MT S VN C S

V PHNG TRNH NGHIM NGUYN

Tc Gi : Ph Thi Thun10 chuyn Ton THPT chuyn TH - Bnh Thun

Trong chng trnh ton THCS v THPT th phng trnh nghim nguynvn lun l mt ti hay v kh i vi hc sinh. Cc bi ton nghimnguyn thng xuyn c mt ti cc k thi ln, nh, trong v ngoi nc.Trong bi vit ny ti ch mun cp n cc vn c bn ca nghimnguyn (cc dng; cc phng php gii) ch khng i su (v vn hiu bitc hn). Ti cng s khng ni v phng trnh Pell (v n c nhiu trong ccsch) v phng trnh Pythagore; Fermat (cng c nhiu trong sch; khinim rt n gin) Ch : cc bn c th tm c thm cun "phng trnhv bi ton nghim nguyn" ca thy V Hu Bnh.

Phng Php 1: p Dng Tnh Chia Ht

Dng 1: phng trnh dng ax+ by = c

V d 1: Gii phng trnh nghim nguyn sau:

2x+ 25y = 8 (1)

Gii: C th d dng thy y chn. t y = 2t. Phng trnh (1) trthnh: x+ 25t = 4.

T ta c nghim phng trnh ny:x = 4 25ty = 2tt Z

Ch : Ta cn c cch th 2 tm nghim ca phng trnh trn. lphng php tm nghim ring gii phng trnh bc nht 2 n. Ta davo nh l sau: Nu phng trnh ax+ by = c vi (a; b) = 1 c 1 tp nghiml (x0; y0) th mi nghim ca phng trnh nhn t cng thc:

x = x0 + bty = y0 at

t Znh l ny chng minh khng kh (bng cch th trc tip vo phng

trnh) Da vo nh l ny ; ta ch cn tm 1 nghim ring ca phng trnhax+by = c . i vi cc phng trnh c h s a; b; c nh th vic tm nghimkh n gin nhng vi cc phng trnh c a; b; c ln th khng d dng

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cht no . Do ta phi dng n thut ton Euclide (cc bn c th tmc cc sch ; ti s khng ni nhiu v thut ton ny) . Ngoi ra cn cthm phng php hm Euler .

Dng 2: a v phng trnh c s:

V d 2: Gii phng trnh nghim nguyn sau:

2x+ 5y + 3xy = 8 (2)

Gii:(2) x(2 + 3y) + 5y = 8 3[x(2 + 3y) + 5y] = 24 3x(2 + 3y) + 15y = 24 3x(2 + 3y) + (2 + 3y) 5 = 34 (3x+ 5)(3y + 2) = 3434 = 17.2 = 34.1Lp bng d dng tm c nghim phng trnh trn.V d 3: Gii phng trnh nghim nguyn sau:

x2 + 2y2 + 3xy 2x y = 6 (3)

Gii:(3) x2 + x(3y 2) + 2y2 y + a = 6 + aa l 1 s cha bit; a s c xc nh sau.Xt phng trnh: x2 + x(3y 2) + 2y2 y + a = 0 = (3y 2)2 4(2y2 y + a) = y2 8y + 4 4aChn a = 3 = y2 8y + 16 = (y 4)2 x1 = y 1; x2 = 2y + 3T ta c phng trnh c s: (x+ y + 1)(x+ 2y 3) = 3Dng 3: Phng php tch cc gi tr nguyn

V d 4: Gii phng trnh nghim nguyn sau:

xy x y = 2 (4)Gii:(4) x(y 1) = y + 2 x = y+2

y1 x = 1 + 3

y1 (y 1)|3Phng Php 2: Phng Php La Chn Modulo (hay cn gi l xt s

d tng v)

2

Trc tin ta c cc tnh cht c bn sau: 1 s chnh phng chia 3 d0, 1; chia 4 d 0, 1 ; chia 8 d 0, 1, 4

V D 5: Gii phng trnh nghim nguyn sau:

x2 + y2 = 2007 (5)

Gii:x2 0; 1(mod4) y2 0; 1(mod4) V T = x2 + y2 0; 1; 2(mod4)Cn V P = 2007 3(mod4) Do phng trnh trn v nghim.C th m rng thm cho nhiu modulo nh 5; 6; v m rng cho s

lp phng; t phng; ng phng... Ta n vi v d sau:V d 6: Gii phng trnh nghim nguyn dng sau:

19x + 5y + 1890 = 1975430

+ 1993 (6)

Gii:D thy V T 19x(mod5).Mt khc: 19x = (20 1)x (1)x(mod5)x chn th 19x 1(mod5); x l th 19x 1 4(mod5) V T 1; 4(mod5)Cn V P 1993 3(mod5) (v l)Do phng trnh trn v nghim.Ch : Nhiu bi ton nghim nguyn trong thi v ch ton cc nc

i khi phi xt n modulo khc ln ; ta xt n v d sau:V D 7:(Balkan1998) Gii phng trnh nghim nguyn sau:

m2 = n5 4 (7)Gii:m2 0; 1; 3; 4; 5; 9(mod11)n5 4 6; 7; 8(mod11) (v l)Do phng trnh ny v nghim.Ch 3 dng; tht ngn gn v p phi khng no. Ni chung xt

modulo hiu qu cn phi ty thuc vo s nhy bn ca ngi lm ton.Ni thm: i vi cc phng trnh nghim nguyn c s tham gia ca

cc s lp phng th modulo thng dng l modulo9 v x3 0; 1; 8(mod9)(hy t chng minh).

Ta xt V D sau.V D 8: Gii phng trnh nghim nguyn sau:

x3 + y3 + z3 = 1012 (8)

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Gii:Da vo nhn xt trn : (8)x3 + y3 + z3 0; 1; 2; 3; 6; 7; 8(mod9)Cn 1012 4(mod9) (v l).Do phng trnh trn v nghim .Phng Php 3: Dng Bt ng Thc

Dng 1: i vi cc phng trnh m cc bin c vai tr nh nhau th

ngi ta thng dng phng php sp xp th t cc bin.V D 9: Gii phng trnh nghim nguyn dng sau:

x+ y + z = 3xyz (9)

Gii:Khng mt tnh tng qut c th gi s 1 x y z 3xyz = x+ y + z 3z xy 1 x = 1; y = 1 z = 1Nghim phng trnh l (1; 1; 1)Dng 2: i vi cc phng trnh nghch o cc bin ta cng c th

dng phng php ny (nu vai tr cc bin cng nh nhau ). Cch giikhc dnh cho:

V D 9: Chia 2 v phng trnh trn cho xyz ta c: 1xy+ 1

yz+ 1

zx= 3

Gii:Khng mt tnh tng qut c th gi s 1 x y z 1

xy+ 1

yz+ 1

zx= 3 3

x2

x2 1 x = 1 y = 1 vz = 1. Ta xt n mt v d tip theo thy s hiu qu ca phng

php ny:V D 10: Gii phng trnh nghim nguyn dng sau:

Gii:1x+ 1

y+ 1

z= 1

Khng mt tnh tng qut c th gi s:1 x y z 1

x+ 1

y+ 1

z= 1 3

x x 3.

Ln lt th: x = 1; phng trnh v nghim nguyn.Xt x = 2 1

y+ 1

z= 1

2 2

y y 4

Mt khc y x = 2 y {2; 3; 4}.Ta th y ln lt;y = 2 phng trnh v nghim nguyn;

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y = 3 z = 6y = 4 => z = 4Xt x = 3 1

y+ 1

z= 2

3 2

y y 3Mc khc y x = 3 y = 3 z = 3.Vy nghim phng trnh l (2; 3; 6); (2; 4; 4); (3; 3; 3) v cc hon v.

Dng 3: p Dng Cc Bt ng Thc C in.

V D 11: Gii phng trnh nghim nguyn dng sau:

x6 + z3 15x2z = 3x2y2z (y2 + 5)3 (10)Gii:(10) (x2)3 + (y2 + 5)3 + z3 = 3x2z(5 + y2)p Dng BDT Cauchy cho 3 s; ta c V T V PDu = xy ra x2 = y2 + 5 = zT phng trnh x2 = y2 + 5 (x y)(x+ y) = 5(phng trnh c s; d dng tm c x; y ri tm ra z).p s: nghim phng trnh l (x; y; z) = (3; 2; 9)

Ghi ch: Vic p Dng BDT vo bi ton nghim nguyn rt t dng vn dng BDT rt d b "l" nu ngi ra khng kho lo. Tuy nhincng c mt vi trng hp dng BDT kh hay. Ta n vi V D sau.

V D 12: Gii phng trnh nghim nguyn dng sau vi x; y; z l cc

s i 1 khc nhau.

x3 + y3 + z3 = (x+ y + z)2 (11)

Gii:p dng BDT quen thuc sau:x3+y3+z3

3 (x+y+z

3)3

x3 + y3 + z3 = (x+ y + z)2 (x+y+z)39

x+ y + z 9. V x; y; z khc nhau x+ y + z 1 + 2 + 3 = 6 x+ y + z {6; 7; 8}Ln lt th cc gi tr ca x+ y + z ta tm c x; y; zp s: (1; 2; 3) v cc hon v .

Dng 4: p dng tnh n iu ca bi ton. Ta ch ra 1 hoc 1 vi gi

tr ca bin tho phng trnh ri chng minh l nghim duy nht.V D 13: Gii phng trnh nghim nguyn dng sau

3x + 4x = 5x (12)

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Gii:(12) (3

5)x + (4

5)x = 1

x = 1; phng trnh v nghim nguynx = 2; tho mn.x 3 (3

5)x < (3

5)2 (4

5)x < (4

5)2

(35)x + (4

5)x < 1.

Do x = 2 l nghim duy nht ca phng trnh. Cn phng trnhny th sao nh: (

3)x + (

4)x = (

5)x

Bng cch tng t; d dng nhn ra x = 4 l nghim duy nht.Ni thm : i vi phng trnh trn; ta c bi ton tng qut hn. Tm

cc s nguyn dng x; y; z tho: 3x + 4y = 5z.p s n gin l x = y = z = 2 nhng cch gii trn v tc dng vi

bi ny. gii bi ny th hu hiu nht l xt modulo (cc phng trnhcha n m th phng php tt nht vn l xt modulo). Phn ny chni thm nn chng ta tm thi khng gii bi ton ny by gi m s lidp khc.

Dng 5: Dng iu kin hoc 0 phng trnh bc 2 c nghim.V D 14: Gii phng trnh nghim nguyn sau:

x2 + 2y2 = 2xy + 2x+ 3y (13)

Gii:(13) x2 2x(y 1) + 2y2 3y = 0 = (y + 1)2 (2y2 3y) = y2 + 5y + 1 0Gii bt phng trnh trn khng kh; d dng suy ra c:292

y 52

292

Do y nguyn nn d dng khoanh vng c gi tr ca y v th chn. Nichung th phng php ny c dng khi () c dng f(x) = ax2+bx+c(hoc f(y)) vi h s a < 0 .

Cn khi a > 0 th dng phng php ni n trong v d 3 a vphng trnh c s 1 cch nhanh chng.

Phng Php 4: Phng php chn hay ta c th gi n bng 1 ci tn

khc l p hn l phng php nh gi. Phng php nh gi c bn davo 2 nhn xt sau:

1. Khng tn ti n Z tho a2 < n2 < (a+ 1)2 vi a Z2. Nu a2 < n2 < (a+ 2)2 vi a;n Z th n = a+ 1.

Ta n vi V D sau:

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V D 15: Gii phng trnh nghim nguyn sau:

x4 + x2 + 1 = y2 (14)

Xt hiu (x2 + 1)2 y2 = x2 0 (x2 + 1)2 y2Xt hiu y2 (x2)2 = x2 + 1 > 0 => y2 > (x2)2 (x2)2 < y2 (x2 + 1)2Theo nhn xt trn y2 = (x2 + 1)2Th vo phng trnh ban u x4 + x2 + 1 = (x2 + 1)2 x2 = 0 x = 0Nhn xt trn c th m rng vi s lp phng; ta n vi v d tip

theo:V D 16: Gii phng trnh nghim nguyn sau :

x3 y3 = 2y2 + 3y + 1 (15)

Gii:Bng cch trn ta c c : (y 1)3 < x3 (y + 1)3suy ra hoc x = y hoc x = y + 1 ln lt xt x = y;x = y + 1 ta tm

c cc nghim phng trnh l: (1;1); (1; 0)Phng Php 5: Dng tnh cht ca s chnh phng .

Dng 1: Trc tin ta n vi 1 mnh sau :

xy = z2 vi (x; y) = 1 th x = k2

y = t2

kt = z

Chng minh mnh ny khng kh ; ta chng minh bng phn chng:Gi s x; y khng l s chnh phng nn trong phn tch thnh c nguynt ca x hoc y tn ti 1 s cha t nht 1 c nguyn t p vi s m l .Gi s l x . V (x; y) = 1nn y khng cha tha s p => z2 cng cha thas p vi s m l (v l tri vi iu kin z2 l s chnh phng) . By gi tan vi 1 v d .

V D 17: Gii phng trnh nghim nguyn sau:

2x4 + 3x2 + 1 y2 = 0 (16)

Gii:(16) (x2 + 1)(2x2 + 1) = y2

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R rng (x2 + 1; 2x2 + 1) = 1

{

x2 + 1 = t2

2x2 + 1 = z2

T phng trnh x2+1 = t2 => (tx)(t+x) = 1 (phng trnh c s)T tm c nghim phng trnh . p s: (x; y) = (0; 1)

Dng 2:

Ta c mnh th 2:Nu n; t l cc s nguyn tho n(n+1) = t2 th hoc n = 0 ; hoc n+1 = 0Chng minh mnh ny khng kh:Gi s n 6= 0;n+ 1 6= 0 => t 6= 0 n2 + n = t2

4n2 + 4n = 4t2

(2n+ 1)2 = 4t2 + 1Dng phng php chn: (2t)2 < (2n+ 1)2 < (2t+ 1)2

V l do mnh c chng minh.By gi p dng mnh trn ; ta n vi v d sau .V D 18: Gii phng trnh nghim nguyn sau:

x2 + 2xy + y2 + 5x+ 5y = x2y2 6 (17)

Gii:(17) (x+ y + 2)(x+ y + 3) = x2.y2Suy ra hoc x+ y + 2 = 0 hoc x+ y + 3 = 0. Phng trnh ny vn cn

nhng cch gii khc nhng iu ti mun nhn mnh chnh l vic dngmnh trn gip cho li gii bi ton tr nn ngn gn hn .

Phng Php 6: Li v hn (hay cn gi l phng php xung thang).

Phng php ny dng chng minh mt phng trnh f(x; y; z; ) no ngoi nghim tm thng x = y = z = 0 th khng cn nghim no khc.Phng php ny c th c din gii nh sau: Bt u bng vic gi s(x0; y0; z0; ) l nghim ca f(x; y; z; ) . Nh nhng bin i ; suy lun shc ta tm c 1 b nghim khc (x1; y1; z1; ) sao cho cc nghim quan hvi b nghim u tin bi 1 t s k no . V D: x0 = k.x1; y0 = k.y1; . Ri li t b (x2; y2; z2; ) tho x1 = k.x2; y1 = k.y2; . Qu trnh ctip tc dn n: x0; y0; z0; chia ht cho ks vi s l 1 s t nhin tu .iu ny xy ra x0 = y0 = z0 = = 0 . r rng hn ta xt mt VD .

V D 19: Gii phng trnh nghim nguyn sau:

x2 + y2 = 3y2 (18)

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Gii:Gi (x0; y0; z0) l 1 nghim ca phng trnh trn . Xt theo modulo 3 .

Ta chng minh x0; y0 u chia ht cho 3 . Tht vy ; r rng v phi chia

ht cho 3 => x20 + y20

...3 Ta c: x20 0; 1(mod3) y20 0; 1(mod3) Do => x20 + y

20

...3 x0; y0 u chia ht cho 3 . t x0 = 3.x1; y0 = 3.y1 . Th

vo v rt gn: 3(x21+ y21) = z

20 R rng z0

...3 . t z0 = 3.z1 . Th vo v rtgn: x21 + y

21 = 3z

21 Do nu (x0; y0; z0) l 1 nghim ca phng trnh trn

th (x1; y1; z1) cng l 1 nghim . Tip tc l lun nh trn th x1; y1; z1 u

chia ht cho 3 . Ta li tm c nghim th 2 l (x2; y2; z2) vi x2; y2; z2...3.

Tip tc v ta dn n: x0; y0; z0...3k . iu ch xy ra x0 = y0 = z0 = 0

.V D 20: Gii phng trnh nghim nguyn sau:

x2 + y2 + z2 = 2xyz (Korea1996) (19)

Gii:Gi s (x0; y0; z0) l 1 nghim ca phng trnh trn . x20 + y20 + z20 =

2.x0.y0.z0R rng x20 + y

20 + z

20 chn (do 2.x0.y0.z0 chn) nn c 2 trng hp xy

ra.Trng Hp 1: c 2 s l ; 1 s chn. Khng mt tnh tng qut gi

s x0; y0 l ; z0 chn. Xt theo modulo 4 th: x20 + y

20 + z

20 2(mod4) Cn

2.x0.y0.z0...4 (do z0 chn) (v l)

Trng Hp 2: 3 s u chn. t x0 = 2.x1; y0 = 2.y1; z0 = 2.z1 th vo

v rt gn ta c: x21 + y21 + z

21 = 4.x1.y1.z1 lp lun nh trn ta li c

x1; y1; z1 chn.

Qu trnh li tip tc n: x0; y0; z0...2k vi k N

iu xy ra x0 = y0 = z0 = 0. Tm li nghim phng trnh l(x; y; z) = (0; 0; 0)

Phng Php 7: Nguyn Tc Cc Hn hay cn gi l Nguyn L Khi

u Cc Tr. V mt hnh thc th phng php ny khc vi phng phpli v hn nhng v tng s dng th nh nhau ; u chng minh 1 phngtrnh khng c nghim khng tm thng. Phng php bt u bng vicgi s (x0; y0; z0; ) l nghim ca f(x; y; z; ) vi iu kin rng bucvi b (x0; y0; z0; ). V D nh x0 nh nht hoc x0+ y0+ nh nht...Bng nhng php bin i s hc ta tm c 1 b nghim khc (x1; y1; )tri vi nhng iu kin rng buc trn. V d khi chon b (x0; y0; z0; )vi x0 nh nht ta li tm c b (x1; y1; z1; ) tho x1 < x0. T dnn phng trnh cho c nghim l (0; 0; 0; .0). Ta hy xt 1 v d.

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V D 21: Gii phng trnh nghim nguyn sau:

8x4 + 4y4 + 2z4 = t4 (20)

8x4 + 4y4 + 2z4 = t4

Gii:Gi s (x0; y0; z0; t0) l 1 nghim phng trnh trn vi iu kin x0

nh nht.T phng trnh t chn. t t = 2.t1 Th vo v rt gn ta c:

4x40 + 2y40 + z

40 = 8t

41

R rng z0 chn. t z0 = 2z1 2x40 + y40 + 8z41 = 4t41Tip tc y0 chn. t y0 = 2y1 x40 + 8y41 + 4z41 = 2t41V d thy x0 cng chn. t x0 = 2x1 8x41 + 4y41 + 2z41 = t41Nhn vo phng trnh trn r rng (x1; y1; z1; t1) cng l 1 nghim phng

trnh trn v d thy x1 < x0 (v l do ta chn x0 nh nht)Do phng trnh trn c nghim duy nht (0; 0; 0; 0)Ch : Ta cng c th chn b (x0; y0; z0; t0) tho x0 + y0 + z0 + t0 nh

nht ; l lun tng t v d thy x1 + y1 + z1 + t1 < x0 + y0 + z0 + t0 t cng dn n kt lun bi ton.

Phng Php 8: S Dng Mt Mnh C Bn Ca S Hc. Trc

tin ta n vi bi ton nh sau:Cho p l s nguyn t c dng p = k 2t + 1 vi t nguyn dng ; k l s

t nhin l.

Chng minh rng nu x2t+ y2

t ...p th x...p; y

...pChng minh:

Gi s x...p th r rng y

...p Theo Fermat nh: xp1 1(modp) yp1 1(modp) p = k2t + 1 nn {

xk.2t 1(modp)

yk.2t 1(modp)

xk.2t + yk.2t 2(modp) (21)Mt khc do k l nn theo hng ng thc a2n+1 + b2n+1: xk.2

t+ yk.2

t=

(x2t+ y2

t).A) (A l 1 s no )

R rng:

xk.2t

+ yk.2t 0(modp) (do gi thit x2t + y2t ... p) (22)

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Do theo (21); (22) ta c iu phi chng minh.Xt 1 trng hp nh ca bi ton trn:

Khi t = 1; v k l nn k = 2s+ 1 p = 4s+ 3Lc ta c mnh sau:

p l s nguyn t c dng p = 4s + 3. Khi nu x2 + y2...p th x

...p; y...p

Mnh ht sc n gin ny li l 1 cng c v cng hiu qu i vi nhiubi ton kh.

V D 22: (bi ton Lebesgue)

Gii phng trnh nghim nguyn sau:x2 y3 = 7 (y l 1 trng hp nh ca phng trnh Mordell)Ghi ch: Phng trnh Mordell l phng trnh c dng x2+k = y3(k; x; y

Z); bi ton trn l trng hp phng trnh Mordell vi k = 7Gii:Trc tin ta c b nh sau:Mi s nguyn c dng A = 4t+3 u c t nht 1 c nguyn t c dng

p = 4s+ 3Chng Minh: Gi s A khng c c nguyn s no c dng p = 4s + 3

=> A = (4t1 + 1)(4t1 + 1) = 4.(4t1.t2 + t1 + t2) + 1 = 4h + 1(v l) Do A c 1 c dng 4t1 + 3 Nu 4t1 + 3 l s nguyn t th b c chngminh. Nu 4t1 + 3 l hp s. L lun tng t ta li c 4t1 + 3 c 1 c cdng 4t2 + 3. Nu 4t2 + 3 li l hp s th lai tip tc. V qu trnh trn lhu hn nn ta c iu phi chng minh. Quay li bi ton. x2 = y3 + 7

Xt y chn y3 + 7 7(mod8) x2 7(mod8) (v l do x2 0; 1; 4(mod8))

Xt y l vit li phng trnh: x2 + 1 = y3 + 8 x2 + 1 = (y + 2)(y2 2y + 4)Nu y = 4k + 1 y + 2 = 4k + 3Nu y = 4k + 3 y2 2y + 4 = (4k + 3)2 2(4k + 3) + 4 = 4h+ 3Do y lun c 1 c dng 4n+ 3 v theo b trn th 4n+ 3 lun c

t nht 1 c nguyn t p = 4s+ 3 x2 + 1...p = 4s+ 3Theo mnh trn x...p; 1...p ( v l) Do phng trnh trn v nghim.V D 23:

Gii phng trnh nghim nguyn sau: x2+5 = y3 (phng trnh Mordellvi k = 5)

Gii:Xt y chn y3 0(mod8) x2 + 5 0(mod8) x2 (mod8) (v l do x2 0; 1; 4(mod8))Xt y l Nu y = 4k + 3 y3 3(mod4)

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x2 + 5 3(mod4) x2 2(mod4) (v l x2 0; 1; (mod4))Nu y = 4k + 1 Vit li phng trnh x2 + 4 = y3 1 x2 + 4 = (y 1)(y2 + y + 1)R rng y2 + y + 1 = (4k+)2 + (4k + 1) + 1 = 4t+ 3Do y3 1 c t nht 1 c nguyn t p = 4s+ 3 x2 + 4...p = 4s + 3 4...p p = 2 (v l) Do phng trnh trn v

nghim.V cui cng thy thm s hiu qu ca mnh ny ; ta hy n vi

bi ton ca Euler .V D 24:

Gii phng trnh nghim nguyn sau: 4xy x y = z2Nhng trc ht hy xem li gii ca Euler nhn nhn ra s gi tr

ca mnh trn:Gi s pt c tp nghim (x; y; z) = (a; b; c) vi c l gi tr nh nht ca

z. Suy ra 4ab a b = c2 16ab 4a 4b = 4c2 (16ab 4a) (4b 1) 1 = 4c2 (4a 1)(4b 1) 1 = 4c2 (*)Cng vo 2 v (*) :4(4a 1)2 8(4a 1)cTa c: (4a 1)(4b 1) 1 + (4(4a 1)2 8(4a 1)c) = 4c2 + 4(4a

1)2 8(4a 1)c (4a 1)(4b 1 + 4(4a 1) 8c) = 4(c (4a 1))2 (4a 1)[4(b+ 4a 1 2c) 1) = 4(c (4a 1))2 (**)Vy nu pt (*) c nghim l (a; b; c) th pt (*) cng c nghim l (a; b+

4a 1 2c; c (4a 1))V c l gi tr nh nht ca z Suy ra nghim z = |c (4a 1)| > c |c (4a 1)|2 > c2 pt() > () 4(c (4a 1))2 = (4a 1)(4b 1 + 4(4a 1) 8c) > 4c2 = (4a

1)(4b 1) 1 (4a 1)(4b 1 + 4(4a 1) 8c) > (4a 1)(4b 1) (4b 1 + 4(4a 1) 8c) > 4b 1 4(4a 1) 8c > 0 4a 1 > 2c (1)V a, b c vai tr nh nhau nn ta cng cm c 4b 1 > 2c (2)T (1) v (2) 4a 1 2c+ 1; 4b 1 2c+ 1 pt (*) : 4c2 = (4a 1)(4b 1) (2c+ 1)2 1 4c2 4c2 + 4c c 0 (v l)

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Vy pt ny v nghimNhng nu dng mnh trn th li gii ngn gn hn nhiu:4xy x y = z2 4(4xy x y) = 4z2 16xy 4x 4y = 4z2 (16xy 4x) (4y 1) = 4z2 + 1 (4x 1)(4y 1) = 4z2 + 1 = (2z)2 + 12R rng 4x 1; 4y 1 u c dng 4t+ 3.Tht vy: 4x 1 = 4(x 1) + 3; 4y 1 = 4(y 1) + 3Do (4x 1)(4y 1) c t nht 1 c nguyn t p = 4s+ 3 z2 + 1...p = 4s+ 3 1...p (v l) Do phng trnh trn v nghim.Cc dng c bn ca phng trnh v nh nghim nguyn mnh gii

thiu ht. Vic sp xp cc dng ; phng php l theo ch ca mnh nnt nhiu s sai st. Sau y l phn ni thm v cc phng trnh v nhsiu vit v phng trnh khc (kin thc s si nn mnh ni cng s thi)

u tin l phng trnh dng m : Nh ni th phng trnh dng

m thng c phng php chung l xt Modulo (nhng khng phi l lunlun)

Ta n vi cc V D c bn:V D 25: Gii phng trnh nghim nguyn sau:

2x + 7 = y2 (x Z; y Z) (23)Gii:x = 0: phng trnh v nghim x = 1 y = 3Xt x 2 2x 0(mod4)7 3(mod4) y2 = 2x + 7 3(mod4) (v l do y2 0; 1; 4(mod4))Nghim phng trnh l (x; y) = (1; 3); (1;3)V D 26: Gii phng trnh nghim nguyn sau:

2x + 21 = y2(x Z; y Z) (24)Gii:Xt x l . t x = 2k + 1 2x = 2.4k = 2(3 + 1)k 2(mod3) 2x + 21 2(mod3) (do 21...3) y2 2(mod3) (v l) (do y2 0; 1(mod3))Xtx chn. t x = 2k

13

22k + 1 = y2 y2 22k = 1 (y 2k)(y + 2k) = 1Phng trnh c s; qu n gin. p s (x; y) = (2; 5); (2;5)V D 26: Gii phng trnh nghim nguyn dng sau:

2x + 2y + 2z = 2336 vi x < y < z (Vit Nam 1982) (25)

Gii:(25) 2x(1 + 2yx + 2zx) = 2336 = 25.73R rng 1 + 2yx + 2zx l{

2x = 25

1 + 2yx + 2zx = 73

x = 5.1 + 2yx + 2zx = 73 2yx(1 + 2zy) = 72 = 23.9L lun nh trn 2yx = 23 y = 8 z = 11Nghim phng trnh l (x; y; z) = (5; 8; 11)Ch : Vi cch gii trn ta c th x p phng trnh dng ny: 2x +

2y + 2z = 2n (x y z;n N)p s: (x; y; z) = (n 2;n 2;n 1)V d 27: Gii phng trnh nghim nguyn dng sau:

5x3 = y3 + 317 (26)

Gii: Trong phng trnh ny c s tham gia ca s lp phng v nh ni phn phng php la chn modulo th trong bi ny ; modulo taxt s l modulo 9 y = 0 ; phng trnh v nghim nguyn .

y = 1 x = 4y 2 3y 0(mod9)317 2(mod9) 5x3 = y3 + 317 2(mod9) (v l v 5x3 0; 5; 4(mod9))Ta n vi cc bi ton kh hnV D 28: Gii phng trnh nghim nguyn dng sau:

xy = yx (27)

Gii:

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R rngx = y l 1 nghim. Xt x 6= y.Khng mt tnh tng qut gi s x < y xxy = y x yx = y do y nguyn nn x yx nguyn. y...x.t y = tx,Th vo ta c: xt = tx xt1 = tR rng t 2 (v gi s x 6= y) t = 2 x = 2 y = 4 t 3; lc

r rng x 2Ta chng minh: xt1 > t Do x 2 nn ta ch vic chng minh: 2t1 > t .Ta cm quy np theo t. t = 3 ; ng .Gi s khng nh ng vi t = k tc l 2k1 > kTa cm khng nh ng vi t = k+1 tc l chng minh 2k > k+1 . Rt

n gin ; theo gi thit quy np th: 2k1 > k 2k > 2k > k+1 (do k > 1)Do phng trnh v nghim vi t 3 Kt lun: nghim phng trnh

l (x; y) = (a; a); (2; 4); (4; 2) vi a ZCh : Ta c th gii phng trnh theo cch khc .Nhng trc ht ; ta

cn chng minh mnh sau: an...bn a...b

Ta chng minh phn thun ; phn o l iu hin nhin . Trong phntch a; b ra dng chun tc th s nguyn t p c ly tha tng ng l s; t.Do trong phn tch an; bn ra dng chun tc th s nguyn t p c ly

tha tng ng l ns;nt. V an...bn ns nt s t V p c chn tu

nn a...b

Quay li vi bi ton . Ta ch xt trng hp x 6= y Khng mt tnh tngqut gi s x > y . t x = y + t xy = yy+t xy...yy

x...y x = ty .Ri lm tng t nh trnV D 29: Gii phng trnh nghim nguyn khng m sau

2x 3y = 1 (28)

Gii:Xt theo modulo 3Vit li phng trnh 2x 1 = 3yXt: y = 0 => x = 1. Xt y 1 3y 0(mod3) 2x 1 0(mod3)Mt khc: 2x 1 = (3 1)x 1 (1)x 1(mod3) x chn ( v x chn th (1)x 1 = 0 0(mod3)

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t x = 2k 22k 1 = 3y (2k 1)(2k + 1) = 3y

2k + 1 = 3u

2k 1 = 3vu+ v = y

(2k + 1) (2k 1) = 2 = 3u 3v 2 + 3v = 3uNu u = 0 3v = 1 (v l)Nu u 1 3u 0(mod3) 2 + 3v...3 v = 0 u = 1. y = 1 x = 2 Kt lun: nghim

phng trnh l (x; y) = (0; 1); (2; 1)V D 30: Gii phng trnh nghim nguyn khng m sau:

3x + 4y = 5z (29)

Bi ton ny c cp trong phn trc v y l li gii ca n:Xt theo modulo 3 3x + 4y = 3x + (3 + 1)y 1(mod3) 5z = (6 1)z (1)z 1(mod3) z chn . t z = 2a 3x + 4y = 52a 3x = 52a (2y)y = (5a 2y)(5a + 2y)Do c 2 trng hp xy ra:Trng Hp 1: {

5a 2y = 3m5a + 2y = 3n

(m;n 1) iu ny khng xy ra v (5a2y)+(5a+2y) = 2.5a = 3m+3n...3Nhng 2.5a th khng chia ht cho 3

Trng Hp 2: {5a 2y = 15a + 2y = 3x

(5a 2y) + (5a + 2y) = 2.5a = 3x + 1 1(mod3) 2.5a = 2.(6 1)a 2.(1)a 1(mod3)Do a l . Ta c: 5a 2y = 1 1(mod3) (6 1)a (3 1)y (1)a (1)y 1(mod3)Do a l nn r rng y chn . t a = 2k + 1; y = 2t 52k+1 + 22t = 3xNu t = 1 y = 2 x = z = 2Nu t 2 22t 0(mod8)

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Ta c 52k+1 = 5.25k = 5.(24 + 1)k 5(mod8) V T 5(mod8)Tuy nhin xt modulo 8 cho v phi . Nu x chn ; x = 2s 3x = 9s =

(8 + 1)s 1(mod8)Nu x l ; x = 2s+ 1 3x = 3.9s 3(mod8)T ta c V P = 3x 1; 3(mod8) cn V T 5(mod8) (v l)Kt lun: nghim ca phng trnh l (x; y; z) = (2; 2; 2)V D 31: Gii phng trnh nghim nguyn dng:

2x + 5y = 19z (30)

Gii:19z = (18 + 1)z 1(mod3) 2x + 5y = (3 1)x + (6 1)x (1)x + (1)y 1(mod3) x; y l . t x = 2k + 1 2x = 2.4k = 2(5 1)k 2.(1)k(mod5)Nu k chn 2x 2(mod5)Nu k l 2x 3(mod5) V T = 2x + 5y 2; 3(mod5)Cn V P = 19z = (20 1)z (1)z 1; 4(mod5) V l do phng

trnh trn v nghim .Bi ton vi cc nghim nguyn t

V D 31 : Tm n N :

1. n4 + n2 + 1 l s nguyn t

2. n5 + n+ 1 l s nguyn t

3. n4 + 4n l s nguyn t

Gii:

1. n4+n2+1 = (n2+n+1)(n2n+1) l s nguyn t

p s: n = 1; p = 5V D 32:

Tm s nguyn t p 5p2 + 1 l s nguyn tGii:p chn p = 2 5p2 + 1 = 21 ( khng tho) p l 5p2 + 1 chn nn

l hp s . Vy khng tn ti s p tho iu kin trn .V D 33:

Tm cc s nguyn t x; y; z tho: xy + 1 = z2

Gii:Xt x l xy + 1 = z2 chn z = 2 xy = 3 (khng tn ti x; y tho)Xt x chn x = 2 2y + 1 = z2Nu y l . t y = 2k + 1 2y = 2.4k = 2.(3 + 1)k 2(mod3) 2y + 1 0(mod3) z2...3 z = 3 y = 3Nu y chn y = 2 5 = z2 (v l) . Kt lun: nghim ca phng trnh l (x; y; z) = (2; 3; 3)T bi ton trn hn chng ta d dng hnh dung l li gii bi ton sau:

Tm cc s nguyn t x; y; z tho: xy + 1 = zCc Phng Trnh chng minh v s nghim:

V D 34: Chng minh rng phng trnh x3+ y3 = z4 c v s nghim.

Gii:Ta xy dng nghim ca phng trnh ny .

x4 + y3 = 2a + 2a = 2a+1 Chn z = 2a+17 Do x; y; z nguyn nna...3

a...4

a+ 1...7

Gii h trn ta c a = 84t+48 Kt lun: Phng trnh c v s nghimc dng: (x; y; z) = (221t+12; 228t+16; 212t+7)

V D 36: Chng minh rng phng trnh 7x2 + 3y7 = z11 c v s

nghim .Gii:

t a = 2.7 = 14 R rng tn ti v s s n an+ 1...11

Tht vy ; xt phng trnh 14n+ 1 = 11k (r rng c v s nghim)Ch 7 + 3 = 10 Do phng trnh c v s nghim c dng: x =

10an2 ; y = 10

an7 ; 10

an+111

Do an...2; an

...7; an+ 1...11 nn x; y; z nguyn.

Cn vi phng trnh ny th sao nh: 7x2+3y2 = 6.z11 Rt n gin Taa v phng trnh V d trn 7.610x2 + 3.610y2 = (6.z)11

Sau y l phn cc bi tp ; mnh s xp cc bi tp khng theo tngdng v cc bn phi xc nh dng ca n c phng n x l thch hp.

Phng trnh vi tp Z:

1. 3x+ 7y = 9

2. 25x+ 7y = 16

3. x2 + 3xy y2 + 2x 3y = 54. 2x2 + 3y2 + xy 3x 3 = y5. x41 + x

42 + .....+ x

414 = 1599

6. x2 + y2 = 16z + 6

7. x! + y! = (x+ y)!

8. x! + y! = z!

9. 19x3 17y3 = 5010. 5x3 + 11y3 + 13z3 = 0

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11. x2 = y3 + 16

12. x2 + y2 = 6(z2 + t2)

13. xy 2y 3x+ x2 = 314. 5(x2 + xy + y2) = 7(x+ 2y)

15. x3 y3 xy = 1516. x2 + xy + y2 = x+ y

17. 1 + x+ x2 + x3 = y2

18. x3 + y3 = (x+ y)2

19. y3 x3 = 2x+ 120. x4 + x2 + 4 = y2 y21. x2 + y2 = 7z2

22. x2 = y3 + 16

23. x2 + y2 + z2 = x2y2 (Hn Quc 1988)

24. 6(6x2 + 3y2 + z2) = 5t2

25. 19x2 + 28y2 = 2001

26. x2 + xy + y2 = 2x+ y

27. x2y2 = z2(z2 x2 y2) (Bulgari 1998)28. n4 + 2n3 + 2n2 + 2n = 1 = y2

29. y3z2 + (y3 2xy)z + x(x y) = 030. x4 + x2 y2 + y + 10 = 031. 2(x+ y + z) + 9 = 3xyz

32. x4 + (x+ 1)4 = y2 + (y + 1)2

33. y3 = x3 + 2x+ 1

34. (x 2)4 x4 = y3

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35. x4 2y2 = 1Tp N

36. xyz = 2(x+ y + z)

37. x+ y + z + t = xyzt

38. x1 + x2 + ....+ x12 = x1.x2...x12

39. xn + x2n = y2n

40. m3 + n3 = n3

41. x2 + y2 = 2011(10 z)42. 2n + 122 = z2 9

Cc bi Ton vi s nguyn t:

43. Tm x x4 + 4x l s nguyn t

44. 1x+ 1

y= 2

p

45. (p 1)! + 1 = pn (p nguyn t ; n N)46. p(p+ 1) + q(q + 1) = n(n+ 1) p; q;n nguyn t .

47. p2 = 8q + 1 (p; q nguyn t)

Cc bi ton kh:

48. (APMO) Tm n nguyn dng phng trnh sau c nghim xn +(2 + x)n + (2 x)n = 0

49. Chng minh rng phng trnh sau c v s nghim: (Brazil 1990)a3 + 1990b3 = c4

50. (Rumani 2001) x3 + y3 + z3 = n.x2y2z2 (n;x; y; z N).51. (a b)2 = k(4ab 1) (a; b Z)52. 1! + 2! + ....+ x! = yz x; y; z 153. Cho n Z CMR nu A = 2 + 228n2 + 1 l s nguyn th A l s

chnh phng

54. (Nga 1996) 2xx = yy + zz x; y; z N

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55. Chng minh rng phng trnh sau c v s nghim a2b2+b2c2+c2a2 =x2 a; b; c; x Z

56. 3x + 4y = 7z (x; y; z Z)57. 5x = 1 + 2y x; y N58. 3x + 3y = 6z x; y N59. m! + 48 = 48(m+ 1)n

60. (a2 + b2)c = (ab)1999 (a; b; c N)61. (a2 + b2)m = (ab)n(a; b;m;n N)62. 28x = 19y + 87z (x; y; z Z)63. (Sng Tc) Chng minh rng phng trnh a5 + b1980 = c19 c v s

nghim .

64. (Sng Tc) 2n 1 = 343x3 (x;n N)65. (IMO 2006) 1 + 2x + 22x+1 = y2

66. Tm n phng trnh c nghim (x+y+u+v)2 = n2xyuz (x; y;u; v;n n)

67. x2 + y2 + z2 = 1975304 x y z x; y; z Z

68. x2+y2

x+yl s nguyn v x

2+y2

x+y

...1978 . CMR x = y

69. k =n+ 1 +

n 1 k;n N

70. x4 + y4 = z4 x; y; z N

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