Phng trnh nghiem nguyen 2011 1Oeo o Qaooq Toan 3A HSP Hue c oa Phuong trnh nghiem nguyen (phuong trnh giai duoc tren tap so nguyen) |amotbophanquantrongcuasohoc.L|chsugiaicacphuongtrnh nghiemnguyen|amotcuochanhtrnh|ykyvadayhapdandoivoi nhung ai yeu thch toan hoc. Han cac ban deu da nghe qua cau chuyen thu v| ve |nh |y Iermat! Mot bai toan tuong chung nhu rat don gian: "xn + yn =zn khong co nghiem nguyen duong neu n3"! a khien cho ca gioi toan hoc phai dau dau suot 3 the ki va mai cho den cuoi the ki 20 th |oi giai tronven cho bai toan moi duoctm ra tren co so tinh hoa cua toan hoc hien dai. Bai toan Iermat do |a mot phan trong chuoi cac phuong trnh nghiem nguyen iophantine mot nha toan hoc Hy Lap songoA|exandriavaotheki!!!,nguoichuyennghiencuuvephuong trnh nghiem nguyen. Tu thoi cua iophantine, nguoi ta da tu dat ra cau hoi:Cchoqkhcoqmctthaottcooehaoqeqloltoteoeoelcolphacoq teioh 7lcphootloer Va do chnh |a noi dung cua bai toan Hi|bert thu I0 noi tieng, da duoc nha toan hoc Hi|bert ( nha toan hoc |on nhat cua moi thoi dai) de ra trong so 23 bai toan cho toan hoc the k 20 ta i ai hoi toan hoc quoc te dau the k 20. Va bai toan do da duoc nha toan hoc nguoi Mga Yuri Mati]asievich giai nam I970. Cau tra |oi cho bai toan thu I0|a: !Tuyvay,doivoimotso|opphuongtrnh iophantine,tacothesudungmotsocachsaudegiaichung: +Sudungcactnhchatchiahetdethuheptaphopnghiemcothe. + ung cac uoc |uong ve do |on cua nghiem de thu hep tap hop nghiem cothe.Thongthngtahaysdungentnhchatnghiemcctr (nho nhat hay |on nhat theo mot quan he nao do). Trong pham vi cua mot bai tap |on nay, toi se gioi thieu mot so phuong trnhiophantinevacachgiaichung.Vatruockhibatdauvoicac phuongtrnhdo,taseditmhieudoinetvecachgiaiphuongtrnh iophantine da thuc tren cung nhung kho khan trong viec giai phuong trnh iophantine. Phng trnh nghiem nguyen 2011 2Oeo o Qaooq Toan 3A HSP Hue I.Phng trnh DIOPHANTINE mt n. Chng ta bt u phn ny vi phng trnh Diophantine mt n (hay cn gi l phng trnh a thc) c dng: 10 1 1... 0m mm ma x a x a x a+ + + + =(1) Vi *m N e , ia e 0, i m = . Nu tn ti s nguynxtha mn phng trnh (1) th khi : 1 20 1 1( ... )m mm ma x a x a x a + + + = Vyxl mt c ca am. Nu0ma = th ta d dng tm cxbng cch lit k ccc(huhn)nguynmvnguyndngcaam ri lnltthvo phngtrnh(1)tmragitrphhp.Nu0ma = thrrng0 x = lmt nghim ca phng tnh (1), lc ta xt phng trnh: 1 20 1 1... 0m mma x a x a + + + =i vi phng trnh ny ta li tip tc cch gii trn bng cch phn loi am-1=0 hay khc 0. Nu am-1=0 th ta li tip tc xt tip mt phng trnh bc m-2 v lp li qu trnh gii trn. V bc m cho trc l c nh nn qu trnh ny l hu hn, do vy ta s tm c tt c cc nghim nguyn ca phng trnh (1). Ta xt v d sau: Bi ton : Gii phng trnh sau trn tp : 72 0 x x ++ =(*) Gii Nul mt nghim nguyn ca (*) th2 x . Vy cc gi tr mc th nhn l { } 2, 1,1, 2 . Ln lt thay vo phng trnh (*) ta thy ch c =-1 l tha mn phng trnh (*). Vy phng trnh cho c nghim nguyn duy nht l . Phng php nu trn ni chung l khng kh (ch gm cc tnh ton s cp) tm nghim nguyn ca mt phng trnh a thc c h s nguyn, ngay c khi a thc c bc rt cao. Trng hp ny khc bit hn so vi cc l thuyt v nh l ca i s v nh chng ta bit , cng thc nghim ca phng trnh bc ba v bc bn rt phc tp v cc phng trnh bc cao hn th khng c mt cng thc nghim chung no gii chng. Phng trnh nghiem nguyen 2011 3Oeo o Qaooq Toan 3A HSP Hue Mtvntraynal :Lmthnotmttcccnghimhutca phng trnh (1)? Chng ta gii quyt vn nh sau : Gi s krs=(vi s l cc s t nhin khc 0, k l mt s nguyn v (k,s)=1) l mt nghim hu t khc 0 ca (1). Gi s v tri ca phng trnh (1) l mt a thc bc m th 00 a = . Lc , t phng trnh (1), th

vo ta c : 1 2 10 1 21 2 10 1 1( ... )( ... )m m m mmm m m mm ma k a k a k s as sas a k a k s a s k = + + += + + + Thai phng trnh nytac 0ms a k v mmk as .V(s,k)=1 nn 0s a v mk a .T y l c phng php chung tm cc nghim hu t ca phng trnh (1) nh sau : +Litkcccnguynmvdngcaamvcccstnhincaa0. +Litkccs kxs= viklccaamvslcnguyndngcaa0, (k,s)=1. +Ln lt thvo phng trnh (1) tm ra gi tr tha mn. Talulstnticccscaamva0lhuhnnnvicgiiphng trnh (1) trn tp hu t bng phng php trn s thu c nghim sau hu hn bc th . Ta xt mt v d sau r hn v thut ton ny : Bi ton : Gii phng trnh sau trn :5 43 4 2 0 x x x + = (**) Gii Gi sl mt nghim hu t ca (**), lc

vi{ } 2, 1,1, 2 k = , s={ } 1 . Ln lt kim tra cc gi tr trn vo (**) ta c=1 l nghim hu t duy nht ca phng trnh (**). II.Mt s vn lin quan n cc phng trnh Diophantine hai hay nhiu n. Diylmtscuhictraclinquannvictmnghim nguyn ca phng trnh hai hay nhiu n : +Liuphngtrnhangxtctntinghimnguynhaykhng ? +S cc nghim nguyn ca phng trnh ang xt l hu hn hay v hn ? +Tm tt c nghim nguyn ca phng trnh ang xt ? Phng trnh nghiem nguyen 2011 4Oeo o Qaooq Toan 3A HSP Hue l nhng cu hi m khng phi bt k ai trong chng ta c th c cu tr li. Chngtakhngbit,vdnhphngtrnh 3 3 330 x y z + + = ,ctntinghim nguynhaykhng ?Hayphngtrnh 3 3 33 x y z + + = rrngcbnnghim nguynl( ) ( ) ( ) ( ) ( ) { }, , 1,1,1 , 4, 4, 5, 4, 5, 4, 5, 4, 4 xyz = nhngcphilttc nghimnguynca phngtrnhny?Trlichonhngcuhinykhkhn khng km vic xc nh cc ch s thp phn ca ! Dthyphngtrnh 3 3 32 x y z + + = cvsnghimnguyn,vdnh ( )3 3 2, , (1 6 ,1 6 , 6 ) xyz n n n = + vinnguyndnglmthnghimcaphng trnh.Tuynhinchngtalikhngbitttcccnghimnguyncaphng trnh ny. Tri li ta c th chng minh phng trnh 3 3 34 x y z + + = khng c bt k nghim nguynno.Thtvy,xtphngtrnh 3 3 3x y z k + + = ,rrngnu(x,y,z)l nghim ca phng trnh th 3 3 3x y z + +v k chia cho 9 c cng s d. Ta c nu 9 x q r = +vi{ } 0,1,..., 7,8 r = ,qe , lc 39 ' ' x q r = +vi{ } ' 0,1,8 r = ,' q e . Tng tchoy,tacc 3 39 '' '' x y q r + = + vi{ } '' 0,1, 2, 7,8 r = ,'' q e .Vcuicngp dng cho z ta thu c ng thc : 3 3 39 x y z p s + + = +vi{ } 0,1, 2, 3, 6, 7,8 s = ,pe . Vy nu k chia 9 d 4 hoc 5 th chc chn phng trnh v nghim nguyn. Vy phng trnh 3 3 34 x y z + + =v nghim trn. Thmmtv d na,phng trnh 3 3 36 x y z + + = nhchngta bit l ccc nghimnguynnhl( ) { } , , ( 1, 1, 2), ( 43, 58, 65), ( 55, 235, 236) xyz = nhng chng ta li khng bit s cc nghim nguyn ca phng trnh ny c hu hn hay khng ? ikhinhngkhkhncavictmnghimnguynchxutphttcctnh ton thun ty. Ngha l chng ta bit phng php tm nghim nguyn nhng cc tnh ton lin quan khi thc hin li qu di. V d nh ngi ta c th chng minhphngtrnh 2932 1 xy = cnghimnguyn(x,y)vix>1,y>1(xemthm cch chng minh 2932 1 l hp s chng X [1]) nhng chng ta khng th tm ra nghim . R rng l c phng php c th gii phng trnh ny: Chng tachia2293-1chomtsnobhnn,tiptcnhvychonkhitm c s c phn d bng 0. Nhng cc php tnh li qu di v phc tp nu thc hin bng cc cng c tnh ton n gin. Mt khc, chng ta khng c bt k phng php no cho php xc nh phng trnh 3 3 330 x y z + + =c nghim nguyn hay khng sau khi thc hin cc php ton di dng. D dng chng minh c rng phng trnh ny v nghim trn v 30 chia 9 d 4. Phng trnh nghiem nguyen 2011 5Oeo o Qaooq Toan 3A HSP Hue III.Mt s phng trnh Diophantine c bit. Trc khi nghin cu mt s phng trnh Diophantine c bit, chng ta trang b mt nh l sau vic chng minh n gin hn: nhl1:Gisrnga,blhaistnhinnguyntcngnhausaocho tch ca chng l ly tha bc n ca mt s t nhin, tc l a.b = cn vi c, n l cc s t nhin. Khi bn thn a, b cng l cc ly tha bc n. Chng minh : t (a, c) =d. Khi a = da1, c = dc1 vi (a1, c1) = 1. Theo gi thit a.b = cn chng tacda1b=dnc1nhaya1b=dn-1c1n.Nhngvd|av(a,b)=1nn(d,b)=1 (dn-1, b) = 1. Phng trnh a1b = dn-1c1n cho ta b| dn-1c1n. Nh vy ta c b| c1n. Mt khc cng t phng trnh trn ta c c1n| a1b kt hp vi gi thit (a1, c1) = 1 ta c c1n| b.Nh vy ta chng minh c b| c1n v c1n| b, tc l c1n = b. Khi a1 = dn-1 nn a = a1d = dn. Nh vy ta i n kt lun rng mi s a, b l mt ly tha bc n ca mt s t nhin. T nh l 1, ta c h qu sau: Hqu:Gisk,nvclccstnhinva1,a2,,aklmtdyst nhin nguyn t snh i v a1. a2ak = cn. Khi mi ai (i = 1,.., k) l mt ly tha bc n. H qu trn d dng chng minh t nh l 1 thng qua phng php quy np. 3. 1Phng trnh

. Mc ny chng ta s xt mt phng trnh ba n bc hai rt c bit:

(2) c gi l phng trnh phng trnh Pythagorean. Nh chng ta bit, y l mt phng trnh c bit quan trng trong lng gic v hnh hc gii tch. V mt trng hp c bit ca n, khi x = y, lin quan n mt chng minh n gin nht ch ra s tn ti ca s v t. Chngta s tmttcccnghimnguyncaphng trnh(2).Rrng(2)c nghim tm thng (0,0,0). Nu x v y u khc 0 th chng ta ch vic xt cc gi trx,ynguyndng vvic iduccnghimnguyndngkhngnh hng g n phng trnh. Nu (x,y,z) lmt nghim nguyn ca phng trnh th ta gi (x,y,z) l mt b s Pythagorean. Mt nghim ca phng trnh (2) c gi l mt nghim c s nu cc s x,y,z nguyn dng v nguyn t cng nhau. Ta d dng chng minh c nu (x,y,z) l nghim c s th chng nguyn t snh i. Tht vy, nu tn ti s nguyn t p l c ca x v y. Theo tnh chia ht ta c x2+y2 chia ht cho p, tc l z2chia ht cho p hay z chia ht cho p. Vy (x,y,z )= p. V l v (x,y,z) = 1. Phng trnh nghiem nguyen 2011 6Oeo o Qaooq Toan 3A HSP Hue Nu, , q l mt nghim c s ca (2) v d l mt s t nhin ty , khi : , , x d y d z d q = = = (3) Cng l mt nghim ca (2). Tht vy, v 2 2 2 q + =nn khi nhn d2 vo hai v ta c c iu cn chng minh. Ngcli,nu(x,y,z)lmtnghimnguyndngca(2)thtd=(x,y,z). Nud>1tht, ,x y zd d d q = = = thhinnhin( , , ) q lmtbs Pythagorean(theotnhchtcchunglnnhtth, , q nguyntcngnhau tng i mt). Chng ta gi nghim (x,y,z) ca (2) thuc vo lp th d nu (x,y,z) = d. Nh vy tmttcccnghimnguyndngca(2)thucvolpthdtachcn nhndvottcccnghimcsca(2).Do,khngmttnhtngqut, chngta gii hnvictmccnghimnguynca(2)v tmccnghimcs ca (2). Ta c th tm nghim c s ca (2) bng nh l sau : nhl2 :Bs(x,y,z)viylschnlnghimcscaphngtrnh

khi v ch chi tn ti cc s t nhin m, n khng cng tnh chn l sao cho (m,n)=1, m > n v2 2 2 2, 2 , x m n y mnz m n = = = +Chng minh | | Trc ht ta thy rng nu (x,y,z) l mt nghim c s ca (2) th mt trong hai s x , y l chn v s cn li l l. Tht vy, nu c x v y u chn th x2 +y2 chn, do z2 chn nn z chn, do (x,y,z) 2mu thun vi gi thit y l nghimcs.Nux,yul,tabiudinx=2k+1,y=2h+1.Khix2 = 4k2+4k+11 (mod 4) vy2=4h2+4h+11 (mod 4) nn x2+y22 (mod 4) hay z22 (mod 4). Ta li c z2 l s chn (do x,y u l) nn z chn hay z = 2a. Do z2 = 4a2 0(mod 4) v l vi lp lun va ri. Vy x, y khng cng chn hoc cng l.Theo lp lun trn, ta c th gi s y l s chn (do tnh i xng ca x2+y2). Khi x v z l cc s l. Phng trnh (2) c th vit li : ( )( )2y z x z x = + (4) R rng z+x v z-x l tng v hiu ca hai s l khc nhau nn chng l cc s chn.Theoth2, 2 z x az x b + = = (5)via,blccstnhin.Do , z a bx a b = + = T cc phng trnh ny d dng suy ra a, b nguyn t cng nhau. V nu gi s(a, b) = d >1 th z = kd, x = hd vi h,d l cc s t nhin. Dn ti y2=z2-x2 =(k2-h2)d2 hay d2 l c ca y2 hay d l c ca y. iu ny v l v (x,y,z) = 1. Phng trnh nghiem nguyen 2011 7Oeo o Qaooq Toan 3A HSP Hue Theo gi thuyt, v y chn nn y = 2c vi c l s t nhin. T (4) v (5) ta c 2. c ab = .(6) Nhng v (a,b) = 1 nn theo nh l 1 ta c a = m2, b =n2 vi m, n l s t nhin v (m,n) = 1. T ta c:2 2 2 2, z a b m n x a b m n = + = + = = v v c2 = ab = m2n2 nn c = mn, do y = 2c =2mn. Vy ta ch ra c (x,y,z) l nghim c s ca (2) v y l s chn th: 2 2 2 2, 2 , x m n y mnz m n = = = + (7) vi m,n nguyn dng v (m,n) =1.Hin nhin m > n v x nguyn dng. Hn th na mt trong hai s m, n l chn, s cn li l l. Tht vy, r rng m, n khng th cng chn v (n,m) = 1. V m, n cng khng th ng thi l bi nu th th x, y, z cng chn (do(7)) mu thun vi gi thit (x, y, z) =1. Do vy2 mnnn y =2mn chia ht cho 4. | |:Gi s m, n l cc s t nhin sao cho m > n v (m,n) =1 v mt trong hai s m, n chn, s cn li l. Chng ta d dng thy rng : 2 2 2 2 2 2 2( ) (2 ) ( ) m n mn m n + = + (8) Do ta ch cn chng minh (x, y, z) = 1 vi2 2 2 2, 2 , x m n y mnz m n = = = +hay 2 22 , 2 m z x n z x =+ = . (9) Gi s (x, y ,z) = d >1. Khi d khng th l s chn v x, y l cc s l. Theo (9) th d l c chung ca c m2 v n2 (do d l s l). M v (m, n) =1 nn(m2, n2)= 1. Vy d =1. V l. Do (x, y, z) =1. pdngnh l 2taddngtmcccnghimcscaphngtrnh(2) thngquaccsm,nthamniukincanhl.Ccsm,ncgil nghimsinhcaphngtrnh(2).Vhinnhintac x zy+lmtdngca phn s ti gin mn. lit k ra mt cch y v c h thng cc nghim c s ca (2),ta ln lt cho m nhn cc gi tr 2,3,4v khi ta cho n nhn cc gi tr nguyn t cng m, b hn m v chn (l)khi m l (chn). Nh cp t u, nhn c tt c cc nghim ca (2) ta ch vic nhn cc nghim c s ca n vi cc s t nhin 1,2,3. Bn cnh , vic i v tr x v y trong cc nghim trn cng s cho ta cc nghim ca (2). Phng trnh nghiem nguyen 2011 8Oeo o Qaooq Toan 3A HSP Hue Theo cng thc (8), vic th cc sm, n thch hp vo (7) ch cho chng ta cc nghimcsca(2),ngayckhiichvtrcaxvythcngvnl nghimcs.Vdnhvinghim(9,12,15)ca(2)thkhngtntist nhin m, n no nguyn t cng nhau v m > n sao cho9 = m2-n2 , 12 = 2 mn, 15 = m2+n2. Tht vy, ta ch cn xt cc phn tch ca l15 =12+14 = 22+11=32+6 u khng c dng m2+n2. Mt cch tng qut hn, ta c th xem nghim ca (2) c dng sau : ( )2 2 2 2( ) , 2 , ( ) 10 x m n l y mnl z m n l = = = + vi m, n < m, l l cc s t nhin, m v n tha mn cc iu kin nh l 2.Vy(2)c nghiml(x,y,z)v(y,x,z)sinhratcng thc(10).Tuynhin, cnlurngcngthcnycthcnghimtrngnhaungayckhim,n,l khc nhau. V d, vi (m, n, l) = (2, 1, 4) = ( 4, 2, 1) cng cho mt nghim ca (2) l ( x, y, z) = (12, 16, 20)hay b s (m, n, l) = (8, 4, 1) = ( 4, 2, 4) cng cng cho nghim l ( x, y, z) = (48, 64, 80).Ddng thyrng nghim(3,4,5) l nghimnguyn dng nhnhtca(2) v chng l cc s t nhin lin tip. D dng chng minh y l nghim duy nht ctnhchtny.Nun-1,n,n+1lccstnhinlintipthamnphng trnh (2) th n2 = 4n hay n = 4. Do ch c duy nht mt nghim (3,4,5) c tnh cht ny. 3. 2Phng trnh

Trongccnghimca

taddngnhnrachainghimtha mn phng trnh:

l (3,4,5) v (21,20,29). Ta s chng minh s nghim nguyn ca phng trnh dng ny l v hn. Ta d dng c c phng trnh 2 2 2( 1) x x z + + =do iu kin1 x y = (tc x v y l hai s t nhin lin tip). Khi : 2 22 22 2 2 22 22(3x+2z+1)+ (3x+2z+2)=18x+24xz+8z+18x+12z+5 +(x+l)+16x+16x+24xz+8z+12z+4=16x 9 4 24 8 16=(4x+3z+2)xz xz z x=+ + + + + Nhvy,nuchotrcmtbsPythagoreanthamniukin1 x y = thtasnhncccbsPythagoreancngctnhchtbngbiuthc trn.ngin,taxtbs(3,4,5),lc3.3+2.5+1=20,20+1=21v Phng trnh nghiem nguyen 2011 9Oeo o Qaooq Toan 3A HSP Hue 4.3+3.5+2=29 hay 3.20+2.29+1=119, 119+1=120 v 4.20+3.29+2=169. Chng ta lit k ra 6 b s u tin nhn c t phng php trn nh sau: 3 20 119 696 4059 23660 4 21 120 697 4060 23661 5 29 169 985 5741 33461 Vn t ra l phng php nyc cho ra tt c ccb s Pythagorean c tnh cht trn hay khng? chnh l ni dung ca nh l sau: nh l 3 : Nu cc s t nhin x, z tha mn phng trnh 2 2 2( 1) x x z + + =(12) V nu x > 3 th khi ta c: 0 03 2 1, 3 4 2 x x z z z x = + = (13) l cc s t nhin tha mn phng trnh:2 2 20 0 0( 1) x x z + + = (14) vi z0 < z. Chng minh T (13) ta c :2 2 2 2 20 0 0 02 2 20( 1) 2 2 1 18 8 24 18 12 516 9 24 16 12 4x x x x x z xz x zz x z xz x z + + = + + = + + += + + + T (12) ta li c 2 22 2 1 z x x = + +nn:2 2 2 216 9 24 16 12 4 18 8 24 18 12 5 x z xz x z x z xz x z + + + = + + + Do ,t (15) ta d dng suy ra (14). Ta tip tc chng minh rng cc s x0, z0 l cc s t nhin v z0 < z, tc l chng minh: 3 2 1 0 x z + > v 0 3 4 2 z x z < + < + (16) Ta ch rng x > 3 nn x2 > 3x = 2x + x > 2x +3, khi theo (12) ta c : 2 2 2 22 2 24 8 8 4 9 8 49 8 4 (2 3) 9 6 1 (3 1)z x x x x xx x x x x x= + + = + +< + + + = + + = + Do 2z < 3x+1 v v x > 0,2z < 4x + 2 nn z < 2x +1.V cng theo (12) v x > 0 nn:2 2 2 29 18 18 9 16 16 4 (4 2) z x x x x x = + +> + + = +Nn 3z > 4x+2. Vn cui cng, gis tn ti b s Pythagorean (x, x+1, z) khc bit vi b s (xn, xn+1, zn) c nh ngha theo nh l 3. Trong cc b s (x, y, z) nh th tn ti mt b (15) Phng trnh nghiem nguyen 2011 10Oeo o Qaooq Toan 3A HSP Hue s m z c gi tr nh nht. Lc r rng x khng th b hn hoc bng 3 v nu nh vythtascbsPythagorean(3,4,5)thucvolp(xn,xn+1,zn). t3 2 1, 3 4 2 u x z v z x = + = (17). Theo nh l 3 th (u, u+1, v) l b s Pythagorean v v < z. Ch rng z l s nh nht trong tt c cc' zca tt c cc b s Pythagorean khc vi cc b c dng (xn, xn+1, zn) v ng nhin tn ti mt n t nhin no u = xn, v = zn v t:1 1 1 13 2 1, 1, 4 3 2n n n nx u v y x z u v+ + + += + + = + = + +Theo (17)ta li c: 113(3 2 1) 2(3 4 2) 14(3 2 1) 3(3 4 2) 2nnx x z z x xz x z z x z++= + + + == + + + = Nhvy(x,x+1,z)thucvolpccbsPythagorean(xn,xn+1,zn)canhl3, iu ny mu thun vi gi thit ban u. Vy ta chng minh c tt c cc b s Pythagoreanchaiphntulintipnhauchnhlccbsthucvolp(xn, xn+1, zn). 3. 3Phng trnh

. Theo bn th phng trnh Pythagorean c tn ti nghim c dng (a2, b2,c) hay khng? Cu tr li nm trong nh l sau: nh l 4: Phng trnh 4 4 2x y z + =(18) khng c nghim trn. Chng minh Gi s tri li phng trnh (18) c nghim nguyn dng v z l s t nhin nh nht saocho 2 4 4z x y = +vix,ylccstnhin.Lc(x,y,z)lmtnghimca phng trnh (18). Ta c (x, y) =1 bi v nu (x, y) = d > 1, ta s c x =dx1 v y=dy1 vi x1, y1 l cc s t nhin nguyn t cng nhau. Lc z2 = d4(x14 + y14)v do vy 4 2d z

nn 2d ztcl 21z dz =viz1lstnhin.Bivyphngtrnh(18)trthnh 4 4 2 21 1 1x +y= z < z ,trivigithit(x,y,z)lnghimnguyndngca(18)cznh nht.Bivy,v(x,y)=1nn(x2,y2)=1,dnnx2,y2,zlmtnghimcsca phng trnh Pythagorean: ( ) ( )2 22 2 2x y z + =(19) Khi , theo nh l 2, tn ti cc s t nhin m, n nguyn t cng nhau, m > n sao cho: 2 2 222 22x m ny mnz m n = == + y ta gi s l y chn , x l (do tnh i xng ca x v y). Ta c mt trong hai s m, n phi chn, s cn li l s l. Nu m chn, n l khi x2+n2= m2 dn n c x v n u l. iu ny v l v (x, n, m) l mt nghim c s ca phng Phng trnh nghiem nguyen 2011 11Oeo o Qaooq Toan 3A HSP Hue trnh Pythagorean (do (x, n, m)=1 v x2+n2=m2) nn x, n khc tnh chn l. Vy m l v n = 2k vi k l s t nhin. V (m, n) = 1 nn (m, k) =1. Do y2 =22mk m y = 2l vi l l s t nhin (do y chn). Vy l2=mk. T ta c m =a2, k =b2 vi a, b l cc s t nhin. Ta c n = 2k = 2b2. p dng nh l 2 cho nghim c s (x, n, m) ta c n = 2m1n1 v m = m12+n12 vi m1, n1 l cc s nguyn t cng nhau. V n = 2b2 nn b2= m1n1 nn ta li c m1, n1 l cc bnh phng, tc l m1=a12,n1 =b12.Mm=a2nna2=m12+n12=a14+b14.Nhngv 2 2 2a a m m n z s = < + =hay a < z. iu ny tri vi gi thit ca z. Vy ta hon thnh chng minh nh l 4. Nhn xt: 1)Xt phng trnh Fermat

. R rng 4 4 4 4 4 2 2( ) x y z x y z + = + =V th t kt qu trn ta c phng trnh trn khng c nghim nguyn dng. 2)Tng qut hn: tt c cc phng trnh Fermat n n nx y z + =vi n = 2s, s 2 khng c nghim ngun dng. Tht vy: ( ) ( ) ( )2 2 24 4 42 2 2 2 2 2s s s s s sx y z x y z + = + =Theonhnxt1th 4 4 4x y z + = khngcnghimnguyndngnntaciu phi chng minh. 3. 4Phng trnh

. Cng vic ca phn ny l tm tt c cc nghim nguyn ca phng trnh: 2 2 2 2x y z t + + = (20) Trctintarngtnhthaitrong3sx,y,zphichn.Gisiu khng ng, tc l c duy nht mt s chn hoc c ba s u l. Trong trng hp th nht, khng mt tnh tng qut, ta gi s x chn, y v z u l. Lc x2 chiacho8d0hoc4cny2vz2chiacho8uchosdl1.Do 2 2 2x y z + +chia cho 8 d 2 hoc 6. Ta ch rng t2 l mt s chn nn s d ca n khi chia cho 8 l 0 hoc 4. Do ta c iu mu thun. Trong trng hp th hai ta thy 2 2 2x y z + +chia cho 8 s cho s d l 3 v lc t2 l mt s l nn c s d l 1 khi chia cho 8. Vy gi s ca ta c mu thun.Nh vy ta c th gi s y v z cng chn. Vy: y = 2l, z = 2m (21) vi m, l l cc s t nhin. T (20) ta thy rng t > x. t t x = u (22), ta nhn c s t nhin u m: Phng trnh nghiem nguyen 2011 12Oeo o Qaooq Toan 3A HSP Hue 2 2 2 22 2 22 2 2( ) 4 42 4 44 4 2x u x l mxu u l mu l m xu+ = + + + = + = +

V v phi ca (23) l tng cc s chn nn u chn, tc tn ti s t nhin n sao cho u = 2n (24). Th vo (23) ta c : 2 2 22 2 2l m nn l m nx xn+ = + = (25) Th (22) vo biu thc trn ta c: 2 2 22l m nt x u x nn+ += + = + =Bncnh,rngxlstnhinnnt(25)tac 2 2 2l m n + > .Nhvy chng ta chng minh c rngtt c cc nghim nguyn dngca (20) s nhn c t cng thc : 2 2 2 2 2 2, 2 , 2 ,l m n l m nx y l z m tn n+ + += = = = (26) vi m, n, l l cc s t nhin, n l c ca 2 2l m +v nh hn 2 2l m + . By gi chng ta quan tm n iu ngc li c ng khng? Tc l nu tn ti cc s l, m, n tha mn (26), khi ta nhn c cc gi tr x, y, z, t c phi l cc nghim ca (20) khng ? T gi thit d dng thy rng x, y, z, t l cc s t nhin. Kh t (26) thay vo (20) ta c ngay x, y, z,t l cc nghim ca (20), tc l:( )2 22 2 2 2 2 222(2 ) 2l m n l m nl mn n| | | | + + ++ + = ||\ . \ . Cng t (26) chng ta c: 2yl = , 2zm = , 2t xn=nn l, m, n xc nh duy nht mt b nghim (x, y, z, t). T kt qu trn ta c nh l sau : nh l 5 : Tt c cc nghim nguyn dng ca phng trnh 2 2 2 2x y z t + + = vi y, z l cc s chn, nhn c t cc biu thc sau: 2 2 2 2 2 2, 2 , 2 ,l m n l m nx y l z m tn n+ + += = = =vi l, m l cc s t nhin ty , n l c ca 2 2l m +v nh hn 2 2l m + . Mi nghim nguyn ca phng trnh ny u nhn c t cc cng thc ny. (23) Phng trnh nghiem nguyen 2011 13Oeo o Qaooq Toan 3A HSP Hue nh l 5 khng nhng ni ln c s tn ti nghim nguyn ca phng trnh (20) m cn ch ra cho ta phng php tm cc nghim nguyn ca n. D dng thy rng loi i nhng nghim c cng x, y, z (tc l hon i 3 phn t u ca nghim ta vn nhn c mt nghim ca (20)), chng ta c th loi di nhng cp l, m m l < m v ch ly nhng n lm cho x l s l. nhn li nhng nghim loi, ta ch cn nhn cc nghim ca (20) c x l vi cc ly tha ca3. 5 Phng trnh . Gi s rng x, y, z, t l cc s t nhin tha mn phng trnh xy = zt v t (x, z) =a 1. Khi x = ac, z= ad, vi c, d l cc s t nhin nguyn t cng nhau. Do ta c acy = adt hay cy = dt v v (c, d) =1 nn d|y hay y = bd, vi b l s t nhin. Khi t = bc. iu ny cho thy rng nu cc s t nhin x, y, z, t tha mn phng trnh xy=zt th s tn ti cc s t nhin a, b, c, d sao cho (c, d) =1 v x = ac, y = bd, z = ad, t = bc. Hin nhin iu ngc li cng ng, tc l nu cho trc cc s t nhin a, b, c, d th ta xc nh c x, y, z, t bng cc cng thc trn th khi xy = zt. T ta c nh l sau : nh l 6 : Tt c cc nghim nguyn dng ca phng trnh xy=zt c cho bi cng thc: x = ac, y = bd, z = ad, t = bc vi a, b, c, d l cc s t nhin ty . Hn na iu ny vn ng nu thm iu kin (c, d) =1. nhn c cc nghim ca phng trnh xy = zt, chng ta cng c th lm nh sau:btuviccstnhinxvz.Khiv, 1( , ) ( , )x zxz xz| | = |\ .nntng thc. .( , ) ( , )x zy txz xz= tac ( , )zyxzhay.( , )zy uxz= ,khi.( , )xt uxz= .Do,c nh x, z, u v t ( , )uzyxz= , ( , )uxtxz= th ta nhn c mt nghim (x, y, z, t) ca phng trnh xy=zt. Nh vy, tt c cc nghim (x, y, z, t) ca phng trnh xy=zttrncchobicngthc ( , )uzyxz= , ( , )uxtxz= vix,y,ulccstnhin chotrc.Mtchnhlnux,y,z,tlmtnghimnguyndngca phng trnh xy=ztth ( )( )( ), ,, , ,xz x txxyz t=. Ddngchngminhcrngttcccnghimnguyndngcaphng trnh xy = z2 l cc b s (x, y, z) c cho theo biu thc x = u2t, y = v2t, z = uvt vi u, v, t l cc s t nhin ty . Chng ta c th b sung thm iu kin (u, v) =1 th cng thc trn s cho nghim duy nht. Phng trnh nghiem nguyen 2011 14Oeo o Qaooq Toan 3A HSP Hue Tngt,cngthcchonghimnguyncaphngtrnhxy=z3cara nh sau: 2 3 2 3, , x uv t y u vw z uvtw = = = vi u, v, t, w l cc s t nhin ty . Cui cng, ta xt mt dng c bit ca phng trnh dng xy=ztl 1 1n mi ii ix y= ==[ [vi m, n l cc s t nhin. Lc nghim nguyn ca phng trnh ny c cng thc nh sau : 1 2 11 2 1 1 2 1. ....( ... , ... )mnm my y yx tx x x y y y = , 1 2 11 2 1 1 2 1....( ... , ... )mnm mx x xy tx x x y y y =Vi 1 2 1 1 2 1, ,..., , , ,...,m mx x x y y y , t l cc s t nhin xc nh. 3. 6 Phng trnh

. ChngtasdngchngminhcaJ.Celchrarngphngtrnhtrnv nghim trn . Gisrngphngtrnh

2

(27)cnghimnguyn dng v (x,y,z) l nghim ca (27) c z t gi tr nh nht. Nu (x,y) = d > 1 th x = dx1, y = dy1 v khi theo (27) ta c d4|z2 hayd2|z nn z = d2z1 (x1, y1, z1 l ccs nguyn dng. Chia 2 v phng trnh (27) cho d4 ta nhn c phng trnh 4 2 2 4 41 1 1 1 19 27 x x y y z + + = . iu ny tri vi gi thit ca z. Vy (x, y) = 1.Nu2|x thyls lvphng trnh(27)chota 4|27y2-z2nnzl s l.iu ny khng th v theo (27) z2 l tng ca mt s chn vi 2 s l nn phi l s chn. Vy x l s l. Nu 2y th t (27) cho ta 8|z2-5 (v x, y l nn t x = 2k + 1, y = 2h + 1 ta s nhncvtrisdl5khichiacho8)miunylkhngthvbnh phng ca mt s c s d l 0, 1 hoc 4 khi chia cho 8, do z2-5 chia 8 d 3, 4 hoc 7. Vy x l s l, y l s chn. Nu 3|x th r rng ta c 27|z2 nn 9|z. Khi ta c 81|27y4 nn 3|y,mu thun vi gi thit (x, y) = 1. Vy (3, x) = 1. Chng ta c th chng minh rng (x,z) = 1. Tht vy, nu (x,z) = d th theo (27) ta c d|27y4. Khi (x,3y) = 1 nn (d,27y4) = 1 hay d = 1. t y = 2y1. Phng trnh (27) tng ng vi phng trnh : Phng trnh nghiem nguyen 2011 15Oeo o Qaooq Toan 3A HSP Hue 2 24 2 21 1 127 ( 9 )( 9 )2 2z x z xy y y+ = + (28) V v tri (28) dng nn tng v tch v phi cng dng. t d1 l c chung lnnhtcavtrivvphi.Khi 2 41 127 d y nn 21 19 d y vdovy 21 ( , ) d x z . Nhng (x, z) = 1 nn (x2,z) = 1 hay d1 = 1. Khi : 2 22 4 2 41 1 19 27 , 9 ,2 2z x z xy a y b y ab+ + = = =(29) hoc 2 22 4 2 41 1 19 , 9 27 ,2 2z x z xy a y b y ab+ + = = =(30). via,bnguyntcngnhau.T(29)tac 2 2 2 4 418 27 x a b a b + = hay 2 4 4 2 227 18 x b a a b + = vy 2 43 x b + hay 431 b + (vbnhphngmtskhngchia ht cho 3 c s d l 1 khi chia cho 3). M iu ny l khng th v ly tha bc 4 ca b c dng 3k + r vi r = 0 hoc 1, do vy b4 + 1 = 3k + s vi s = 1 hoc 2. Vy ta ch cn xt phng trnh (30). T (30) ta c 2 2 2 4 418 27 x a b a b + = (31). T ta c a hoc b chn (do x l). Nu a chn th 4 2 2 2 418 278 4 a x a b b k = + + = +(do x v b l nn x4, b4 c dng 8k + 1 cn 18a2b28 (do a chn), 27 = 8.3+3 nn 2 2 2 418 27 x a b b + +chia 8 d 4) m iu ny khng th v a4 phi c dng 8h do a chn. Vy b chn nn ta c: 2 24 2 29 9272 2 2 2a x a xb b b| || | + = | |\ .\ . t: 2 22 29 9,2 2 2 2a x a xd b b| | + = |\ . Ta c d22|27b4 nn d2|9b2 v d2|x. Nh vy d2|(9b2,x) v v (3y,x) = 1 nn d2 = 1.Nu 22902 2a xb < tha2 , 22902 2a xb > v nguyn t cng nhau nn ta c: 2 22 4 2 49 9, 27 ,2 2 2 2a x a xb m b n b mn = = =

Phng trnh nghiem nguyen 2011 16Oeo o Qaooq Toan 3A HSP Hue vi m, n nguyn dng.Nh vy 2 4 2 2 49 27 a m m n n = + +v a y1 y z. Nh vy (m,n,a) cng l mt nghim ca (27), tri vi gi thit (x,y,z) l nghim c z nh nht. Nh vy phng trnh (27) khng c nghim trn . 3. 7Phng trnh

. Gi s rng phng trnh

2

c nghim nguyn x, y, z sao cho x v 0. Chng ta c th gi s rng (,)v trong trng hp (x, y) =d > 1, ta t x = dx1, y = dy1, khi d3|2z3 nn d| dn n z = dz1. Vy x13+y13=z13 vi (x1, y1)=1.T

2

v (x, y ) =1 ta c x, y l cc s l, nn,2 2x y x yu v+ = =l cc s nguyn. Hn na v (x, y)= 1 v x = u+ v, y = u v nn (u, v) = 1. Ta cng c (u+v)3+(u-v)3=2z3 u(u2+3v2)=z3vpdngxy,z0,tathuc 2 2( )4x y zuvz= 0. Ta xt hai trng hp: Nu(u,3)=1thv(u, v)=1 nntac(u,u2+3v2)=1.V thtntiz1,z2 nguyntcngnhausaochou=z13,u2+3v2=z23.Doz23-z16=3v2hay 2 2 2 2 22 1 2 1 2 1( ) ( ) 3 3 z z z z z z v( + = . t 22 1( ) t z z = , khi t (z1, z2)=1 ta c (t, z1)=1 v 2 2 4 21 1( 3 4 ) 3 t t tz z v + + = . iu ny dn n 3|t hay t = 3t1 v 2 2 4 21 1 1 1 1(9 9 3 ) t t t z z v + + = , t ta c 3|v hay v = 3v1. V (z1, 3) =1 nn 2 2 41 1 1 19 9 3 t t z z + + khng chia ht cho 9, khiv9|v2nntanhnc3|t1hayt1=3t2.Ttalic 2 4 22 2 2 1 1 1(27 9 ) t t tz z v + + =vi (t2, z1) = 1 (do (t, z1)=1) v 2 42 2 2 1 1( , 27 9 ) 1 t t tz z + + = . V th tn ti cc s t nhin b, c sao cho 22t b = , 4 2 2 4 21 127 9 b bz z c + + = . Cc s b, 1zl cc s t nhin nn khi b = 0 th t2 = 0 dn n t = 0, khi z2 =z12. Bn cnh d v (z1, z2) =1 nn z1 = 1, z2 =1, suy ra v = 0. Vy x =y, mu thun vi gi thit xy. mt khc nu z1=0 th u = 0, khi 3v2 = z23 dn ti v = 0 l iu khng th (do xy). Nh vy chng ta i n kt lun phng trnh 4 2 2 4 21 127 9 b bz z c + + =c nghim nguyn dng m y l iu khng th. Nu 3|u hay u = 3u1, khi v (u,v)=1 nn (3, v) =1. Khi , tu(u2+3v2)=z3

ta c z = 3z1 v2 2 31 1 1(3 ) 3 u u v z + = . V (v, 3) = 1 nn 3|u1 nn ta li c u1 = 3u2 , t suy ra ( )2 2 32 2 127 u u v z + =. Nhng v (u2, v) = 1 nn ( )2 22 2, 27 1 u u v + = , do ta li Phng trnh nghiem nguyen 2011 17Oeo o Qaooq Toan 3A HSP Hue c a, b cho 32, u a =2 2 3227u v b + = vi (a, b) = 1, (b, 3) = 1 ( do (v,3) = 1). Khi tactac6 2 327a v b + = .tt=b3a2tanhnc(t,3)=1v 2 2 4 2( 9 27 ) t t a t a v + + = .V(a,b)=1nn(a,t)=1,kthpvi(t,3)=1talithu c (t, 2 2 49 27 t a t a + + ) =1. Tip tc t t = a12, 2 2 2 419 27 b t a t a = + +vi (a1, b1 ) = 1. Nh vy t 2 2 4 2( 9 27 ) t t a t a v + + = ta c 4 2 2 4 21 1 19 27 a a a a b + + =vi a1 0 , a 0 v nu a1 = 0 th t = 0 tri vi gi thit (t, 3)=1 v nua = 0 th u = 0, tt yu z = 0 tri vigisz0.Nhvychngtaintrnghpphngtrnh 4 2 2 4 21 1 19 27 a a a a b + + =c nghim nguyn dng m y l iu khng th. Nh vy chng ta hon tt chng minh cho nh l sau : nh l 8 : Phng trnh

khng c nghim nguyn (x, y, z) nu x y v z 0. iu ny cho thy lp phng ca 3 s t nhin khng th lp thnh cp s cng. Nuty=1thddngthyrngccphngtrnh 3 32 1 x z = khngc nghim(x,z)nguynngoitrccnghim(-1,-1),(1,0)(chophngtrnh 3 32 1 x z = ) v nghim ( 1, 1), (-1, 0) (cho phng trnh 3 32 1 x z = ). H qu 1 : Khng tn ti s tam gic c dng l lp phng ca mt s t nhin. Chng minh Nh ta bit s tam gic th m c cng thc ( 1)2m m+ . Gi s tn ti s tam gic th x > 1sao cho 3( 1)2xxn+=hay 3( 1) 2 xx n + = vi n l s tnhin.Nuxchnthx=2k,klstnhin.Khi 3(2 1) k k n + = vv(k, 2k+1) = 1 nnc hai sa, b t nhin nguyn t cng nhau sao cho a3 = k,b3 = 2k+1.Lc tac phng trnh b3- 2a3 =1,mtheo nhnxt trn th phng trnh ny khng c nghim a0.Nuxlth=2k-1vik > (v >) v (2k -1)k = n3 vi (2k-, k) . Nh v chng ta li i ti h qu tn ti a, b ngun dng v ngun t cng nhau sao cho 2k-1 = a3, k = b3. Nh vy ta c phng trnh 2b3- a3 v nghim khi b0. Vy ta chng minh khng tn ti x tha mn gi thit phn chng. H qu 2 : Phng trnh

khng c nghim nguyn dng no khc nghim (x,y) = (3,2). Chng minh Phng trnh nghiem nguyen 2011 18Oeo o Qaooq Toan 3A HSP Hue Gi s tn ti s t nhin x3vsao cho

hay

. Nh vy ta c

( )( ). Nu x l s chn khi ta c ((x-1),(x+1)) = 1, do tn ti cc s nguyn dng a, b nguyn t cng nhau sao cho (x-1) = a3, (x+1)=b3.Khi(b-a)(b2+ab+a2)=b3-a3=2.Dob2+ab+a2|2miunyl khng th v b2+ab+a2 13+1.1+13 =3. Vy x phi l s l. Ta vit x = 2k +1 vi k l s t nhin > 1 ( v nuk = 1 th x = 3). Lc y3 = x2-1 phi l s chn, tc y = 2n. Khi k(k+1) = 2n3 vi k ls t nhin, tri vi h qu 1. Vy gi s phn chng l sai. H qu 2 c chng minh. Cuicngtachrngphngtrnh

khngcnghimnguyn khc (0,0,0). Ta c th chng minh n bng cch dng b sau : B :Tt c nghim nguyn ca phng trnh 3 2 23 s a b = + vi (a, b) = 1, s l s l c cho theo cng thc sau : 2 23 s o | = + , 3 29 a o o| = , 2 33 3 b o | | = Vi cc s nguyn , tha mn a b (mod 2) v (, 3) = 1. Vic chng minh b cc bn c th xem trang 415 [1]. Chng ta s s dng b trn chng minh phng trnh

(32) v nghim trn . Tht vy, gi s (x, y, z) l nghim ca (32). Hin nhin ta c th chn (x, y, z) sao cho0 xyz = c gi tr nh nht. Ta d dng chng minh c x, y, z nguyn t snhi(nud=(x,y)thd|x3+y3hayd|z3,vy, ,x y zd d d| | |\ .cnglnghimca (32), mu thun gi thit (x, y, z) l nghim c0 xyz =nh nht). Do vy c x, y, z khng th cng chn v c duy nht l mt s chn. Gi s rng z chn, khi x v y l v x y, x + y l cc s chn, do vy c th t: x + y = 2u, x y = 2w hay x = u + w, y = u w. V (x, y) = 1 nn (u, w) = 1 v u w(mod 2) (do x, y cng l). Th vo phng trnh (32), ta c:2 2 32( 3 ) u u w z + = . Nu (u, 3) = 1 th t u w(mod 2) ta c 2 2(2, 3 ) 1 u u w + = , do tn ti cc s nguyn s, t nguyn t cng nhau v 32u t = , 2 2 33 u w s + =R rng s l s l. Theo b trn,phng trnh 2 2 33 u w s + =c nghim : 2 23 s o | = + , 3 29 u o o| = , 2 3w 3 3 o | | = Phng trnh nghiem nguyen 2011 19Oeo o Qaooq Toan 3A HSP Hue Viccsnguyn,thamnab(mod2)v(,3)=1.Khi 32 2 ( 3 )( 3 ) t u o o | o | = = + vchngkhkhngnhnra2o , ( 3 ) o | ,( 3 ) o | + nguyntsnhi.Dovytatiptcc 32o o = , 33 o | t = , 33 o | + = v 3 3 3o t = + .Vytac ( ) , , t o lmtnghimca(32).Nhng 3 32 0 t u x y ot = = = + =v 3x y xyz xyz + s < . iu ny mu thun vi gi thit ban u ca (x, y, z). Nu 3|u hay u = 3v, khi (32) tr thnh 2 2 318(3 w) v v z + = . S dng gi thit 3v w (mod 2) v (3v, w) = 1, ta thu c 2 2(18 , 3 w) 1 v v + = , bi vy tn ti cc s nguyn t, s sao cho (s, t) = 1 v 18v = t3, 3v2+w2 = s3. D thy s l s l v theo b trn ta c phng trnh 3v2+w2 = s3 c nghim l2 23 s o | = + , 3 2w 9 o o| = , 2 33 3 v o | | = . Khi ( )318 27.2 ( ) t v | o | o | = = + .Nhchngminhtrn,talic2 , , | o | o | + nguyntsnhinntatiptcc 32| o = , 3o | t = , 3o | + = . D thy 3 3 3 o t = +nn ( ) , , o t l nghim ca (32). Tuy nhin,33 2 2 1027 3 9 9t v ux y ot = = = = + =V 3 19x y xyz xyz + s