc nht mt nBc ca phng trnh ng dBi tpng dng nh l Euler gii phng trnh ng dBi tp

H thng d v nh l Thng d Trung HoaMt s k hiu s dng trong bi vitH thng dnh l thng d Trung HoaBi tp ngh & gi p s

Mt s bi ton s hc hay trn VMFm3+17 4 ...3nc(ac+1)2=(5c+2)(2c+b)

Ti liu tham kho

vuihoc24h.vn-chuyendesohocvmf split 5 (1)

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