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Chuyen De So Hoc VMF

Oct 29, 2015

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  • Chuyn

    S HC

    Din n Ton hc

  • Chuyn

    S HC

    Ch bn

    Trn Quc Nht Hn [perfectstrong]Trn Trung Kin [Ispectorgadget]

    Phm Quang Ton [Phm Quang Ton]L Hu in Khu [Nesbit]inh Ngc Thch [T*genie*]

    c 2012 Din n Ton hc

  • Li gii thiu

    Bn c thn mn,

    S hc l mt phn mn quan trng trong ton hc gn b vichng ta xuyn sut qu trnh hc Ton t bc tiu hc n trung hcph thng. Chng ta c tip xc vi S hc bt u bng nhngkhi nim n gin nh tnh chia ht, c chung ln nht, bi chungnh nht... gip lm quen d dng hn vi s k diu ca nhng cons cho n nhng vn i hi nhiu t duy hn nh ng d, snguyn t, cc phng trnh Diophantine m ni ting nht l nh lln Fermat..., u u t tm vi m n v m, t cu b lp mt bib 4 chia ht cho 2 n Gio s thin ti Andrew Wiles (ngi giiquyt bi ton Fermat), chng ta u c th thy c hi th ca Shc trong .

    S hc quan trng nh vy nhng l thay s chuyn vit v n likhng nhiu nu em so vi kho tng s cc bi vit v bt ngthc trn cc din n mng. Xut pht t s thiu ht cng nh k nim trn mt nm Din n Ton hc khai trng trangch mi (16/01/2012 - 16/01/2013), nhm bin tp chng ti cng vinhiu thnh vin tch cc ca din n chung tay bin son mtchuyn gi n bn c.

    Chuyn l tp hp cc bi vit ring l ca cc tc gi Nguyn MnhTrng Dng (duongld) , Nguyn Trn Huy (yeutoan11), NguynTrung Hiu (nguyentrunghieua), Phm Quang Ton (Phm QuangTon), Trn Nguyn Thit Qun (L Lawliet), Trn Trung Kin (Is-pectorgadget), Nguyn nh Tng (tungc3sp)... cng s gp sc

    i

  • ii

    gin tip ca nhiu thnh vin tch cc trn Din n Ton hc nhNguyen Lam Thinh, nguyenta98, Karl Heinrich Marx, TheGunner, perfectstrong...

    Kin thc cp trong chuyn tuy khng mi nhng c th gipcc bn phn no hiu su hn mt s khi nim c bn trong S hccng nh trao i cng cc bn nhiu dng bi tp hay v kh t cp d n cc bi ton trong cc k thi Hc sinh gii quc gia, quc t.

    Chuyn gm 7 chng. Chng 1 cp n cc khi nim v cv Bi. S nguyn t v mt s bi ton v n c gii thiu trongchng 2. Chng 3 ni su hn v Cc bi ton chia ht. Phngtrnh nghim nguyn,Phng trnh ng d c phc ha trongcc chng 4 v 5. H thng d v nh l Thng d Trung Hoas c gi n chng ta qua chng 6 trc khi kt thc chuyn bng Mt s bi ton s hc hay trn VMF chng 7.

    Do thi gian chun b gp rt ni dung chuyn cha c u ttht s t m cng nh c th cn nhiu sai st trong cc bi vit,chng ti mong bn c thng cm. Mi s ng h, ng gp, phbnh ca c gi s l ngun ng vin tinh thn to ln cho ban bintp cng nh cho cc tc gi nhng phin bn cp nht sau cachuyn c tt hn, ng gp nhiu hn na cho kho tng hcthut ca cng ng ton mng. Chng ti hi vng qua chuyn nys gip cc bn tm thm c cm hng trong s hc v thm yu vp ca nhng con s. Mi trao i gy xin gi v a ch email :[email protected].

    Trn trng,Nhm bin tp Chuyn S hc.

    Din n Ton hc Chuyn S hc

  • Mc lc

    i Li gii thiu

    1Chng 1

    c v Bi

    1.1 c s, c s chung, c s chung ln nht 1

    1.2 Bi s, bi s chung, bi s chung nh nht 4

    1.3 Bi tp ngh 6

    9Chng 2

    S Nguyn T

    2.1 Mt s kin thc c bn v s nguyn t 9

    2.2 Mt s bi ton c bn v s nguyn t 13

    2.3 Bi tp 19

    2.4 Ph lc: Bn nn bit 24

    29Chng 3

    Bi ton chia ht

    3.1 L thuyt c bn 29

    3.2 Phng php gii cc bi ton chia ht 31

    57Chng 4

    Phng trnh nghim nguyn

    iii

  • iv Mc lc

    4.1 Xt tnh chia ht 574.2 S dng bt ng thc 744.3 Nguyn tc cc hn, li v hn 86

    89Chng 5

    Phng 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 gii phng trnhng d 96

    5.7 Bi tp 101

    103Chng 6

    H thng d v nh l Thng d Trung Hoa

    6.1 Mt s k hiu s dng trong bi vit 1036.2 H thng d 1046.3 nh l thng d Trung Hoa 1176.4 Bi tp ngh & gi p s 125

    129Chng 7

    Mt s bi ton s hc hay trn VMF

    7.1 m3 + 17...3n 129

    7.2 c(ac+ 1)2 = (5c+ 2)(2c+ b) 136

    141 Ti liu tham kho

    Din n Ton hc Chuyn S hc

  • Chng

    1c v Bi

    1.1 c s, c s chung, c s chungln nht 1

    1.2 Bi s, bi s chung, bi s chungnh nht 4

    1.3 Bi tp ngh 6

    Nguyn Mnh Trng Dng (duongld)Nguyn Trn Huy (yeutoan11)

    c v bi l 2 khi nim quan trng trong chng trnh s hc THCS.Chuyn ny s gii thiu nhng khi nim v tnh cht c bn vc, c s chung, c chung ln nht, bi, bi s chung, bi chungnh nht. Mt s bi tp ngh v cc vn ny cng s c cp n cui bi vit.

    1.1 c s, c s chung, c s chung ln nht

    Trong phn ny, chng ti s trnh by mt s khi nim v c s,c s chung v c s chung ln nht km theo mt vi tnh cht cachng. Mt s bi tp v d cho bn c tham kho cng s c ara.

    1.1.1 nh ngha

    nh ngha 1.1 S t nhin d 6= 0 c gi l mt c s ca s tnhin a khi v ch khi a chia ht cho d. Ta ni d chia ht a, k hiu d|a.Tp hp cc c ca a l: U(a) = {d N : d|a}. 4

    1

  • 2 1.1. c s, c s chung, c s chung ln nht

    Tnh cht 1.1 Nu U(a) = {1; a} th a l s nguyn t. nh ngha 1.2 Nu U(a) v U(b) c nhng phn t chung th nhngphn t gi l c s chung ca a v b. Ta k hiu:

    USC(a; b) = {d N : (d|a) (d|b)}= {d N : (d U(a)) (d U(b))}.

    Tnh cht 1.2 Nu USC(a; b) = {1} th a v b nguyn t cng nhau.nh ngha 1.3 S d N c gi l c s chung ln nht ca a v b(a; b Z) khi d l phn t ln nht trong tp USC(a; b). K hiu cchung ln nht ca a v b l UCLN(a; b), (a; b) hay gcd(a; b). 4

    1.1.2 Tnh cht

    Sau y l mt s tnh cht ca c chung ln nht:

    Nu (a1; a2; . . . .; an) = 1 th ta ni cc s a1; a2; . . . ; an nguynt cng nhau.

    Nu (am; ak) = 1,m 6= k, {m; k} {1; 2; . . . ;n} th ta ni cca1; a2; . . . ; an i mt nguyn t cng nhau.

    c USC(a; b) th(a

    c;b

    c

    )=

    (a; b)

    c.

    d = (a; b)(a

    d;b

    d

    )= 1.

    (ca; cb) = c(a; b). (a; b) = 1 v b|ac th b|c. (a; b) = 1 v (a; c) = 1 th (a; bc) = 1. (a; b; c) = ((a; b); c). Cho a > b > 0

    Nu a = b.q th (a; b) = b. Nu a = bq + r(r 6= 0) th (a; b) = (b; r).

    Din n Ton hc Chuyn S hc

  • 1.1. c s, c s chung, c s chung ln nht 3

    1.1.3 Cch tm c chung ln nht bng thut ton Euclide

    tm (a; b) khi a khng chia ht cho b ta dng thut ton Euclidesau:

    a = b.q + r1 th (a; b) = (b; r1).

    b = r1.q1 + r2 th (b; r1) = (r1; r2).

    rn2 = rn1.qn1 + rn th (rn2; rn1) = (rn1; rn).

    rn1 = rn.qn th (rn1; rn) = rn.

    (a; b) = rn.

    (a; b) l s d cui cng khc 0 trong thut ton Euclide.

    1.1.4 Bi tp v d

    V d 1.1. Tm (2k 1; 9k + 4), k N. 4Li gii. Ta t d = (2k 1; 9k + 4). Theo tnh cht v c s chungta c d|2k1 v d|9k+ 4. Tip tc p dng tnh cht v chia ht ta lic d|9(2k 1) v d|2(9k + 4). Suy ra d|2(9k + 4) 9(2k 1) hay d|17.Vy (2k 1; 9k + 4) = 1. V d 1.2. Tm (123456789; 987654321). 4Li gii. t b = 123456789; a = 987654321. Ta nhn thy a v b uchia ht cho 9.Ta li c :

    a+ b = 1111111110

    =1010 10

    9.

    9a+ 9b = 1010 10(1.1)

    Mt khc :

    10b+ a = 9999999999= 1010 1. (1.2)

    Chuyn S hc Din n Ton hc

  • 4 1.2. Bi s, bi s chung, bi s chung nh nht

    Tr (1.2) v (1.1) v theo v ta c b8a = 9. Do nu t d = (a; b)th 9

    ...d.M a v b u chia ht cho 9, suy ra d = 9.

    Da vo thut ton Euclide, ta c li gii khc cho V d 1.2 nh sau :Li gii. 987654321 = 123456789.8+9 th (987654321; 123456789) =

    (123456789; 9).

    123456789 = 9.1371421.

    (123456789; 987654321) = 9.

    V d 1.3. Chng minh rng dy s An =1

    2n(n + 1), n N cha

    nhng dy s v hn nhng s i mt nguyn t cng nhau. 4Li gii. Gi s trong dy ang xt c k s i mt nguyn t cngnhau l t1 = 1; t2 = 3; . . . ; tk = m(m N). t a = t1t2. . . tk. Xt shng t2a+1 trong dy An:

    t2a+1 =1

    2(2a+ 1)(2a+ 2)

    = (a+ 1)(2a+ 1) tk

    Mt khc ta c (a+ 1; a) = 1 v (2a+ 1; a) = 1 nn (t2a+1; a) = 1.Do t2a+1 nguyn t cng nhau vi tt c k s {t1; t2; . . . tk}. Suy rady s An cha v hn nhng s i mt nguyn t cng nhau.

    1.2 Bi s, bi s chung, bi s chung nh nht

    Tng t nh cu trc trnh by phn trc, trong phn nychng ti cng s a ra nhng nh ngha, tnh cht c bn ca bis, bi s chung, bi s chung nh nht v mt s bi tp v d minhha.

    Din n Ton hc Chuyn S hc

  • 1.2. Bi s, bi s chung, bi s chung nh nht 5

    1.2.1 nh ngha

    nh ngha 1.4 S t nhin m c gi l mt bi s ca a 6= 0 khiv ch khi m chia ht cho a hay a l mt c s ca m. 4

    Nhn xt. Tp hp cc bi s ca a 6= 0 l:B(a) = {0; a; 2a; . . . ; ka}, k Z.

    nh ngha 1.5 S t nhin m c gi l mt bi s ca a 6= 0 khiv ch khi m chia ht cho a hay a l mt c s ca m 4

    nh ngha 1.6 Nu 2 tp B(a) v B(b) c phn t chung th cc phnt chung gi l bi s chung ca a v b. Ta k hiu bi s chungca a v b: BSC(a; b).

    nh ngha 1.7 S m 6= 0 c gi l bi chung nh nht ca a vb khi m l phn t dng nh nht trong tp BSC(a; b). K hiu :BCNN(a; b), [a; b] hay lcm(a; b). 4

    1.2.2 Tnh cht

    Mt s tnh cht ca bi chung ln nht:

    Nu [a; b] = M th(M

    a;M

    b

    )= 1.

    [a; b; c] = [[a; b]; c].

    [a; b].(a; b) = a.b.

    1.2.3 Bi tp v d

    V d 1.4. Tm [n;n+ 1;n+ 2]. 4Li gii. t A = [n;n + 1] v B = [A;n + 2]. p dng tnh cht[a; b; c] = [[a; b]; c], ta c: B = [n;n+ 1;n+ 2].D thy (n;n+ 1) = 1, suy ra [n;n+ 1] = n(n+ 1).

    Chuyn S hc Din n Ton hc

  • 6 1.3. Bi tp ngh

    Li p dng tnh cht [a; b] =a.b

    (a; b)th th

    [n;n+ 1;n+ 2] =n(n+ 1)(n+ 2)

    (n(n+ 1);n+ 2)

    .Gi d = (n(n+ 1);n+ 2). Do (n+ 1;n+ 2) = 1 nn

    d = (n;n+ 2)= (n; 2).

    Xt hai trng hp:

    Nu n chn th d = 2, suy ra [n;n+ 1;n+ 2] = n(n+ 1)(n+ 2)2

    .

    Nu n l th d = 1, suy ra [n;n+ 1;n+ 2] = n(n+ 1)(n+ 2) .

    V d 1.5. Chng minh rng [1; 2; . . . 2n] = [n+ 1;n+ 2; . . . ; 2n]. 4Li gii. Ta thy c trong k s nguyn lin tip c mt v ch mt schia ht cho k. Do bt trong cc s {1; 2; . . . ; 2n} u l c ca mts no trong cc s {n + 1;n + 2; . . . ; 2n}. Do [1; 2; . . . n; 2n] =[n+ 1;n+ 2; . . . ; 2n].

    1.3 Bi tp ngh

    Thay cho li kt, chng ti xin gi n bn c mt s bi tp ngh luyn tp nhm gip cc bn quen hn vi cc khi nim v cctnh cht trnh by trong chuyn .

    Bi 1. a. Cho A = 5a + 3b;B = 13a + 8b(a; b N) chng minh(A;B) = (a; b).

    b. Tng qut A = ma+nb;B = pa+qb tha mn |mqnp| =1 vi a, b,m, n, p, q N. Chng minh (A;B) = (a; b).

    Bi 2. Tm (6k + 5; 8k + 3)(k N).

    Din n Ton hc Chuyn S hc

  • 1.3. Bi tp ngh 7

    Bi 3. T cc ch s 1; 2; 3; 4; 5; 6 thnh lp tt c s c su ch s(mi s ch vit mt ln). Tm UCLN ca tt c cc s .

    Bi 4. Cho A = 2n+ 1;B =n(n+ 1)

    2(n N). Tm (A;B).

    Bi 5. a. Chng minh rng trong 5 s t nguyn lin tip bao gicng chn c mt s nguyn t cng nhau vi cc scn li.

    b. Chng minh rng trong 16 s nguyn lin tip bao gi cngchn c mt s nguyn t cng nhau vi cc s cn li.

    Bi 6. Cho 1 m n(m;n N).a. Chng minh rng (22

    n 1; 22n + 1) = 1.b. Tm (2m 1; 2n 1).

    Bi 7. Cho m,n N vi (m,n) = 1. Tm (m2 + n2;m+ n).Bi 8. Cho A = 2n+3;B = 2n+1+3n+1(n N);C = 2n+2+3n+2(n

    N). Tm (A;B) v (A;C).

    Bi 9. Cho su s nguyn dng a; b; a; b; d; d sao cho (a; b) = d; (a; b) =d. Chng minh rng (aa; bb; ab; ab) = dd.

    Bi 10. Chng minh rng dy s Bn =1

    6n(n+ 1)(n+ 2)(n N) cha

    v hn nhng s nguyn t cng nhau.

    Bi 11. Chng minh rng dy s 2n 3 vi mi n N v n 2 chady s v hn nhng s nguyn t cng nhau.

    Bi 12. Chng minh dy Mersen Mn = 2n 1(n N) cha dy s vhn nhng s nguyn t cng nhau.

    Bi 13. Chng minh rng dy Fermat Fn = 22n

    + 1(n N) l dy snguyn t cng nhau.

    Bi 14. Cho n N;n > 1 v 2n 2 chia ht cho n. Tm (22n ; 2n 1).

    Chuyn S hc Din n Ton hc

  • 8 1.3. Bi tp ngh

    Bi 15. Chng minh rng vi mi n N, phn s 21n+ 114n+ 3

    ti gin.

    Bi 16. Cho ba s t nhin a; b; c i mt nguyn t cng nhau. Chngminh rng (ab+ bc+ ca; abc) = 1.

    Bi 17. Cho a; b N. Chng minh rng tn ti v s n N sao cho(a+ n; b+ n) = 1.

    Bi 18. Gi sm;n N(m n) tha mn (199k1;m) = (19931;n).Chng minh rng tn ti t(t N) sao cho m = 1993t.n.

    Bi 19. Chng minh rng nu a;m N; a > 1 th(am 1a 1 ; a 1

    )=

    (m; a 1).Bi 20. Tm s nguyn dng n nh nht cc phn s sau ti gin:

    a.1

    n1996 + 1995n+ 2,

    b.2

    n1996 + 1995n+ 3,

    c.1994

    n1996 + 1995n+ 1995,

    d.1995

    n1996 + 1995n+ 1996.

    Bi 21. Cho 20 s t nhin khc 0 l a1; a2; . . . an c tng bng Sv UCLN bng d. Chng minh rng UCLN ca S a1;S a2; . . . ;S an bng tch ca d vi mt c no ca n 1.

    Din n Ton hc Chuyn S hc

  • Chng

    2S Nguyn T

    2.1 Mt s kin thc c bn v snguyn t 9

    2.2 Mt s bi ton c bn v s nguynt 13

    2.3 Bi tp 192.4 Ph lc: Bn nn bit 24

    Nguyn Trung Hiu (nguyentrunghieua)Phm Quang Ton (Phm Quang Ton)

    2.1 Mt s kin thc c bn v s nguyn t

    2.1.1 nh ngha, nh l c bn

    nh ngha 2.1 S nguyn t l nhng s t nhin ln hn 1, ch c 2c s l 1 v chnh n. 4

    nh ngha 2.2 Hp s l s t nhin ln hn 1 v c nhiu hn 2c. 4

    Nhn xt. Cc s 0 v 1 khng phi l s nguyn t cng khng phil hp s. Bt k s t nhin ln hn 1 no cng c t nht mt cs nguyn t.

    nh l 2.1 Dy s nguyn t l dy s v hn.

    9

  • 10 2.1. Mt s kin thc c bn v s nguyn t

    Chng minh. Gi s ch c hu hn s nguyn t l p1; p2; p3; ...; pn;trong pn l s ln nht trong cc nguyn t.Xt s N = p1p2...pn + 1 th N chia cho mi s nguyn t pi(i = 1, n)u d 1 (*)Mt khc N l mt hp s (v n ln hn s nguyn t ln nht l pn)do N phi c mt c nguyn t no , tc l N chia ht cho mttrong cc s pi (**).Ta thy (**) mu thun (*). Vy khng th c hu hn s nguyn t.

    nh l 2.2 Mi s t nhin ln hn 1 u phn tch c ra thas nguyn t mt cch duy nht (khng k th t cc tha s).

    Chng minh. * Mi s t nhin ln hn 1 u phn tch c ra thas nguyn t:Tht vy: gi s iu khng nh trn l ng vi mi s m tho mn:1 < m < n ta chng minh iu ng n n.Nu n l nguyn t, ta c iu phi chng minh.Nu n l hp s, theo nh ngha hp s, ta c: n = a.b (vi a, b < n)Theo gi thit quy np: a v b l tch cc tha s nh hn n nn n ltch cu cc tha s nguyn t.* S phn tch l duy nht:Gi s mi s m < n u phn tch c ra tha s nguyn t mtcch duy nht, ta chng minh iu ng n n:Nu n l s nguyn t th ta c iu phi chng minh. Nu n l hps: Gi s c 2 cch phn tch n ra tha s nguyn t khc nhau:

    n = p.q.r....n = p.q.r....

    Trong p, q, r..... v p, q, r.... l cc s nguyn t v khng c snguyn t no cng c mt trong c hai phn tch (v nu c stho mn iu kin nh trn, ta c th chia n cho s lc thngs nh hn n, thng ny c hai cch phn tch ra tha s nguyn tkhc nhau, tri vi gi thit ca quy np).Khng mt tnh tng qut, ta c th gi thit p v p ln lt l cc snguyn t nh nht trong phn tch th nht v th hai.V n l hp s nn n > p2 v n > p2. Do p 6= p n > p.p

    Din n Ton hc Chuyn S hc

  • 2.1. Mt s kin thc c bn v s nguyn t 11

    Xt m = n pp < n c phn tch ra tha s nguyn t mt cchduy nht ta thy:

    p|n p|n pp hay p|mKhi phn tch ra tha s nguyn t ta c: m = n pp = pp.P.Q... viP,Q P ( P l tp cc s nguyn t). pp|n pp|p.q.r... p|q.r... p l c nguyn t ca q.r...M p khng trng vi mt tha s no trong q, r... (iu ny tri viga thit quy np l mi s nh hn n u phn tch c ra tha snguyn t mt cch duy nht).Vy, iu gi s khng ng. nh l c chng minh.

    2.1.2 Cch nhn bit mt s nguyn t

    Cch 1

    Chia s ln lt cho cc nguyn t t nh n ln: 2; 3; 5; 7...

    Nu c mt php chia ht th s khng nguyn t.

    Nu thc hin php chia cho n lc thng s nh hn s chia m ccphp chia vn c s d th s l nguyn t.

    Cch 2

    Mt s c hai c s ln hn 1 th s khng phi l s nguyn t.

    Cho hc sinh lp 6 hc cch nhn bit 1 s nguyn t bng phngphp th nht (nu trn), l da vo nh l c bn:

    c s nguyn t nh nht ca mt hp s A l mt s khng vtquA.

    Vi quy tc trn trong mt khon thi gian ngn, vi cc du hiu chiaht th ta nhanh chng tr li c mt s c hai ch s no l

    Chuyn S hc Din n Ton hc

  • 12 2.1. Mt s kin thc c bn v s nguyn t

    nguyn t hay khng.

    H qu 2.1 Nu c s A > 1 khng c mt c s nguyn t no t2 n

    A th A l mt nguyn t.

    2.1.3 S cc c s v tng cc c s ca 1 s

    Gi s: A = px11 .px22 ......pnxn; trong : pi P;xi N; i = 1, n

    Tnh cht 2.1 S cc c s ca A tnh bng cng thc:

    T (A) = (x1 + 1)(x2 + 1).....(xn + 1)

    V d 2.1. 30 = 2.3.5 th T (A) = (1 + 1)(1 + 1)(1 + 1) = 8. Kim tra:(30) = {1; 2; 3; 5; 6; 10; 15; 30} nn (30) c 8 phn t. 4Tnh cht 2.2 Tng cc c mt s ca A tnh bng cng thc:

    (A) =

    ni=1

    pxi+1i 1pi 1

    2.1.4 Hai s nguyn t cng nhau

    nh ngha 2.3 Hai s t nhin c gi l nguyn t cng nhau khiv ch khi chng c c chung ln nht (CLN) bng 1. 4Tnh cht 2.3 Hai s t nhin lin tip lun nguyn t cng nhau.

    Tnh cht 2.4 Hai s nguyn t khc nhau lun nguyn t cng nhau.

    Tnh cht 2.5 Cc s a, b, c nguyn t cng nhau khi v ch khi (a, b, c)= 1.

    nh ngha 2.4 Nhiu s t nhin c gi l nguyn t snh i khichng i mt nguyn t cng nhau. 4

    Din n Ton hc Chuyn S hc

  • 2.2. Mt s bi ton c bn v s nguyn t 13

    2.1.5 Mt s nh l c bit

    nh l 2.3 (Dirichlet) Tn ti v s s nguyn t p c dng:p = ax+ b (x, a, b N, a, b l 2 s nguyn t cng nhau). Vic chng minh nh l ny kh phc tp, tr mt s trng hp cbit, chng hn c v s s nguyn t dng: 2x 1; 3x 1; 4x+ 3; 6x+5; . . .

    nh l 2.4 (Tchebycheff-Betrand) Trong khong t s t nhinn n s t nhin 2n c t nht mt s nguyn t (n > 2).

    nh l 2.5 (Vinogradow) Mi s l ln hn 33 l tng ca 3 snguyn t.

    2.2 Mt s bi ton c bn v s nguyn t

    2.2.1 C bao nhiu s nguyn t dng ax + b

    V d 2.2. Chng minh rng: c v s s nguyn t c dng 3x 1.4Li gii. Mi s t nhin khng nh hn 2 c 1 trong 3 dng: 3x; 3x+1hoc 3x 1 Nhng s c dng 3x (vi x > 1) l hp s Xt 2 s c dng 3x+ 1: l s 3m+ 1 v s 3n+ 1.

    Xt tch (3m + 1)(3n + 1) = 9mn + 3m + 3n + 1. Tch ny cdng: 3x+ 1

    Ly mt s nguyn t p bt c dng 3x 1, ta lp tch ca pvi tt c cc s nguyn t nh hn p ri tr i 1 ta c: M =2.3.5.7....p 1 = 3(2.5.7....p) 1 th M c dng 3x 1.C 2 kh nng xy ra:

    1. Kh nng 1: M l s nguyn t, l s nguyn t c dng3x 1 > p, bi ton c chng minh.

    Chuyn S hc Din n Ton hc

  • 14 2.2. Mt s bi ton c bn v s nguyn t

    2. Kh nng 2:M l hp s: Ta chia M cho 2, 3, 5, ...., p u tnti mt s d khc 0 nn cc c nguyn t ca M u lnhn p, trong cc c ny khng c s no c dng 3x+1 (chng minh trn). Do t nht mt trong cc c nguynt ca M phi c dng 3x (hp s) hoc 3x+ 1

    V nu tt c c dng 3x+1 thM phi c dng 3x+1 ( chngminh trn). Do , t nht mt trong cc c nguyn t ca Mphi c dng 3x 1, c ny lun ln hn p.

    Vy: C v s s nguyn t dng 3x 1.

    V d 2.3. Chng minh rng: C v s s nguyn t c dng 4x+ 3.4Li gii. Nhn xt. Cc s nguyn t l khng th c dng 4x hoc4x + 2. Vy chng ch c th tn ti di 1 trong 2 dng 4x + 1 hoc4x+ 3.Ta s chng minh c v s s nguyn t c dng 4x+ 3.

    Xt tch 2 s c dng 4x+ 1 l: 4m+ 1 v 4n+ 1.Ta c: (4m+1)(4n+1) = 16mn+4m+4n+1 = 4(4mn+m+n)+1.

    Vy tch ca 2 s c dng 4x+ 1 l mt s cng c dng 4x+ 1.

    Ly mt s nguyn t p bt k c dng 4x+ 3, ta lp tch ca 4pvi tt c cc s nguyn t nh hn p ri tr i 1 khi ta c:N = 4(2.3.5.7.....p) 1. C 2 kh nng xy ra

    1. N l s nguyn t N = 4(2.3.5.7....p) 1 c dng 4x 1.Nhng s nguyn t c dng 4x 1 cng chnh l nhng sc dng 4x+ 3 v bi ton c chng minh.

    2. N l hp s. Chia N cho 2, 3, 5, ...., p u c cc s dkhc 0. Suy ra cc c nguyn t ca N u ln hn p.

    Cc c ny khng th c dng 4x hoc 4x+ 2 (v l hp s).Cng khng th ton cc c c dng 4x+ 1 v nh th N phic dng 4x + 1. Nh vy trong cc c nguyn t ca N c tnht 1 c c dng 4x 1 m c ny hin nhin ln hn p.

    Din n Ton hc Chuyn S hc

  • 2.2. Mt s bi ton c bn v s nguyn t 15

    Vy: C v s s nguyn t c dng 4x 1 (hay c dng 4x+ 3).

    Trn y l mt s bi ton chng minh n gin ca nh l Dirichlet:C v s s nguyn t dng ax+ b trong a, b, x N, (a, b) = 1.

    2.2.2 Chng minh s nguyn t

    V d 2.4. Chng minh rng: (p 1)! chia ht cho p nu p l hp s,khng chia ht cho p nu p l s nguyn t. 4Li gii. Xt trng hp p l hp s: Nu p l hp s th p l tch

    ca cc tha s nguyn t nh hn p v s m cc lu tha nykhng th ln hn s m ca chnh cc lu tha y cha trong

    (p 1)!. Vy: (p 1)!...p (pcm).

    Xt trng hp p l s nguyn t: V p P p nguyn t cngnhau vi mi tha s ca (p 1)! (pcm).

    V d 2.5. Cho 2m 1 l s nguyn t. Chng minh rng m cng ls nguyn t. 4Li gii. Gi s m l hp s m = p.q (p, q N; p, q > 1)Khi : 2m1 = 2pq1 = (2p)q1 = (2p1)((2p)q1+(2p)q2+.....+1)v p > 1 2p 1 > 1 v (2p)q1 + (2p)q2 + .....+ 1 > 1Dn n 2m 1 l hp s :tri vi gi thit 2m1 l s nguyn t.Vy m phi l s nguyn t (pcm)

    V d 2.6. Chng minh rng: mi c nguyn t ca 1994!1 u lnhn 1994. 4Li gii. Gi p l c s nguyn t ca 1994! 1Gi s p 1994 1994.1993.....3.2.1...p 1994!...p.M 1994! 1...p 1...p (v l)Vy: p > 1994 (pcm).

    V d 2.7. Chng minh rng: n >2 th gia n v n! c t nht 1 snguyn t (t suy ra c v s s nguyn t). 4

    Chuyn S hc Din n Ton hc

  • 16 2.2. Mt s bi ton c bn v s nguyn t

    Li gii. V n > 2 nn k = n! 1 > 1, do k c t nht mt c snguyn t p. Tng t bi tp 3, ta chng minh c mi c nguynt p ca k u ln hn k.Vy: p > n n < p < n! 1 < n! (pcm)

    2.2.3 Tm s nguyn t tha mn iu kin cho trc

    V d 2.8. Tm tt c cc gi tr ca s nguyn t p : p + 10 vp+ 14 cng l s nguyn t. 4Li gii. Nu p = 3 th p+ 10 = 3 + 10 = 13 v p+ 14 = 3 + 14 = 17u l cc s nguyn t nn p = 3 l gi tr cn tm.Nu p > 3 p c dng 3k + 1 hoc dng 3k 1

    Nu p = 3k + 1 th p+ 14 = 3k + 15 = 3(k + 5)...3

    Nu p = 3k 1 th p+ 10 = 3k + 9 = 3(k + 3)...3

    Vy nu p > 3 th hoc p+ 10 hoc p+ 14 l hp s : khng tha mnbi. Vy p = 3.

    V d 2.9. Tm k N trong 10 s t nhin lin tip:

    k + 1; k + 2; k + 3; ....k + 10

    c nhiu s nguyn t nht. 4Li gii. Nu k = 0: t 1 n 10 c 4 s nguyn t: 2; 3; 5; 7.Nu k = 1: t 2 n 11 c 5 s nguyn t: 2; 3; 5; 7; 11.Nu k > 1: t 3 tr i khng c s chn no l s nguyn t. Trong 5s l lin tip, t nht c 1 s l bi s ca 3 do , dy s c t hn 5s nguyn t.Vy vi k = 1, dy tng ng: k + 1; k + 2, .....k + 10 c cha nhiu snguyn t nht (5 s nguyn t).

    V d 2.10. Tm tt c cc s nguyn t p : 2p+p2 cng l s nguynt. 4Li gii. Xt 3 trng hp:

    Din n Ton hc Chuyn S hc

  • 2.2. Mt s bi ton c bn v s nguyn t 17

    p = 2 2p + p2 = 22 + 22 = 8 6 P p = 3 2p + p2 = 23 + 32 = 17 P

    p > 3 p 6 ...3. Ta c 2p + p2 = (p2 1) + (2p + 1).V p l 2p + 1...3 v p2 1 = (p+ 1)(p 1)...3 2p + p2 6 P

    Vy c duy nht 1 gi tr p = 3 tho mn.

    V d 2.11. Tm tt c cc s nguyn t p sao cho: p|2p + 1. 4Li gii. V p P : p|2p + 1 p > 2 (2; p) = 1Theo nh l Fermat, ta c: p|2p1 1. M

    p|2p + 1 p|2(2p1 1) + 3 p|3 p = 3Vy: p = 3.

    2.2.4 Nhn bit s nguyn t

    V d 2.12. Nu p l s nguyn t v 1 trong 2 s 8p+ 1 v 8p 1 ls nguyn t th s cn li l s nguyn t hay hp s? 4Li gii. Nu p = 2 8p+ 1 = 17 P; 8p 1 = 15 6 P Nu p = 3 8p 1 = 23 P; 8p 1 = 25 6 P Nu p > 3, xt 3 s t nhin lin tip: 8p 1; 8p v 8p+ 1. Trong

    3 s ny t c 1 s chia ht cho 3. Nn mt trong hai s 8p + 1v 8p 1 chia ht cho 3.

    Kt lun: Nu p P v 1 trong 2 s 8p+ 1 v 8p 1 l s nguyn tth s cn li phi l hp s.

    V d 2.13. Nu p 5 v 2p + 1 l cc s nguyn t th 4p + 1 lnguyn t hay hp s? 4Li gii. Xt 3 s t nhin lin tip: 4p; 4p + 1; 4p + 2. Trong 3 s tc mt s l bi ca 3.M p 5; p P nn p c dng 3k + 1 hoc 3k + 2

    Nu p = 3k + 1 th 2p+ 1 = 6k + 3...3: (tri vi gi thit)

    Chuyn S hc Din n Ton hc

  • 18 2.2. Mt s bi ton c bn v s nguyn t

    Nu p = 3k+2. Khi 4p+1 = 4(3k+2)+1 = 12k+9...3 4p+1l hp s

    V d 2.14. Trong dy s t nhin c th tm c 1997 s lin tipnhau m khng c s nguyn t no hay khng ? 4Li gii. Chn dy s: (ai) : ai = 1998! + i + 1 (i = 1, 1997) ai

    ...i +1 i = 1, 1997Nh vy: Dy s a1; a2; a3; .....a1997 gm c 1997 s t nhin lin tipkhng c s no l s nguyn t.

    V d 2.15 (Tng qut bi tp 2.14). Chng minh rng c th tmc 1 dy s gm n s t nhin lin tip (n > 1) khng c s nol s nguyn t ? 4Li gii. Ta chn dy s sau: (ai) : ai = (n+ 1)! + i+ 1 ai

    ...i+ 1 i =1, n.Bn c hy t chng minh dy (ai) trn s gm c n s t nhinlin tip trong khng c s no l s nguyn t c.

    2.2.5 Cc dng khc

    V d 2.16. Tm 3 s nguyn t sao cho tch ca chng gp 5 ln tngca chng. 4Li gii. Gi 3 s nguyn t phi tm l a, b, c. Ta c: abc = 5(a+ b+

    c) abc...5V a, b, c c vai tr bnh ng nn khng mt tnh tng qut, gi s:

    a...5 a = 5

    Khi : 5bc = 5(5 + b+ c) 5 + b+ c = bc (c 1)(b 1) = 6

    Do vy:

    {b 1 = 1c 1 = 6

    {b = 2c = 7

    chn{b 1 = 2c 1 = 3

    {b = 3c = 4

    loi

    Vy b s (a; b; c) cn tm l hon v ca (2; 5; 7).

    V d 2.17. Tm p, q P sao cho p2 = 8q + 1. 4

    Din n Ton hc Chuyn S hc

  • 2.3. Bi tp 19

    Li gii. Ta c:

    p2 = 8q + 1 8q = p2 1 = (p+ 1)(p 1) (2.1)

    Do p2 = 8q + 1 : l p2 : l p : l. t p = 2k + 1.Thay vo (2.1) ta c:

    8q = 2k(2k + 2) 2q = k(k + 1) (2.2)

    Nu q = 2 4 = k(k + 1) khng tm c k NVy q > 2. V q P (2, q) = 1.T (2.2) ta c:

    a) k = 2 v q = k + 1 k = 2; q = 3. Thay kt qu trn vo (2.2)ta c: p = 2.2 + 1 = 5

    b) q = k v 2 = k + 1 q = 1 :loi.

    Vy (q; p) = (5; 3).

    2.3 Bi tp

    2.3.1 Bi tp c hng dn

    Bi 1. Ta bit rng c 25 s nguyn t nh hn 100. Tng ca 25 snguyn t nh hn 100 l s chn hay s l?

    HD :Trong 25 s nguyn t nh hn 100 c cha mt s nguynt chn duy nht l 2, cn 24 s nguyn t cn li l s l. Do tng ca 25 s nguyn t l s chn.

    Bi 2. Tng ca 3 s nguyn t bng 1012. Tm s nguyn t nh nhttrong ba s nguyn t .

    HD: V tng ca 3 s nguyn t bng 1012, nn trong 3 snguyn t tn ti t nht mt s nguyn t chn. M snguyn t chn duy nht l 2 v l s nguyn t nh nht. Vys nguyn t nh nht trong 3 s nguyn t l 2.

    Chuyn S hc Din n Ton hc

  • 20 2.3. Bi tp

    Bi 3. Tng ca 2 s nguyn t c th bng 2003 hay khng? V sao?

    HD: V tng ca 2 s nguyn t bng 2003, nn trong 2 snguyn t tn ti 1 s nguyn t chn. M s nguyn tchn duy nht l 2. Do s nguyn t cn li l 2001. Do2001 chia ht cho 3 v 2001 > 3. Suy ra 2001 khng phi l snguyn t.

    Bi 4. Tm s nguyn t p, sao cho p+ 2; p+ 4 cng l cc s nguynt.

    Bi 5. Cho p v p + 4 l cc s nguyn t (p > 3). Chng minh rngp+ 8 l hp s.

    HD: V p l s nguyn t v p > 3, nn s nguyn t p c 1trong 2 dng:

    Nu p = 3k+ 2 th p+ 4 = 3k+ 6 = 3(k+ 2) p+ 4...3 vp+ 4 > 3. Do p+ 4 l hp s: tri bi.

    Nu p = 3k+ 1 th p+ 8 = 3k+ 9 = 3(k+ 3) p+ 8...3 vp+ 8 > 3. Do p+ 8 l hp s.

    Bi 6. Chng minh rng mi s nguyn t ln hn 2 u c dng 4n+1hoc 4n 1.

    Bi 7. Tm s nguyn t, bit rng s bng tng ca hai s nguynt v bng hiu ca hai s nguyn t.

    HD: Gi s a, b, c, d, e l cc s nguyn t v d > e. Theo bi:

    a = b+ c = d e ()T (*) a > 2 nn a l s nguyn t l b+ c; d e l s l.Do b, d l cc s nguyn t b, d l s l c, e l s chn. c = e = 2 (do c, el s nguyn t) a = b + 2 = d 2 d = b+ 4.Vy ta cn tm s nguyn t b sao cho b + 2 v b + 4 cng lcc s nguyn t.

    Din n Ton hc Chuyn S hc

  • 2.3. Bi tp 21

    Bi 8. Tm tt c cc s nguyn t x, y sao cho: x2 6y2 = 1.Bi 9. Cho p v p + 2 l cc s nguyn t (p > 3). Chng minh rng

    p+ 1...6.

    2.3.2 Bi tp khng c hng dn

    Bi 1. Tm s nguyn t p sao cho cc s sau cng l s nguyn t:

    a) p+ 2 v p+ 10.

    b) p+ 10 v p+ 20.

    c) p+ 10 v p+ 14.

    d) p+ 14 v p+ 20.

    e) p+ 2 v p+ 8.

    f) p+ 2 v p+ 14.

    g) p+ 4 v p+ 10.

    h) p+ 8 v p+ 10.

    Bi 2. Tm s nguyn t p sao cho cc s sau cng l s nguyn t:

    a) p+ 2, p+ 8, p+ 12, p+ 14

    b) p+ 2, p+ 6, p+ 8, p+ 14

    c) p+ 6, p+ 8, p+ 12, p+ 14

    d) p+ 2, p+ 6, p+ 8, p+ 12, p+ 14

    e) p+ 6, p+ 12, p+ 18, p+ 24

    f) p+ 18, p+ 24, p+ 26, p+ 32

    g) p+ 4, p+ 6, p+ 10, p+ 12, p+ 16

    Bi 3. Cho trc s nguyn t p > 3 tha

    a) p+ 4 P. Chng minh rng: p+ 8 l hp s.b) 2p+ 1 P. Chng minh rng: 4p+ 1 l hp s.c) 10p+ 1 P. Chng minh rng: 5p+ 1 l hp s.

    Chuyn S hc Din n Ton hc

  • 22 2.3. Bi tp

    d) p+ 8 P. Chng minh rng: p+ 4 l hp s.e) 4p+ 1 P. Chng minh rng: 2p+ 1 l hp s.f) 5p+ 1 P. Chng minh rng: 10p+ 1 l hp s.g) 8p+ 1 P. Chng minh rng: 8p 1 l hp s.h) 8p 1 P. Chng minh rng: 8p+ 1 l hp s.i) 8p2 1 P. Chng minh rng: 8p2 + 1 l hp s.j) 8p2 + 1 P. Chng minh rng: 8p2 1 l hp s.

    Bi 4. Chng minh rng:

    a) Nu p v q l hai s nguyn t ln hn 3 th p2 q2...24.b) Nu a, a+ k, a+ 2k(a, k N) l cc s nguyn t ln hn

    3 th k...6.

    Bi 5. a) Mt s nguyn t chia cho 42 c s d r l hp s. Tm sd r.

    b) Mt s nguyn t chia cho 30 c s d r. Tm s d r bitrng r khng l s nguyn t.

    Bi 6. Tm s nguyn t c ba ch s, bit rng nu vit s theoth t ngc li th ta c mt s l lp phng ca mt st nhin.

    Bi 7. Tm s t nhin c 4 ch s, ch s hng nghn bng ch shng n v, ch s hng trm bng ch s hng chc v s vit c di dng tch ca 3 s nguyn t lin tip.

    Bi 8. Tm 3 s nguyn t l cc s l lin tip.

    Bi 9. Tm 3 s nguyn t lin tip p, q, r sao cho p2 + q2 + r2 P.Bi 10. Tm tt c cc b ba s nguyn t a, b, c sao cho abc < ab +

    bc+ ca.

    Bi 11. Tm 3 s nguyn t p, q, r sao cho pq + qp = r.

    Din n Ton hc Chuyn S hc

  • 2.3. Bi tp 23

    Bi 12. Tm cc s nguyn t x, y, z tho mn xy + 1 = z.

    Bi 13. Tm s nguyn t abcd tha ab, ac l cc s nguyn t v b2 =cd+ b c.

    Bi 14. Cho cc s p = bc + a, q = ab + c, r = ca + b(a, b, c N) lcc s nguyn t. Chng minh rng 3 s p, q, r c t nht hai sbng nhau.

    Bi 15. Tm tt c cc s nguyn t x, y sao cho:

    a) x2 12y2 = 1b) 3x2 + 1 = 19y2

    c) 5x2 11y2 = 1d) 7x2 3y2 = 1e) 13x2 y2 = 3f) x2 = 8y + 1

    Bi 16. Chng minh rng iu kin cn v p v 8p2 + 1 l cc snguyn t l p = 3.

    Bi 17. Chng minh rng: Nu a2b2 l mt s nguyn t th a2b2 =a+ b.

    Bi 18. Chng minh rng mi s nguyn t ln hn 3 u c dng 6n+1hoc 6n 1.

    Bi 19. Chng minh rng tng bnh phng ca 3 s nguyn t ln hn3 khng th l mt s nguyn t.

    Bi 20. Cho s t nhin n 2. Gi p1, p2, ..., pn l nhng s nguyn tsao cho pn n+1. t A = p1.p2...pn. Chng minh rng trongdy s cc s t nhin lin tip: A + 2, A + 3, ..., A + (n + 1),khng cha mt s nguyn t no.

    Bi 21. Chng minh rng: Nu p l s nguyn t th 2.3.4...(p 3)(p2) 1...p.

    Chuyn S hc Din n Ton hc

  • 24 2.4. Ph lc: Bn nn bit

    Bi 22. Chng minh rng: Nu p l s nguyn t th 2.3.4...(p 2)(p1) + 1

    ...p.

    2.4 Ph lc: Bn nn bit

    Mi s nguyn t c 93 ch s lp thnh cp s cng

    Sau y l mt s nguyn t gm 93 ch s:

    100996972469714247637786655587969840329509324689190041803603417758904341703348882159067229719

    K lc ny do 70 nh ton hc lp c nm 1998 tht kh m nhbi c. H mt nhiu thng tnh ton mi tm c mi s nguynt to thnh mt cp s cng.

    T mc tr chi trong 1 tp ch khoa hc, hai nh nghin cu trngi hc Lyonl (Php) o su tng: Tm 6 s nguyn t sao chohiu 2 s lin tip lun lun nh nhau. iu l d i vi cc chuyngia nhng h mun i xa hn. Cng khng c vn g kh khn ivi mt dy 7 s. H cn s h tr mt cht t c 8 s, mt sh tr hn na t ti 9 s. Cui cng thng 3 nm 1998 c 70 nhton hc t khp trn th gii cng vi 200 my in ton hot nglin tc tm ra 10 s, mi s c 93 ch s, m hiu s ca 2 s lintip lun lun l 210. T s nguyn t trn ch cn thm vo 210 lc s nguyn t th 2....

    K lc c l dng : Theo c tnh ca cc nh khoa hc mun tmc 1 dy 11 s nguyn t th phi mt hn 10 t nm.

    Sinh ba rt t, phi chng sinh i li rt nhiu

    Ta bit rng cc s nguyn t c th xa nhau tu iu ny th hin bi tp:

    Din n Ton hc Chuyn S hc

  • 2.4. Ph lc: Bn nn bit 25

    Bi ton 2.1. Cho trc s nguyn dng n tu . Chng minh rngtn ti n s t nhin lin tip m mi s trong chng u l hp s.4Vy nhng, cc s nguyn t cng c th rt gn nhau. Cp s (2, 3)l cp s t nhin lin tip duy nht m c hai bn u l s nguynt. Cp s (p, q)c gi l cp s sinh i, nu c 2 u l s nguynt v q = p + 2. B 3 s (p, q, r) gi l b s nguyn t sinh ba nuc 3 s p,q,r u l cc s nguyn t v q = p+ 2; r = q + 2.

    Bi ton 2.2. Tm tt c cc b s nguyn t sinh ba? 4y l mt bi ton d, dng phng php chng minh duy nht tatm ra b (3, 5, 7) l b ba s nguyn t sinh ba duy nht, cc b 3 sl ln hn 3 lun c 1 s l hp s v n chia ht cho 3.T bi ton 2.2 th bi ton sau tr thnh mt gi thuyt ln ang chcu tr li.

    D on 2.1 Tn ti v hn cp s sinh i.

    S hon ho (hon ton) ca nhng ngi Hy Lp c i

    Ngi Hy Lp c i c quan nim thn b v cc s. H rt th vpht hin ra cc s hon ho, ngha l cc s t nhin m tng cc cs t nhin thc s ca n (cc c s nh hn s ) bng chnh n.

    Chng hn:

    6 = 1 + 2 + 3 28 = 1 + 2 + 4 + 7 + 14

    Ngi Hy Lp c i bit tm tt c cc s hon ho chn ngha lh lm c bi ton sau y:

    Bi ton 2.3. Mt s t nhin chn n 6= 0 l s hon ho nu v chnu: n = 2m+1(2m 1). Trong m l s t nhin khc 0 sao cho2m 1 l s nguyn t. 4T ta c gi thuyt

    Chuyn S hc Din n Ton hc

  • 26 2.4. Ph lc: Bn nn bit

    D on 2.2 Khng tn ti s hon ho l.

    bi ton 2.3 trn, s nguyn t dng 2m 1 gi l s nguyn tMerseme. Cc s nguyn t Merseme c vai tr rt quan trng. Chon nay ngi ta vn cha bit c hu hn hay v hn s nguyn tMerseme.

    D on 2.3 Tn ti v hn s nguyn t Merseme.

    Nm 1985 s nguyn t ln nht m ngi ta bit l s 21320491 gm39751 ch s ghi trong h thp phn. Gn y 2 sinh vin M tmra mt s nguyn t ln hn na l s 22160911 gm 65050 ch s.

    Ta bit rng vi hc sinh lp 6 th xem s A c t hn 20 ch sc l s nguyn t khng bng cch th xem A c chia ht cho s nonh hn A hay khng, th tm ht cc s nguyn t vi chic mysiu in ton cn hng th k !!!

    David SlowinSky son mt phn mm, lm vic trn my siu inton Gray-2 , sau 19 gi ng tm ra s nguyn t 2756839 1. S nyvit trong h thp phn s c 227832 ch s- vit ht s ny cn 110trang vn bn bnh thng. Hoc nu vit hng ngang nhng s trnphng ch .VnTime Size 14 th ta cn khong 570 m.

    Li Kt

    Thng qua ti ny, chng ta c th khng nh rng: Ton hc cmt trong mi cng vic, mi lnh vc ca cuc sng quanh ta, nkhng th tch ri v lng qun c, nn chng ta phi hiu bit vnm bt c n mt cch t gic v hiu qu.

    Mc ch ca ti ny l trang b nhng kin thc c bn c osu c nng cao v rn luyn t duy ton hc cho hc sinh, to ra nntng tin cy cc em c vn kin thc nht nh lm hnh trang cho

    Din n Ton hc Chuyn S hc

  • 2.4. Ph lc: Bn nn bit 27

    nhng nm hc tip theo.

    Vi iu kin c nhiu hn ch v thi gian, v nng lc trnh nntrong khun kh ti ny phn chia dng ton, loi ton ch c tnhtng i. ng thi cng mi ch a ra li gii ch cha c phngphp, thut lm r rng. Tuy c c gng nhiu nhng chnsg ti tthy trong ti ny cn nhiu hn ch. Chng ti rt mong nhnc nhng kin ng gp ca cc thy c gio cng bn c tonhc tht s c ngha cao p nh cu ngn ng Php vit:

    Ton hc l Vua ca cc khoa hcS hc l N hong

    Chuyn S hc Din n Ton hc

  • Chng

    3Bi ton chia ht

    3.1 L thuyt c bn 293.2 Phng php gii cc bi ton chia

    ht 31

    Phm Quang Ton (Phm Quang Ton)

    Chia ht l mt ti quan trng trong chng trnh S hc ca bcTHCS. i km theo l cc bi ton kh v hay. Bi vit ny xingii thiu vi bn c nhng phng php gii cc bi ton chia ht:phng php xt s d, phng php quy np, phng php ng d,v.v...

    3.1 L thuyt c bn

    3.1.1 nh ngha v chia ht

    nh ngha 3.1 Cho hai s nguyn a v b trong b 6= 0, ta lun tmc hai s nguyn q v r duy nht sao cho

    a = bq + r

    vi 0 r < b.Trong , ta ni a l s b chia, b l s chia, q l thng, r l s d.4Nh vy, khi a chia cho b th c th a ra cc s d r {0; 1; 2; ; |b|}.

    c bit, vi r = 0 th a = bq, khi ta ni a chia ht cho b (hoc a lbi ca b, hoc b l c ca a). Ta k hiu b | a. Cn khi a khng chia

    29

  • 30 3.1. L thuyt c bn

    ht cho b, ta k hiu b - a.

    Sau y l mt s tnh cht thng dng, chng minh c suy ra trctip t nh ngha.

    3.1.2 Tnh cht

    Sau y xin gii thiu mt s tnh cht v chia ht, vic chng minhkh l d dng nn s dnh cho bn c. Ta c vi a, b, c, d l cc snguyn th:

    Tnh cht 3.1 Nu a 6= 0 th a | a, 0 | a.

    Tnh cht 3.2 Nu b | a th b | ac.

    Tnh cht 3.3 Nu b | a v c | b th c | a.

    Tnh cht 3.4 Nu c | a v c | b th c | (ax by) vi x, y nguyn.

    Tnh cht 3.5 Nu b | a v a | b th a = b hoc a = b.

    Tnh cht 3.6 Nu c | a v d | b th cd | ab.

    Tnh cht 3.7 Nu b | a, c | a th BCNN(b; c) | a.

    Tnh cht 3.8 Nu c | ab v UCLN(b, c) = 1 th c | a.

    Tnh cht 3.9 Nu p | ab, p l s nguyn t th p | a hoc p | b.

    T tnh cht trn ta suy ra h qu

    H qu 3.1 Nu p | an vi p l s nguyn t, n nguyn dng thpn | an.

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 31

    3.1.3 Mt s du hiu chia ht

    Ta t N = anan1 . . . a1a0Du hiu chia ht cho 2; 5; 4; 25; 8; 125

    2 | N 2 | a0 a0 {0; 2; 4; 6; 8}5 | N 5 | a0 a0 {0; 5}

    4; 25 | N 4; 25 | a1a08; 125 | N 8; 125 | a2a1a0

    Du hiu chia ht cho 3 v 9

    3; 9 | N 3; 9 | (a0 + a1 + + an1 + an)Mt s du hiu chia ht khc

    11 | N 11 | [(a0 + a2 + ) (a1 + a3 + )]101 | N 101 | [(a1a0 + a5a4 + ) (a3a2 + a7a6 + )]

    7; 13 | N 7; 37 | [(a2a1a0 + a8a7a6 + ) (a5a4a3 + a11a10a9 + )]37 | N 37 | (a2a1a0 + a5a4a3 + + anan1an2)19 | N 19 | (an + 2an1 + 22an2 + + 2na0)

    3.2 Phng php gii cc bi ton chia ht

    3.2.1 p dng nh l Fermat nh v cc tnh cht ca chiaht

    nh l Fermat nh

    nh l 3.1 (nh l Fermat nh) Vi mi s nguyn a v snguyn t p th ap p (mod p). Chng minh. 1. Nu p | a th p | (a5 a).

    2. Nu p - a th 2a, 3a, 4a, , (p 1)a cng khng chia ht cho p.Gi r1, r2, , rp1 ln lt l s d khi chia a, 2a, 3a, , (p1)acho p. th chng s thuc tp {1; 2; 3; ; p 1} v i mt khcnhau (v chng hn nu r1 = r3 th p | (3a a) hay p | 2a,

    Chuyn S hc Din n Ton hc

  • 32 3.2. Phng php gii cc bi ton chia ht

    ch c th l p = 2, m p = 2 th bi ton khng ng). Do r1r2 rp1 = 1 2 3 (p 1). Ta c

    a r1 (mod p)2a r2 (mod p)

    (p 1)a rp1 (mod p)

    Nhn v theo v ta suy ra

    123 (p1)ap1 r1r2 rp1 (mod p) ap1 1 (mod p)V UCLN(a, p) = 1 nn ap a (mod p).

    Nh vy vi mi s nguyn a v s nguyn t p th ap a (mod p).Nhn xt. Ta c th chng minh nh l bng quy np. Ngoi ra, nhl cn c pht biu di dng sau:

    nh l 3.2 Vi mi s nguyn a, p l s nguyn t, UCLN(a, p) =1 th ap1 1 (mod p).

    Phng php s dng tnh cht chia ht v p dng nh lFermat nh

    C s: S dng cc tnh cht chia ht v nh l Fermat nh giiton.

    V d 3.1. Cho a v b l hai s t nhin. Chng minh rng 5a2+15abb2 chia ht cho 49 khi v ch khi 3a+ b chia ht cho 7. 4Li gii. ) Gi s 49 | 5a2 + 15ab b2 7 | 5a2 + 15ab b2 7 |

    (14a2 + 21ab) (5a2 + 15ab b2) 7 | (9a2 + 6ab + b2) 7 |(3a+ b)2 7 | 3a+ b.

    ) Gi s 7 | 3a+ b. t 3a+ b = 7c (c Z. Khi b = 7c 3a. Nhvy

    5a2 + 15ab b2 = 5a2 + 15a(7c 3a) (7c 3a)2= 49(c2 + 3ac a2)

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 33

    chia ht cho 49.Vy 5a2 + 15ab b2 chia ht cho 49 khi v ch khi 3a+ b chia ht cho7.

    V d 3.2. Cho 11 | (16a + 17b)(17a + 16b) vi a, b l hai s nguyn.Chng minh rng 121 | (16a+ 17b)(17a+ 16b). 4Li gii. Ta c theo u bi, v 11 nguyn t nn t nht mt tronghai s 16a+ 17b v 17a+ 16b chia ht cho 11. Ta li c (16a+ 17b) +(17a + 16b) = 33(a + b) chia ht cho 11. Do nu mt trong hai s16a+ 17b v 17a+ 16b chia ht cho 11 th s cn li cng chia ht cho11. Cho nn 121 | (16a+ 17b)(17a+ 16b). V d 3.3. Chng minh rng A = 130 + 230 + + 1130 khng chia htcho 11. 4Li gii. Vi mi a = 1, 2, , 10 th (a, 10) = 1. Do theo nh lFermat b th a10 1 (mod 11) a30 1 (mod 11) vi mi a =1, 2, , 10 v 1130 0 (mod 11). Nh vy

    A 1 + 1 + + 1 10 s 1

    +0 (mod 11)

    10 (mod 11) 11 - A

    V d 3.4. Cho p v q l hai s nguyn t phn bit. Chng minh rngpq1 + qp1 1 chia ht cho pq. 4Li gii. V q nguyn t nn theo nh l Fermat nh th

    pq1 1 (mod q)

    Do pq1 + qp1 1 (mod q)

    V q v p c vai tr bnh ng nn ta cng d dng suy ra

    qp1 + pq1 1 (mod p).

    Cui cng v UCLN(q, p) = 1 nn pq1 + qp1 1 (mod pq) haypq1 + qp1 1 chia ht cho pq.

    Chuyn S hc Din n Ton hc

  • 34 3.2. Phng php gii cc bi ton chia ht

    Bi tp ngh

    Bi 1. Chng minh rng 11a+2b chia ht cho 19 khi v ch khi 18a+5bchia ht cho 19 vi a, b l cc s nguyn.

    Bi 2. Chng minh rng 2a + 7 chia ht cho 7 khi v ch khi 3a2 +10ab 8b2.

    Bi 3. Cho p l s nguyn t ln hn 5. Chng minh rng nu n l st nhin c p 1 ch s v cc ch s u bng 1 th n chiaht cho p.

    Bi 4. Gi s n N, n 2. Xt cc s t nhin an = 11 1 c vitbi n ch s 1. Chng minh rng nu an l mt s nguyn tth n l c ca an 1.

    Bi 5. Gi s a v b l cc s nguyn dng sao cho 2a 1, 2b 1 va+ b u l s nguyn t. Chng minh rng ab + ba v aa + bb

    u khng chia ht cho a+ b.

    Bi 6. Chng minh rng vi mi s nguyn t p th tn ti s nguynn sao cho 2n + 3n + 6n 1 chia ht cho p.

    3.2.2 Xt s d

    C s: chng minh A(n) chia ht cho p, ta xt cc s n dngn = kp+ r vi r {0; 1; 2; ; p 1}.Chng hn, vi p = 5 th s nguyn n c th vit li thnh 5k; 5k +1; 5k + 2; 5k + 3; 5k + 4. Ta th mi dng ny vo cc v tr ca n ril lun ra p s. Sau y l mt s v d

    V d 3.5. Tm k N tn ti n N sao cho

    4 | n2 k

    vi k {0; 1; 2; 3}. 4Li gii. Gi s tn ti k N tn ti n N tha mn 4 | n2 k.Ta xt cc Trng hp: (m N)

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 35

    1. Nu n = 4m th n2 k = 16m2 k chia ht cho 4 khi v ch khi4 | k nn k = 0.

    2. Nu n = 4m 1 th n2 k = 16m2 8m+ 1 k chia ht cho 4khi v ch khi 4 | 1 k nn k = 1.

    3. Nu n = 4m 2 th n2 k = 16m2 16m+ 4 k chia ht cho 4khi v ch khi 4 | k nn k = 0.

    Vy k = 0 hoc k = 1.

    V d 3.6. Chng minh rng vi mi n N th 6 | n(2n+7)(7n+1).4Li gii. Ta thy mt trong hai s n v 7n+ 1 l s chn n N. Do 2 | n(2n+ 7)(7n+ 1). Ta s chng minh 3 | n(2n+ 7)(7n+ 1). Thtvy, xt

    1. Vi n = 3k th 3 | n(2n+ 7)(7n+ 1).

    2. Vi n = 3k + 1 th 2n + 7 = 6k + 9 chia ht cho 3 nn 3 |n(2n+ 7)(7n+ 1).

    3. Vi n = 3k + 2 th 7n + 1 = 21k + 15 chia ht cho 3 nn 3 |n(2n+ 7)(7n+ 1).

    Do 3 | n(2n+7)(7n+1) m (2, 3) = 1 nn 6 | n(2n+7)(7n+1) n N.

    V d 3.7. (HSG 9, Tp H Ch Minh, vng 2, 1995) Cho x, y, z l ccs nguyn tha mn

    (x y)(y z)(z x) = x+ y + z (3.1)

    Chng minh rng 27 | (x+ y + z). 4Li gii. Xt hai trng hp sau

    Chuyn S hc Din n Ton hc

  • 36 3.2. Phng php gii cc bi ton chia ht

    1. Nu ba s x, y, z chia ht cho 3 c cc s d khc nhau th cchiu xy, yz, zx cng khng chia ht cho 3. M 3 | (x+y+z)nn t (3.1) suy ra v l .

    2. Nu ba s x, y, z ch c hai s chia cho 3 c cng s d th trong bahiu xy, yz, zx c mt hiu chia ht cho 3. M 3 - (x+y+z)nn t (3.1) suy ra v l.

    Vy x, y, z chia cho 3 c cng s d, khi x y, y z, z x u chiaht cho 3. T (3.1) ta suy ra 27 | (x+ y + z), ta c pcm.

    Bi tp ngh

    Bi 1. i) Tm s t nhin n 7 | (2n 1).ii) Chng minh rng 7 - (2n + 1) n N.

    Bi 2. Chng minh rng vi mi s nguyn a th a(a6 1) chia htcho 7.

    Bi 3. Tm n 13 | 32n + 3n + 1.

    Bi 4. Chng minh rng vi mi a, b N th ab(a2b2)(4a2b2) lunlun chia ht cho 5.

    Bi 5. Chng minh rng 24 | (p 1)(p + 1) vi p l s nguyn t lnhn 3.

    Bi 6. Chng minh rng khng tn ti s nguyn a a2 + 1 chia htcho 12.

    Bi 7. Chng minh rng vi mi s nguyn x, y, z nu 6 | x+ y+ z th6 | x3 + y3 + z3.

    Bi 8. Cho ab = 20112012, vi a, b N. Hi tng a+ b c chia ht cho2012 hay khng ?

    Bi 9. S 3n+2003 trong n l s nguyn dng c chia ht cho 184khng ?

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 37

    Bi 10. Cho cc s nguyn dng x, y, z tha mn x2 + y2 = z2. Chngminh rng xyz chia ht cho 60.

    Bi 11. Cho cc s nguyn dng x, y, z tha mn x2+y2 = 2z2. Chngminh rng x2 y2 chia ht cho 84.

    Bi 12. Cho n > 3, (n N). Chng minh rng nu 2n = 10a+b, (0 1 ta c nn + 5n2 11n+ 5 chia ht cho (n 1)2.

    Bi 6. (HSG 9 Tp H Ni, vng 2, 1998) Chng minh rng 1997 | mvi m,n N tha mn

    m

    n= 1 1

    2+

    1

    3 1

    4+ + 1

    1329 1

    1330+

    1

    1331.

    Bi 7. Chng minh rng 32n+1 + 2n+2 chia ht cho 7 vi mi n N.

    Chuyn S hc Din n Ton hc

  • 44 3.2. Phng php gii cc bi ton chia ht

    Bi 8. Chng minh rng 20032005 + 20172015 chia ht cho 12.

    Bi 9. Cho p l s t nhin l v cc s nguyn a, b, c, d, e tha mna + b + c + d + e v a2 + b2 + c2 + d2 + e2 u chia ht cho p.Chng minh rng s a5 + b5 + c5 + d5 + e5 5abcde cng chiaht cho p.

    Bi 10. (Canada Training for IMO 1987)K hiu:

    1 3 5 (2n 1) = (2n 1)!!2 4 6 (2n) = (2n)!!.

    Chng minh rng (1985)!! + (1986)!! chia ht cho 1987.

    Bi 11. Chng minh rng s 22225555 + 55552222 chia ht cho 7.

    Bi 12. Cho k l s nguyn dng sao cho s p = 3k + 1 l s nguynt v

    1

    1 2 +1

    3 4 + +1

    (2k 1)2k =m

    n

    vi hai s nguyn dng nguyn t cng nhau m v n.Chngminh m chia ht cho p.

    (Tp ch Mathematics Reflections, ng bi T.Andreescu)

    3.2.4 Xt ng d

    nh ngha v mt s tnh cht

    nh ngha 3.2 Cho a, b l cc s nguyn v n l s nguyn dng. Tani, a ng d vi b theo modun n v k hiu a b (mod n) nu a vb c cng s d khi chia cho n. 4Nh vy a n (mod n) n | (a b). V d: 2012 2 (mod 5).

    Tnh cht (bn c t chng minh)Cho a, b, c, d, n l cc s nguyn.

    Tnh cht 3.10a a (mod n),a b (mod n) b a (mod n),a b (mod n), b c (mod n) a c (mod n).

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 45

    Tnh cht 3.11

    {a b (mod n)c d (mod n)

    {a c b d (mod n)ac bd (mod n)

    Tnh cht 3.12 a b (mod n) ak bk (mod n), k 1. Tnh cht 3.13 a b (mod n) ac bc (mod mc), c > 0 Tnh cht 3.14 (a+ b)n bn (mod a), (a > 0). Tnh cht 3.15 Nu d l c chung dng ca a, b v m th a b(mod m) th

    a

    d bd

    (modm

    d).

    Tnh cht 3.16 a b (mod m), c l c chung ca a v b, (c,m) = 1th

    a

    c bc

    (mod m).

    Phng php ng d thc gii cc bi ton chia ht

    C s: S dng cc tnh cht v nh ngha trn gii cc bi tonchia ht.

    V d 3.17. Chng minh rng vi mi s t nhin n th 7 | 8n + 6. 4Li gii. Ta c 8n 1 (mod 7) = 8n + 6 7 0 (mod 7).

    V d 3.18. Chng minh rng 19 | 7 52n + 12 6n. vi mi s nguyndng n. 4Li gii. Ta c 52 = 25 6 (mod 19) = (52)n 6n (mod 19) =7 52n 7 6n (mod 19) = 7 52n + 12 6n 19 6n 0 (mod 19).

    V d 3.19. Vit lin tip cc s 111, 112, , 888 c s A =111112 888. Chng minh rng 1998 | A. 4

    Chuyn S hc Din n Ton hc

  • 46 3.2. Phng php gii cc bi ton chia ht

    Li gii. Ta thy A chn nn 2 | A. Mt khc

    A = 111 1000777 + 112 1000776 + + 888.

    Do 1000k 1 (mod 999), k N nn

    A 111 + 112 + + 888 0 (mod 999).

    Suy ra 999 | A, v (999, 2) = 1 nn 1998 | A.

    V d 3.20. Chng minh rng 7 | 55552222 + 22225555. 4Li gii. Ta c

    2222 4 (mod 7) = 22225555 (4)5555 (mod 7)5555 4 (mod 7) = 55552222 4 (mod 7)

    = 55552222 + 22225555 45555 + 42222 (mod 7)Li c

    45555 + 42222 = 42222 (43333 1)= 42222 (641111 1)

    V 64 1 (mod 7) = 641111 1 0 (mod 7).Do 7 | 55552222 + 22225555

    Bi tp ngh

    Bi 1. Mt s bi tp phng php phn tch c th gii bng phngphp ng d thc.

    Bi 2. Chng minh rng 333555777

    + 777555333

    chia ht cho 10.

    Bi 3. Chng minh rng s 11101967 1 chia ht cho 101968.

    Bi 4. Cho 9 | a3 + b3 + c3, a, b, c Z. Chng minh rng 3 | a b c.Bi 5. Chng minh rng 222333 + 333222 chia ht cho 13.

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 47

    Bi 6. Chng minh rng 9n + 1 khng chia ht cho 100, n N.Bi 7. Chng minh rng vi mi s nguyn khng m n th 25n+3 +

    5n 3n+1 chia ht cho 17.Bi 8. Tm n N sao cho 2n3 + 3n = 19851986.Bi 9. Vit lin tip 2000 s 1999 ta c s X = 19991999 1999.

    Tm s d trong php chia X cho 10001.

    Bi 10. Chng minh rng 100 | 77777

    777 .Bi 11. Cho b2 4ac v b2 + 4ac l hai s chnh phng vi a, b, c N.

    Chng minh rng 30 | abc.

    3.2.5 Quy np

    C s : chng minh mnh ng vi mi s t nhin n p, talm nh sau:

    Kim tra mnh ng vi n = p. Gi s mnh ng vi n = k. Ta i chng minh mnh cng

    ng vi n = k + 1.

    V d 3.21. Chng minh rng A = 4n + 15 1 chia ht cho 9 vi min N. 4Li gii. Vi n = 1 = A = 18 chia ht cho 9.Gi s bi ton ng vi n = k. Khi 9 | 4k+15k1, hay 4k+15k1 =9q vi q N. Suy ra 4k = 9q 15k + 1.Ta i chng minh bi ton ng vi n = k+1, tc 9 | 4k+1+15(k+1)1.Tht vy:

    4k+1 + 15(k + 1) 1 = 4 4k + 15k + 14= 4 (9q 15k + 1) + 15k + 14= 36q 45k + 18

    chia ht cho 9. Ta c pcm.

    Chuyn S hc Din n Ton hc

  • 48 3.2. Phng php gii cc bi ton chia ht

    V d 3.22. (HSG 9 TQ 1978)Chng minh rng s c to bi 3n chs ging nhau th chia ht cho 3n vi 1 n, n N. 4Li gii. Vi n = 1, bi ton hin nhin ng.Gi s bi ton ng vi n = k, tc 3k | aa a

    3n s a

    .

    Vi n = k + 1 ta c:

    aa a 3k+1

    = aa a 3k

    aa a 3k

    aa a 3k

    = aa a 3k

    1 00 0 3k1

    00 0 3k1

    1

    chia ht cho 3k+1. Ta c pcm.

    V d 3.23. Chng minh rng vi mi n N, k l s t nhin l th

    2n+2 | k2n 1

    Li gii. Vi n = 1 th k2n 1 = k2 1 = (k + 1)(k 1). Do k l,nn

    t k = 2m+ 1 vi m N, th khi (k + 1)(k 1) = 4k(k + 1) chiaht cho 23 = 8.Gi s bi ton ng vi n = p, tc 2p+2 | k2p1 hay k2p = q 2p+2 +1vi q N.Ta chng minh bi ton ng vi n = p+ 1. Tht vy

    A = k2p+1 1 = k22p 1 = (k2p)2 1

    =(k2

    p 1) (k2p + 1)= q 2p+2 (2 + q 2p+2)= q 2p+3 (1 + q 2p+1)

    chia ht cho 2p+3. Ta c pcm.

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 49

    Bi tp ngh

    Bi 1. Mt s bi ton cc phng php nu trn c th gii bngphng php quy np.

    Bi 2. Chng minh rng 255 | 16n 15n 1 vi n N.

    Bi 3. Chng minh rng 64 | 32n+3 + 40n 27 vi n N.

    Bi 4. Chng minh rng 16 | 32n+2 + 8n 9 vi n N.

    Bi 5. Chng minh rng 676 | 33n+3 16n 27 vi n N, n 1.

    Bi 6. Chng minh rng 700 | 292n 140n 1 vi n N.

    Bi 7. Chng minh rng 270 | 2002n 138n 1 vi n N.

    Bi 8. Chng minh rng 22 | 324n+1 + 234n+1 + 5 vi n N.

    Bi 9. Chng minh rng s 23n

    + 1 chia ht cho 3n nhng khng chiaht cho 3n+1 vi n N.

    Bi 10. Chng minh rng s 20012n 1 chia ht cho 2n+4 nhng khng

    chia ht cho 2n+5 vi n N.

    Bi 11. Chng minh rng vi mi s t nhin n 2, tn ti mt s tnhin m sao cho 3n | (m3 + 17), nhng 3n+1 - (m3 + 17).

    Bi 12. C tn ti hay khng mt s nguyn dng l bi ca 2007 vc bn ch s tn cng l 2008.

    Bi 13. Chng minh rng tn ti mt s c 2011 ch s gm ton chs 1 v 2 sao cho s chia ht cho 22011.

    Bi 14. Tm phn d khi chia 32ncho 2n+3, trong n l s nguyn

    dng.

    Bi 15. Cho n N, n 2. t A = 77...

    (ly tha n ln). Chng minhrng An + 17 chia ht cho 20.

    Chuyn S hc Din n Ton hc

  • 50 3.2. Phng php gii cc bi ton chia ht

    3.2.6 S dng nguyn l Dirichlet

    Ni dung: Nht 5 con th vo 3 chung th tn ti chung cha t nht2 con.

    nh l 3.3 Nht m = nk + 1 con th vo k chung (k < n) th tnti chung cha t nht n+ 1 con th.

    Chng minh. Gi s khng c chung no cha t nht n+ 1 con th,khi mi chung cha nhiu nht n con th, nn k chung cha nhiunht kn con th, mu thun vi s th l nk + 1.

    nh l 3.4 (p dng vo s hc) Trong m = nk + 1 s c tnht n+ 1 s chia cho k c cng s d.

    Tuy nguyn l c pht biu kh n gin nhng li c nhng ngdng ht sc bt ng, th v. Bi vit ny ch xin nu mt s ng dngca nguyn l trong vic gii cc bi ton v chia ht.

    V d 3.24. Chng minh rng lun tn ti s c dng

    20112011 201100 0chia ht cho 2012. 4Li gii. Ly 2013 s c dng

    2011; 20112011, , 20112011 2011 2012 s 2011

    .

    Ly 2013 s ny chia cho 2012. Theo nguyn l Dirichlet th tn ti hais c cng s d khi chia cho 2012.Gi s hai s l 20112011 2011

    m s 2011

    v 20112011 2011 n s 2011

    (m > n >

    0).= 2012 | 20112011 2011

    m s 2011

    20112011 2011 n s 2011

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 51

    = 2012 | 20112011 2011 mn s 2011

    00 00 n s 2011

    Vy tn ti s tha mn bi.

    V d 3.25. Chng minh rng trong 101 s nguyn bt k c th tmc hai s c 2 ch s tn cng ging nhau. 4Li gii. Ly 101 s nguyn cho chia cho 100 th theo nguyn lDirichlet tn ti hai s c cng s d khi chia cho 100. Suy ra trong101 s nguyn cho tn ti hai s c ch s tn cng ging nhau.

    V d 3.26 (Tuyn sinh 10 chuyn HSPHN, 1993). Cho 5 s nguynphn bit ty a1, a2, a3, a4, a5. Chng minh rng tch

    P = (a1 a2)(a1 a3)(a1 a4)(a1 a5)(a2 a3) (a2 a4)(a2 a5)(a3 a4)(a3 a5)(a4 a5)

    chia ht cho 288. 4Li gii. Phn tch 288 = 25 32.

    1. Chng minh 9 | P : Theo nguyn l Dirichlet th trong 4 sa1, a2, a3 c hai s c hiu chia ht cho 3. Khng mt tnh tngqut, gi s: 3 | a1 a2. Xt 4 s a2, a3, a4, a5 cng c hai s chiu chia ht cho 3. Nh vy P c t nht hai hiu khc nhauchia ht cho 3, tc 9 | p.

    2. Chng minh 32 | P : Theo nguyn l Dirichlet th tng 5 s chotn ti t nht 3 s c cng tnh chn l. Ch c th c hai khnng sau xy ra:

    Nu c t nht 4 s c cng tnh chn l, th t bn s c thlp thnh su hiu khc nhau chia ht cho 2. Do 32 | P .

    Chuyn S hc Din n Ton hc

  • 52 3.2. Phng php gii cc bi ton chia ht

    Nu c 3 s c cng tnh chn l. Khng mt tnh tng qut,gi s ba s l a1, a2, a3. Khi a4, a5 cng cng tnhchn l nhng li khc tnh chn l ca a1, a2, a3. Khi cc hiu sau chia ht cho 2: a1 a2, a1 a3, a2 a3, a4 a5.Mt khc, trong 5 s cho c t nht hai hiu chia ht cho4, cho nn trong 4 hiu a1 a2, a1 a3, a2 a3, a4 a5 ct nht mt hiu chia ht cho 4. Vy 32 | P .

    Ta c pcm.

    V d 3.27. Cho 2012 s t nhin bt k a1, a2, , a2012. Chng minhrng tn ti mt s chia ht cho 2012 hoc tng mt s s chia ht cho2012. 4Li gii. Xt 2012 s

    S1 = a2

    S2 = a1 + a2

    S2012 = a1 + a2 + + a2012

    Trng hp 1: Nu tn ti s Si (i = 1, 2, , 2012) chia ht cho2012 th bi ton chng minh xong.

    Trng hp 2: Nu 2012 - Si vi mi i = 1, 2, , 2012. em 2012s ny chia cho 2012 nhn c 2012 s d. Cc s d nhn gitr thuc tp {1; 2; ; 2011}. V c 2012 s d m ch c 2011gi tr nn theo nguyn l Dirichlet chc chn c hai s d bngnhau. Ga s gi hai s l Sm v Sn c cng s d khi chiacho 2012 (m,n N, 1 n < m 2012) th hiu

    Sm Sn = an+1 + an+2 + + am

    chia ht cho 2012.

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 53

    Nhn xt. Ta c th rt ra bi ton tng qut v bi ton m rngsau:

    Bi ton 3.5 (Bi ton tng qut). Cho n s a1, a2, , an. Chngminh rng trong n s trn tn ti mt s chia ht cho n hoc tng mts s chia ht cho n. 4

    Bi ton 3.6 (Bi ton m rng). (Tp ch Ton Tui Th s 115)Cho n l mt s chuyn dng v n s nguyn dng a1, a2, , an ctng bng 2n 1. Chng minh rng tn ti mt s s trong n s cho c tng bng n. 4

    Bi tp ngh

    Bi 1. Chng minh rng c v s s chia ht cho 201311356

    m trongbiu din thp phn ca cc s khng c cc ch s 0, 1, 2, 3.

    Bi 2. (HSG 9 H Ni, 2006) Chng minh rng tn ti s t nhinn 6= 0 tha mn 313579 | (13579n 1).

    Bi 3. Chng minh rng trong 52 s nguyn dng bt k lun luntm c hai s c tng hoc hiu chia ht cho 100.

    Bi 4. Cho 10 s nguyn dng a1, a2, , a10. Chng minh rng tnti cc s ci {0,1, 1}, (i = 1, 10) khng ng thi bng0 sao cho

    A = c1a1 + c2a2 + + c10a10chia ht cho 1032.

    Bi 5. Chng minh rng tn ti s t nhin k sao cho 2002k 1 chiaht cho 200310.

    Bi 6. Bit rng ba s a, a+ k, a+ 2k u l cc s nguyn t ln hn3. Chng minh rng khi k chia ht cho 6.

    Chuyn S hc Din n Ton hc

  • 54 3.2. Phng php gii cc bi ton chia ht

    3.2.7 Phn chng

    C s: chng minh p - A(n), ta lm nh sau:

    Gi s ngc li p | A(n). Chng minh iu ngc li sai.

    V d 3.28. Chng minh rng vi mi s nguyn n th n2+n+1 khngchia ht cho 9. 4Li gii. Gi s 9 | (n2 +n+ 1). Khi n2 +n+ 1 = (n+ 2)(n 1) + 3chia ht cho 3. Suy ra 3 | n + 2 v 3 | n 1. Nh vy (n + 2)(n 1)chia ht cho 9, tc n2 + n+ 1 chia 9 d 3, mu thun. Ta c pcm.

    Nhn xt. Bi ton ny vn c th gii theo phng php xt s d.

    V d 3.29. Gi s p = k.2t + 1 l s nguyn t l, t l s nguyndng v k l s t nhin l. Gi thit x v y l cc s t nhin mp |(x2

    t+ y2

    t). Chng minh rng khi x v y ng thi chia ht cho

    p. 4Li gii. Gi s tri li p - x, suy ra p - y.Do p l s nguyn t nn theo nh l Fermat nh ta c{

    xp1 1 (mod p)yp1 1 (mod p)

    Theo gi thit th p 1 = k.2t, do {xk.2

    t 1 (mod p)yk.2

    t 1 (mod p)

    T ta cxk.2

    t+ yk.2

    t 2 (mod p). (i)Theo gi thit th

    x2t

    + y2t 0 (mod p).

    Din n Ton hc Chuyn S hc

  • 3.2. Phng php gii cc bi ton chia ht 55

    Do k l nn

    xk.2t

    + yk.2t

    =(x2

    t)k

    +(y2

    t)k ...(x2t + y2t)

    (xk.2

    t+ yk.2

    t) 0 (mod p) (ii)

    T (i) v (ii) suy ra iu mu thun. Vy gi thit phn chng sai. Do x, y ng thi chia ht cho p.

    Bi tp ngh

    Bi 1. Chng minh n2 + n+ 2 khng chia ht cho 15 vi mi n Z.Bi 2. Chng minh n2 + 3n+ 5 khng chia ht cho 121 vi mi n N.Bi 3. Chng minh 9n3 + 9n2 + 3n 16 khng chia ht cho 343 vi

    mi n N.Bi 4. Chng minh 4n3 6n2 + 3n + 37 khng chia ht cho 125 vi

    mi n N.Bi 5. Chng minh n3 + 3n 38 khng chia ht cho 49 vi mi n N.

    Chuyn S hc Din n Ton hc

  • Chng

    4Phng trnh nghimnguyn

    4.1 Xt tnh chia ht 574.2 S dng bt ng thc 744.3 Nguyn tc cc hn, li v hn 86

    Trn Nguyn Thit Qun (L Lawliet)Phm Quang Ton (Phm Quang Ton)

    Trong chng trnh 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 xut hin ti cc k thi ln, nh, trong v ngoinc. Trong bi vit ny ti ch mun cp n cc vn c bn canghim nguyn (cc dng, cc phng php gii) ch khng i nghincu su sc v n. Ti cng khng cp ti phng trnh Pell, phngtrnh Pythagore, phng trnh Fermat v n c nhiu trong cc sch,cc chuyn khc.

    4.1 Xt tnh chia ht

    4.1.1 Pht hin tnh chia ht ca 1 n

    V d 4.1. Gii phng trnh nghim nguyn

    13x+ 5y = 175 (4.1)

    57

  • 58 4.1. Xt tnh chia ht

    Li gii. Gi s x, y l cc s nguyn tha mn phng trnh (4.1). Ta

    thy 175 v 5y u chia ht cho 5 nn 13x...5 x...5 (do GCD(13; 5) = 1).

    t x = 5t (t Z). Thay vo phng trnh (4.1), ta c13.5t+ 5y = 175 13t+ y = 35 y = 35 13t

    Do , phng trnh (4.1) c v s nghim nguyn biu din di dng

    (x; y) = (5t; 35 13t), (t Z)Bi tp ngh

    Bi 1. Gii phng trnh nghim nguyn 12x 19y = 285Bi 2. Gii phng trnh nghim nguyn 7x+ 13y = 65

    Bi 3. Gii phng trnh nghim nguyn 5x+ 7y = 112

    4.1.2 a v phng trnh c s

    V d 4.2. Tm nghim nguyn ca phng trnh

    3xy + 6x+ y 52 = 0 (4.2)Li gii. Nhn xt. i vi phng trnh ny, ta khng th p dngphng php trn l pht hin tnh chia ht, vy ta phi gii nh thno?Ta gii nh sau:

    (4.2) 3xy + y + 6x+ 2 54 = 0 y (3x+ 1) + 2 (3x+ 1) 54 = 0

    (3x+ 1) (y + 2) = 54

    Nh vy, n y ta c x v y nguyn nn 3x+ 1 v y + 2 phi l cca 54. Nhng nu nh vy th ta phi xt n hn 10 trng hp sao?V:

    4 = 1.54 = 2.27 = 3.18 = 6.9

    = (1).(54) = (2).(27) = (3).(18) = (6).(9)

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 59

    C cch no khc khng? Cu tr li l c! Nu ta mt cht ntha s 3x + 1, biu thc ny chia cho 3 lun d 1 vi mi x nguyn.Vi lp lun trn, ta c:

    {3x+ 1 = 1y + 2 = 54

    {x = 0y = 52{

    3x+ 1 = 2y + 2 = 54

    {x = 1y = 56

    V d 4.3. Gii phng trnh nghim nguyn sau:

    2x+ 5y + 3xy = 8 (4.3)

    Li gii. Ta c

    (4.3) x(2 + 3y) + 5y = 8 3x(2 + 3y) + 15y = 24 3x(2 + 3y) + 5(2 + 3y) = 34 (3x+ 5)(3y + 3) = 34

    n y phn tch 34 = 1 34 = 2 17 ri xt cc trng hp. Ch rng 3x+ 5, 3y + 2 l hai s nguyn chia 3 d 2, vn dng iu ny tac th gim bt s trng hp cn xt.

    V d 4.4. Gii phng trnh nghim nguyn

    x2 y2 = 2011 (4.4)

    Li gii. (4.4) (x y)(x + y) = 2011. V 2011 l s nguyn t nnc nguyn ca 2011 ch c th l 1,2011. T suy ra nghim(x; y) l (1006; 1005); (1006;1005); (1006;1005); (1006; 1005).

    V d 4.5. Tm cc s nguyn x, y tho mn iu kin

    x2 + y2 = (x y)(xy + 2) + 9 (4.5)

    Chuyn S hc Din n Ton hc

  • 60 4.1. Xt tnh chia ht

    Li gii. t a = x y, b = xy. Khi (4.5) tr thnh

    a2 + 2b = a(b+ 2) + 9 (a 2)(a b) = 9 (4.6)

    V x, y Z nn a, , a 2, a b u l cc s nguyn. T (4.6) ta c cctrng hp sau:

    {a 2 = 9a b = 1

    {a = 11

    b = 10{x y = 11xy = 10

    (4.7)

    {a 2 = 3a b = 3

    {a = 5

    b = 2{x y = 5xy = 2

    (4.8)

    {a 2 = 1a b = 9

    {a = 3

    b = 6 {x y = 3xy = 6 (4.9)

    {a 2 = 1a b = 9

    {a = 1

    b = 10{x y = 1xy = 10

    (4.10)

    {a 2 = 3a b = 3

    {a = 1b = 2

    {x y = 1xy = 2

    (4.11)

    {a 2 = 3a b = 3

    {a = 1b = 2

    {x y = 1xy = 2

    (4.12)

    D thy cc h (4.7),(4.8),(4.10) khng c nghim nguyn, h (4.9) vnghim, h (4.11) c hai nghim nguyn (1; 2) v (2;1), h (4.12)c hai nghim nguyn (1; 6) v (6; 1).Tm li phng trnh (4.5) c cc cp nghim nguyn (x; y) l (1; 2);(2;1); (1; 6); (6; 1).

    V d 4.6. Tm nghim nguyn ca phng trnh:

    (x2 + 1

    ) (y2 + 1

    )+ 2 (x y) (1 xy) = 4 (1 + xy) (4.13)

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 61

    Li gii. Phng trnh (4.13) tng ng vi:

    x2y2 + x2 + y2 + 1 + 2x 2x2y 2y + 2xy2 = 4 + 4xy (x2 + 2x+ 1)y2 2(x2 + 2x+ 1)y + (x2 + 2x+ 1) = 4 (x+ 1)2(y 1)2 = 4

    [(x+ 1)(y 1) = 2(x+ 1)(y 1) = 2

    Vi (x+ 1)(y 1) = 2 m x, y Z nn ta c cc trng hp sau:

    {x+ 1 = 1y 1 = 2

    {x = 0y = 3

    {x+ 1 = 2y 1 = 1

    {x = 1y = 2

    {x+ 1 = 2y 1 = 1

    {x = 3y = 0

    {x+ 1 = 1y 1 = 2

    {x = 2y = 1

    Vi (x+ 1)(y 1) = 2 , tng t ta cng suy ra c:

    {x+ 1 = 1y 1 = 2

    {x = 2y = 3

    {x+ 1 = 1y 1 = 2

    {x = 0y = 1

    {x+ 1 = 2y 1 = 1

    {x = 1y = 0

    {x+ 1 = 2y 1 = 1

    {x = 3y = 2

    Vy phng trnh cho c cc cp nghim nguyn:

    (x; y) = {(0; 3); (1; 2); (3; 0); (2;1); (2; 3); (0;1); (1; 0); (3; 2)}

    V d 4.7. Tm nghim nguyn ca phng trnh

    x6 + 3x3 + 1 = y4 (4.14)

    Chuyn S hc Din n Ton hc

  • 62 4.1. Xt tnh chia ht

    Li gii. Nhn hai v ca phng trnh (4.14) cho 4, ta c:

    4x6 + 12x3 + 4 = 4y4

    (4x6 + 12x3 + 9) 4y4 = 5 (2x3 + 3)2 4y4 = 5 (2x3 2y2 + 3)(2x3 + 2y2 + 3) = 5.

    Vi lu rng 5 = 1.5 = 5.1 = (1).(5) = (5).(1) v x, y Z nnta suy ra c cc trng hp sau:

    {

    2x3 2y2 + 3 = 12x3 + 2y2 + 3 = 5

    {x3 y2 = 1x3 + y2 = 1

    {x3 = 0y2 = 1

    [x = 0y = 1[x = 0y = 1

    {

    2x3 2y2 + 3 = 12x3 + 2y2 + 3 = 5

    {x3 y2 = 2x3 + y2 = 4

    {x3 = 3y2 = 1 (loi)

    {

    2x3 2y2 + 3 = 52x3 + 2y2 + 3 = 1

    {x3 y2 = 1x3 + y2 = 1

    {x3 = 0y2 = 1 (loi)

    {

    2x3 2y2 + 3 = 52x3 + 2y2 + 3 = 1

    {x3 y2 = 4x3 + y2 = 2

    {x3 = 3y2 = 1

    (loi)

    Vy phng trnh cho c cc cp nghim nguyn:

    (x; y) = {(0; 1); (0;1)}

    Nhn xt. Bi ton ny cng c th gii bng phng php kp.

    V d 4.8. Gii phng trnh nghim nguyn dng:

    1

    x+

    1

    y=

    1

    p(4.15)

    trong p l s nguyn t. 4

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 63

    Li gii.

    (4.15) xy = px+ py (x y)(y p) = p2.

    V p l s nguyn t nn c s nguyn ca p2 ch c th l 1,p,p2.Th ln lt vi cc c trn ta d tm c kt qu. Phn trnh byxin dnh cho bn c.

    Nhn xt. Phng php ny cn hai bc chnh: Phn tch thnh cs v xt trng hp tm kt qu. Hai bc ny c th ni l khngqu kh i vi bn c, nhng xin ni mt s lu thm v bc xttrng hp. Trong mt s bi ton, hng s nguyn v phi sau khiphn tch l mt s c nhiu c, nh vy i hi xt trng hp vtnh ton rt nhiu. Mt cu hi t ra l: Lm th no gim strng hp b xt y? V tr li c cu hi , ta s tham khov d di y.

    V d 4.9. Tm nghim nguyn ca phng trnh:

    x2 + 12x = y2. (4.16)

    Li gii. (thng thng) Phng trnh (4.16) cho tng ng vi:

    (x+ 6)2 y2 = 36 (x+ 6 + y)(x+ 6 y) = 36

    Suy ra x + y + 6, x + 6 y l c ca 36. M s 36 c tt c 18 cnn ta phi xt 18 trng hp tng ng vi

    x+ 6 + y {1;2;3;4;6;9;12;18;36}

    . Kt qu l ta tm c cc cp nghim nguyn (x; y) l

    (0; 0); (12; 0); (16; 8); (16;8); (4; 8); (4;8)

    .

    Nhn xt. ng nh vn m ta nu ra trn, s c qu nhiu xt. Cho nn ta s c cc nhn xt sau thc hin thao tc "siuphm" chuyn t con s 18 xung ch cn 2!

    Chuyn S hc Din n Ton hc

  • 64 4.1. Xt tnh chia ht

    V y c s m chn trong phng trnh nn c th gi s y 0. Khi x + 6 y x + 6 + y, do vy ta loi c tm trng hp v cnli cc trng hp sau:{

    x+ 6 + y = 9

    x+ 6 y = 4 ,{x+ 6 + y = 9x+ 6 y = 4 ,

    {x+ y + 6 = 1x+ y 6 = 36 ,{

    x+ y + 6 = 36

    x y + 6 = 1 ,{x+ y + 6 = 2x y + 6 = 18 ,

    {x+ y + 6 = 18

    x y + 6 = 2 ,{x+ y + 6 = 3x y + 6 = 12 ,

    {x+ y + 6 = 12

    x y + 6 = 3 ,{x+ y + 6 = 6x y + 6 = 6 ,{

    x+ y + 6 = 6

    x+ y 6 = 6 .

    By gi ta c 10 trng hp, ta s tip tc lc b. Nhn thy(x+ y + 6) (x+ 6 y) = 2y nn x+ 6 y v x+ 6 + y c cng tnhchn l, do ta loi thm 6 trng hp, ch cn{

    x+ y + 6 = 18

    x+ y 6 = 2 ,{x+ y + 6 = 2x+ y 6 = 18 ,{

    x+ y + 6 = 6x y + 6 = 6 ,

    {x+ y + 6 = 6

    x+ y 6 = 6.

    Tip tc xt hai phng trnh

    {x+ y + 6 = 6x y + 6 = 6 v

    {x+ y + 6 = 6

    x+ y 6 = 6 ,

    hai phng trnh ny u tm c y = 0. Vy sao khng n ginhn, ta xt y = 0 ngay t u. Phng trnh c dng x(x + 12) = y2,xt hai kh nng:

    Nu y = 0 th x = 0 hoc x = 12. Nu y 6= 0 th x+6+y > x+6y, p dng hai nhn xt trn ta ch

    c hai trng hp:

    {x+ y + 6 = 2x y + 6 = 18 v

    {x+ y + 6 = 18

    x y + 6 = 2 .

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 65

    Phng trnh cho c 6 nghim nguyn

    (x; y) = (16; 8), (0; 0), (12; 0), (16; 8), (4; 8), (4;8)

    Nhn xt. Nh vy bi ton ngn gn, chnh xc nh linh hot trongvic xt tnh chn l, gii hn hai s gim s trng hp cn xt.Ngoi cc cch nh gi trn ta cn c th p dng xt s d tng v nh gi (y cng l mt phng php gii phng trnh nghimnguyn).

    Bi tp ngh

    Bi 1. Th bin i cc bi ton gii phng trnh nghim nguyn phng php Biu th mt n theo n cn li bng phngphp a v c s.

    Bi 2. Tm di cnh mt tam gic vung sao cho tch hai cnhhuyn gp ba ln chu vi tam gic .

    Bi 3. Gii phng trnh nghim nguyn x y + 2xy = 6Bi 4. Gii phng trnh nghim nguyn 2x+ 5y + 2xy = 8

    Bi 5. (Thi HSG lp 9 tnh Qung Ngi nm 2011-2012) Gii phngtrnh nghim nguyn 6x+ 5y + 18 = 2xy

    Bi 6. Tm nghim nguyn (xy 7)2 = x2 + y2

    Bi 7. Tm x, y Z tha mn 2x2 2xy = 5x y 19.Bi 8. Tm nghim nguyn ca phng trnh x2+6xy+8y2+3x+6y =

    2.

    Bi 9. Tm nghim nguyn dng ca phng trnh x3 y3 = xy+ 61Bi 10. Tm nghim nguyn ca phng trnh 4x2y2 = 22 + x(1 + x) +

    y(1 + y)

    Bi 11. Gii phng trnh nghim nguyn x(x+ 1)(x+ 7)(x+ 8) = y2.

    Chuyn S hc Din n Ton hc

  • 66 4.1. Xt tnh chia ht

    Bi 12. Tm nghim nguyn dng ca phng trnh 6x3 xy(11x +3y) + 2y3 = 6 (Tp ch TTT2 s 106).

    Bi 13. Tm nghim nguyn dng ca phng trnh x(x+2y)3y(y+2x)3 = 27 (tp ch THTT s 398).

    Bi 14. Tm nghim nguyn ca phng trnh

    9x2 + 16x+ 96 = 3x16y 24.

    Bi 15. Tm nghim nguyn dng ca phng trnh

    2 +

    x+

    1

    2+

    x+

    1

    4= y

    .

    Bi 16. Tm s nguyn x x2 4x 52 l s chnh phng.

    Bi 17. Gii phng trnh nghim nguyn x2 + 2y2 + 3xy 2x y = 6.

    Bi 18. Gii phng trnh nghim nguyn x2 + 3xy y2 + 2x 3y = 5.

    Bi 19. Gii phng trnh nghim nguyn 2x2 + 3y2 + xy 3x 3 = y.

    Bi 20. (Tuyn sinh vo lp 10 THPT chuyn trng KHTN H Ninm hc 2012-2013) Tm tt c cc cp s nguyn x, y thamn ng thc (x+ y + 1)(xy + x+ y) = 5 + 2(x+ y).

    Bi 21. Gii phng trnh nghim nguyn x42y4x2y24x27y25 = 0.

    (Thi HSG lp 9 tnh Hng Yn nm 2011-2012)

    Bi 22. (Romanian 1999) Chng minh rng phng trnh sau khng cnghim nguyn

    x5 x4y 13x3y2 + 13x2y3 + 36xy4 36y5 = 1937

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 67

    4.1.3 Biu th mt n theo n cn li ri s dng tnh chiaht

    V d 4.10. Tm nghim nguyn ca phng trnh

    2x xy + 3 = 0 (4.17)Li gii. Nhn xt. phng trnh ny ta khng th p dng cc cch bit, vy ta phi lm sao? Ch hn mt xu na ta thy c thbiu din y theo x c ri vn dng kin thc tm gi tr nguyn lp 8 tm nghim nguyn ca phng trnh, th lm theo tng xem sao.

    (4.17) xy = 2x+ 3Nu x = 0 th phng trnh (4.17) cho v nghim nguyn y.Nu x 6= 0 th

    (4.17) y = 2x+ 3x

    = 2 +3

    x

    Nh vy mun y nguyn th ta cn3

    xnguyn hay ni cch khc x l

    c ca 3. Vi mi gi tr nguyn x ta tm c mt gi tr y nguyn.T , ta c b nghim ca (4.17) l

    (x; y) = (3; 1); (1;1); (1; 5); (3; 3)

    V d 4.11 (Thi HSG lp 9 Qung Ngi nm 2011-2012). Tm cc snguyn dng x, y sao cho

    6x+ 5y + 18 = 2xy (4.18)

    Nhn xt. Hng phn tch v nh hng li gii. xc nh cphng php ca dng ny th by gi ta s biu din n x theo y.

    Khng kh vit thnh x =5y 18

    6 2y . Ta dng nh nhn thy biuthc ny rt kh phn tch nh biu thc v d u. Tuy nhin, nu k s thy bn mu l 2y v t l 5y, do ta mnh dn nhn 2vo t xut hin 2y ging nh mu.

    Chuyn S hc Din n Ton hc

  • 68 4.1. Xt tnh chia ht

    Li gii. Ta c

    (4.18) x = 5y 186 2y

    2x = 10y 366 2y

    2x = 66 + 5(6 2y)6 2y =

    666 2y + 5

    2x = 333 y + 5

    Nh vy mun x l s nguyn dng th 3 y l phi l c ca 33.Hay 3 y {1;3;11,33}. Li rng v y 1 nn 3 y 2.Do ch c th 3 y {1;3;11;33}. Ta c bng sau:

    3 y 1 1 3 11 33y 2 4 6 14 36

    x 14 19 8 4 3

    Th li thy cc cp (x; y) nguyn dng tha mn (4.18) l (x; y) =(19; 4), (8; 6), (4; 14), (3; 36).

    Nhn xt. Bi ny ta cng c th s dng phng php a v phngtrnh c s. Cng xin ch vi bn rng li gii trn th ta nhn2 x bin i, do phi c mt bc th li coi gi tr x, y tmc c tha mn (4.18) hay khng ri mi c th kt lun.

    Bi tp ngh

    Bi 1. Gii phng trnh nghim nguyn x2 xy = 6x 5y 8.

    Bi 2. Gii phng trnh nghim nguyn x2 + x+ 1 = 2xy + y.

    Bi 3. Gii phng trnh nghim nguyn x3 x2y + 3x 2y 5 = 0.

    Bi 4. (Vo 10 chuyn THPT HKHTN H Ni nm 2001-2002) Tmgi tr x, y nguyn tha mn ng thc (y 2)x2 + 1 = y2.

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 69

    Bi 5. (Vo 10 chuyn THPT HKHTN H Ni nm 2000-2001) Tmcp s nguyn (x, y) tha mn ng thc y(x 1) = x2 + 2.

    Bi 6. Tm s nh nht trong cc s nguyn dng l bi ca 2007 vc 4 ch s cui cng l 2008.

    Bi 7. Tm nghim nguyn ca phng trnh 5x 3y = 2xy 11.

    4.1.4 Xt s d tng v

    C s phng php. c ngay tiu phng php th chc bn shiu ngay phng php ny ni n vic xt s d tng v cho cngmt s. Vy, ti sao li phi xt v xt nh vy c li ch g trong "cngcuc" gii ton? Hy cng tm hiu qua v d u sau:

    V d 4.12. Tm nghim nguyn ca phng trnh

    x2 + y2 = 2011 (4.19)

    Li gii. Ta c x2; y2 chia 4 c th d 0 hoc 1 nn tng chng chia 4ch c th d 0; 1 hoc 2. Mt khc 2011 chia 4 da 3 nn phng trnh(4.19) v nghim nguyn.

    Nhn xt. Qua v d u ny th ta thy r s d khi chia cho 4 cahai s khc nhau th phng trnh v nghim. Do ta li cng hiuthm mc ch ca phng php ny. Bt m thm t na th phngphp ny ch yu dng cho cc phng trnh khng c nghim nguyn.Cho nn, nu bn bt gp mt phng trnh bt k m bn khng thtm ra c nghim cho phng trnh , th hy ngh n phngphp ny u tin. Cn by gi ta tip tc n vi v d sau:

    V d 4.13 (Balkan MO 1998). Tm nghim nguyn ca phng trnh

    x2 = y5 4 (4.20)

    Li gii. Ta c: x2 0; 1; 3; 4; 5; 9 (mod 11). Trong khi y5 4 6; 7; 8 (mod 11): v l. Vy phng trnh (4.20) v nghim nguyn.

    Chuyn S hc Din n Ton hc

  • 70 4.1. Xt tnh chia ht

    Nhn xt. Mt cu hi na li le ln trong u ta: Lm th no li cth tm c con s 11 m xt ng d c nh? p n ca cuhi ny cng chnh l ci ct li bn vn dng phng php ny, v cng l nhng kinh nghim sau:

    1. i vi phng trnh nghim nguyn c s tham gia ca cc bnhphng th ta thng xt ng d vi 3, 4, 5, 8. C th l:

    a2 0, 1 (mod 3)a2 0, 1 (mod 4)a2 0, 1, 4 (mod 5)a2 0, 1, 4 (mod 8)

    2. i vi cc phng trnh nghim nguyn c s tham gia ca ccs lp phng th ta thng xt ng d vi 9, v x3 0; 1; 8(mod 9) v ng d vi 7, v x3 0, 1, 6 (mod 7).

    3. i vi phng trnh nghim nguyn c s tham gia ca ccly tha bc 4 th ta thng xt ng d vi 8, nh: z4 0, 1(mod 8).

    4. Mt vn cui cng l nh l Fermat: i vi phng trnhnghim nguyn c s tham gia ca cc ly tha c s m l mts nguyn t hay l mt s m khi cng 1 vo s ta c mts nguyn t th ta thng s dng nh l nh Fermat xtng d.

    Trn y l mt s kinh nghim bn thn, cn nu cc bn mun vndng c phng php xt s d ny, yu cu hy ghi nh kinh nghimtrn v tm cch chng minh n. Ngoi ra, nu bn mun m rng tmhiu bit hn na, bn c th tm cc ng d vi ly tha khc nhau(chng hn qua v d 2 ta rt ra c moun 11 cho ly tha bchai, bc nm). Cn by gi, hy th xem kinh nghim trn c hiu qukhng nh!

    V d 4.14 (Bi ton trong tun - diendantoanhoc.net). Chng minhrng phng trnh sau khng c nghim nguyn

    x10 + y10 = z10 + 199

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 71

    Nhn xt. Thng thng cc bi ton khi t cu hi phng trnhc nghim hay khng th thng c cu tr li l khng. Do chng minh phng trnh trn khng c nghim, th ta s tm mt cons sao cho khi chia VT v VP cho con s ny th c hai s d khcnhau.

    Nh vy, cng vic by gi ca ta l tm con s . n s m 10th s khin ta lin tng con s 11 l s nguyn t. Nh vy li giica ta s p dng nh l Fermat nh cho s 11 chng minh hai vphng trnh chia cho 11 khng cng s d.

    Li gii. p dng nh l Fermat nh th

    x10 0, 1 (mod 11)y10 0, 1 (mod 11)z10 0, 1 (mod 11)

    .

    Do x10 + y10 z10 0, 1, 2, 10 (mod 11) m 199 8 (mod 11) nnphng trnh v nghim nguyn.

    V d 4.15 ( thi chn HSG ton quc gia nm 2003 - Bng B).H phng trnh sau c tn ti nghim nguyn hay khng:

    x2 + y2 = (x+ 1)2 + u2 = (x+ 2)2 + v2 = (x+ 3)2 + t2 (4.21)

    Nhn xt. Ta d on phng trnh trn cng s v nghim. Do cn tm mt s v khi chia c 5 v c cc s d khc nhau. biton ny c bnh phng nn ta ngh ti vic s dng cc tnh chtnh: a2 0, 1 (mod 3), a2 0, 1 (mod 4), a2 0, 1, 4 (mod 5), a2 0, 1, 4 (mod 8). bi ton ny, ta s chn 8. By gi ch cn xt tnhd khi chia cho 8.

    Li gii. Gi s phng trnh (4.21) c nghim nguyn (x0, y0, u0, v0, t0),tc l:

    x20 + y20 = (x0 + 1)

    2 + u20 = (x0 + 2)2 + v20 = (x0 + 3)

    2 + t20 (4.22)

    Vi a Z th a2 0, 1, 4 (mod 8). Ta xt cc kh nng sau:

    Chuyn S hc Din n Ton hc

  • 72 4.1. Xt tnh chia ht

    1. Nu x0 0 (mod 4) th x20 + y20 0, 1, 4 (mod 8). Vx0 + 1 1 (mod 8) (x0 + 1)2 1 (mod 8)

    (x0 + 1)2 + u20 1, 2, 5 (mod 8)x0 + 2 2 (mod 4) (x0 + 2)2 4 (mod 8)

    (x0 + 2)2 + v20 0, 4, 5 (mod 8)x0 + 3 3 (mod 4) (x0 + 3)2 1 (mod 8)

    (x0 + 3)2 + t20 1, 2, 5 (mod 8)

    Nhn thy {0, 1, 4} {1, 2, 5} {0, 4, 5} {1, 2, 5} = 0 nn do phng trnh khng c nghim nguyn vi x 0 (mod 4).

    2. Tng t vi x0 1 (mod 4), x0 2 (mod 4) v x0 3 (mod 4)ta cng thc hin tng t v cng cho kt qu phng trnhkhng c nghim nguyn.

    Vy phng trnh (4.21) cho khng c nghim nguyn.

    Nhn xt. V d 4 ta c th tng qut ln:

    V d 4.16. Tm s nguyn dng n ln nht sao cho h phng trnh

    (x+ 1)2 + y21 = (x+ 2)2 + y22 = . . . = (x+ n)

    2 + y2n

    c nghim nguyn. 4y cng chnh l thi chn i tuyn HSG quc gia ton nm 2003- Bng A. Li gii xin ginh cho bn c. Cng xin ni thm mt thanhn rng, phng php xt s d tng v ny, chng ta c tngchng nh n gin, nhng thc cht khng phi th. Dn chng lcc v d trn, u l cc bi ton hay v kh ly t khc cuc thitrong nc v ngoi nc.

    Bi tp ngh

    Bi 1. Cho a thc f(x) c cc h s nguyn. Bit rng f(1).f(2) l sl. Chng minh rng phng trnh f(0) = 0 khng c nghimnghim nguyn.

    Din n Ton hc Chuyn S hc

  • 4.1. Xt tnh chia ht 73

    Bi 2. Tn ti hay khng nghim nguyn ca phng trnh x12 +y12 +z12 = 2

    (372012 + 20141995

    ).

    Bi 3. Gii phng trnh nghim nguyn 312x + 122x + 19972x = y2.

    Bi 4. Gii phng trnh nghim nguyn dng 7z = 2x 3y 1

    Bi 5. Gii phng trnh nghim nguyn dng 2x 3y = 1 + 5z

    Bi 6. Gii phng trnh nghim t nhin 19x+5y +1890 = 1975430

    +1993.

    Bi 7. Gii phng trnh nghim nguyn x3 + y3 + z3 = 1012

    Bi 8. (Tuyn sinh vo lp 10 chuyn Trn Ph, Hi Phng nm hc2012-2013) x4 + y4 + z4 = 2012

    Bi 9. |x y|+ |y z|+ |z x| = 10n 19

    vi mi n N

    Bi 10. Tm nghim nguyn ca phng trnh (2x + 1)(2x + 2)(2x +3)(2x + 4) 5y = 11879

    Bi 11. Tm nghim nguyn ca phng trnh x2 +(x+1)2 +(x+2)2 =y2.

    Bi 12. (Tuyn sinh vo THPT chuyn HKHTN H Ni nm 2011-2012) Chng minh rng khng tn ti b ba s nguyn (x; y; z)tha mn x4 + y4 = 7z4 + 5.

    Bi 13. Gii phng trnh nghim nguyn x41+x42+ = x413+20122015.

    Bi 14. Cho p l s nguyn t l. Chng minh rng phng trnh xp +yp = p [(p 1)!]p khng c nghim nguyn

    Bi 15. Tm nghim nguyn ca phng trnh x2012 y2010 = 7.

    Bi 16. Chng minh rng khng tn ti s nguyn x, y tha mn x5 +y5 + 1 = (x+ 2)5 + (y 3)5.

    Chuyn S hc Din n Ton hc

  • 74 4.2. S dng bt ng thc

    4.2 S dng bt ng thc

    4.2.1 Sp th t cc n

    V d 4.17. Gii phng trnh nghim nguyn dng sau

    1

    x+

    1

    y+

    1

    z= 1 (4.23)

    Li gii. Khng mt tnh tng qut, ta c th gi s

    1 x y z 1x

    +1

    y+

    1

    z= 1 3

    x x 3

    Vi x = 1 th (4.23) khng c nghim nguyn dng.

    Vi x = 2 th 12

    +1

    y+

    1

    z= 1 1

    y+

    1

    z=

    1

    2 2y y 4 Mt

    khc, y x = 2 y {2; 3; 4}. Ta th ln lt cc gi tr ca y Vi y = 2 th (4.23) v nghim nguyn. Vi y = 3 th z = 6. Vi y = 4 th z = 4.

    Vi x = 3, ta c 13

    +1

    y+

    1

    z= 1 1

    y+

    1

    z=

    2

    3 2y y 3 Mt

    khc, do y x = 3 y = 3 z = 3Vy nghim nguyn (x; y; z) ca (4.23) l hon v ca cc b (2; 3; 6);(2; 4; 4); (3; 3; 3).

    Nhn xt. Phng php ny c s dng ch sp th t cc n1 x y z ri gii hn nghim gii.Ta ch s dng phng php sp th t cc n khi vai tr cc n lbnh ng vi nhau. D khi vn dng phng php ny cc bn cnch trnh nhm ln. C th, ta s n vi v d sau:

    V d 4.18. Gii phng trnh nghim nguyn dng

    x+ y + 1 = xyz (4.24)

    Din n Ton hc Chuyn S hc

  • 4.2. S dng bt ng thc 75

    Li gii (Li gii sai). Khng mt tnh tng qut, gi s 1 x y z. Khi x+y+1 3z hay xyz 3z, suy ra xy 3. M z y x 1nn x = y = z = 1.

    Nhn xt. Ci li sai li gii ny l do x, y, z khng bnh ng, nnkhng th sp th t cc n nh trn. Sau y l li gii ng:Li gii. Khng mt tnh tng qut, gi s 1 x y. Ta xt trnghp:

    Nu x = y th

    (4.24) 2y + 1 = y2z y(z 2) = 1

    {y = 1

    yz 2 = 1

    {y = 1

    z = 3

    Nu x < y th t (4.24) suy ra 2y + 1 > xyz. 2y xyz xz 2 xz {1; 2}. Vi xz = 1 x = z = 1, thay vo (4.24) suy ra y + 2 = y

    (v nghim).

    Vi xz = 2 {x = 1

    z = 2hoc

    {x = 2

    z = 1. T y ta tm

    c nghim x = 1, y = 2, z = 2 hoc x = 1, y = 3, z = 1.

    Vy phng trnh c nghim nguyn dng l (1; 1; 3), (1; 2; 2), (2; 1; 2),(2; 3; 1), (3; 2; 1).

    Nhn xt. By gi bn hiu v cch sp xp cc n nh th no.Nhng ti sao bi ny li xt x = y v x < y m li khng i vo phn

    Chuyn S hc Din n Ton hc

  • 76 4.2. S dng bt ng thc

    tch lun nh bi trc. Nu bn rng nu khng phn chia thnhhai trng hp nhu trn th phng trnh (4.24) s thnh 2y+1 y2z,rt kh tip tc phn tch ra nghim. Do vic xt nhu trn lhp l.

    Bi tp ngh

    Bi 1. Gii phng trnh nghim nguyn dng 2(x+y+z)+9 = 3xyz.

    Bi 2. Gii phng trnh nghim nguyn dng xyz = 3(x+ y + z).

    Bi 3. Gii phng trnh nghim nguyn dng 5(x+y+z+ t)+10 =2xyzt

    Bi 4. Gii phng trnh nghim nguyn dng x! + y! = (x+ y)!(K hiu x! l tch cc s t nhin lin tip t 1 n x).

    Bi 5. Tm nghim nguyn dng ca phng trnh x3 +7y = y3 +7x.

    Bi 6. Tm nghim nguyn dng ca phng trnh x1+x2+ +x12 =x1x2 x12.

    Bi 7. Tm tt c cc nghim nguyn dng ca phng trnhx

    y2z2+

    y

    z2x2+

    z

    x2y2= t.

    Bi 8. Tm nghim nguyn dng ca phng trnh x! + y! + z! = u!.

    4.2.2 S dng bt ng thc

    Nhn xt. gii phng trnh ny, ta thng s dng cc bt ngthc quen thuc nh gi mt v ca phng trnh khng nh hn(hoc khng ln hn) v cn li. Mun cho hai v bng nhau th btng thc phi tr thnh ng thc.C th, ta c mt s bt ng thc c bn thng dng:

    1. Bt ng thc Cauchy (hay cn gi l bt ng thc AM-GM):Nu a1, a2, , an l cc s thc khng m th

    a1 + a2 + + ann

    na1a2 an

    Din n Ton hc Chuyn S hc

  • 4.2. S dng bt ng thc 77

    Du ng thc xy ra khi v ch khi a1 = a2 = = an.2. Bt ng thc Bunhiacopxki (hay cn c gi l bt ng thc

    Cauchy - Bunyakovsky - Schwarz): Vi hai b s thc bt k(a1, a2, , an) v (b1, b2, , bn), ta c(

    a21 + a22 + + a2n

    ) (b21 + b

    22 + + b2n

    ) (a1b1 + a2b2 + + anbn)2 .

    ng thc xy ra khi v ch khi tn ti s thc k sao cho ai = kbivi mi i = 1, 2, , n.

    3. Bt ng thc Trebusep (hay cn vit l bt ng thc Chebyshev):Cho dy hu hn cc s thc c sp theo th t a1 a2 an v b1 b2 bn. Khi ta c:

    n(a1b1 +a2b2 + +anbn) (a1 +a2 + +an)(b1 +b2 + +bn)

    Du ng thc xy ra khi v ch khi[a1 = a2 = = anb1 = b2 = = bn .

    By gi ta s cng xem xt mt s v d sau:

    V d 4.19. Gii phng trnh nghim nguyn dng sau:

    x6 + z3 15x2z = 3x2y2z (y2 + 5)3 (4.25)

    Li gii. Nhn xt. phng trnh ny khi mi nhn vo hn a scc bn s c phn ri, khng xc nh c phng php lm, khngvn dng c cc phng php hc. Tuy nhin nu k mtx th ta thy x6 = (x2)3 iu ny c g c bit? Ta thy (x2)3, z3 v(y2 + 5)3 u c cng bc ba v bi cho nguyn dng nn tangh ngay n mt Bt ng thc kinh in: Bt ng thc Cauchyhay cn gi l bt ng thc AM-GM.Ta gii nh sau

    (4.25) (x2)3 + (y2 + 5)3 + z3 = 3x2z(y2 + 5)

    Chuyn S hc Din n Ton hc

  • 78 4.2. S dng bt ng thc

    p dng Bt ng thc AM-GM cho b ba s dng (x2)3, z3 v (y2 +5)3 ta c:

    (x2)3+(y2+5)3+z3 3 3

    (x2)3.(y2 + 5)3.z3 = 3x2z(y2+5) = V P (4.25)

    Du bng ch xy ra khi x2 = y2 + 5 = 5.Mt khc ta c:

    x2 = y2 + 5 (x y)(x+ y) = 5y l mt dng phng trnh nghim nguyn quen thuc ta hc,ti tin chc cc bn u c th d dng gii phng trnh trn, v tx; y trn ta c th tm c z mt cch d dng.p s: Nghim nguyn ca phng trnh (4.25) l (x; y; z) = (3; 2; 9).

    V d 4.20. Tm nghim nguyn ca phng trnh

    (x+ y + z)2 = 3(x2 + y2 + 1)

    Li gii. p dng bt ng thc Bunyakovsky cho hai b s (x, y, 1)v (1, 1, 1) ta c

    (x+ y + 1)2 (12 + 12 + 12)(x2 + y2 + 1) = 3(x2 + y2 + 1)ng thc xy ra khi v ch khi x = y = 1.Vy phng trnh c nghim nguyn l (x, y) = (1, 1).

    Nhn xt. Cc bi Ton v phng trnh nghim nguyn m gii bngcch s dng Bt ng thc rt t dung v rt d b l dng nungi ra khng kho lo. Tuy nhin, ta vn phi thnh tho phngphp ny khng c xem thng n trnh nhng sai lm ng tickhng th sa c.

    Bi tp ngh

    Bi 1. Tm nghim nguyn dng x, y tha mn phng trnh (x2 +1)(x2 + y2) = 4x2y

    Bi 2. Tm nghim nguyn ca phng trnhxy

    z+yz

    x+zx

    y= 3.

    Din n Ton hc Chuyn S hc

  • 4.2. S dng bt ng thc 79

    Bi 3. ( thi tuyn sinh vo i hc Vinh) Tm nghim nguyn caphng trnh

    (x2 + 1)(y2 + 4)(z2 + 9) = 48xyz

    Bi 4. Gii phng trnh nghim nguyn

    4x 2 +

    1y 1 +

    25z 5 = 16

    x 2

    y 1z 5

    Bi 5. Tm nghim nguyn ca h phng trnh

    x2 + z2 = 9

    y2 + t2 = 16

    xt+ yz = 12

    Bi 6. Tm nghim nguyn dng ca phng trnh x3+y36xy+8 =0.

    Bi 7. Tm nghim nguyn ca h phng trnh

    {xy + yz + zx = 12

    x4 + y4 + z4 = 48.

    Bi 8. Cho phng trnh x3 + y3 + z3 = nxyz.

    a, Chng minh rng khi m = 1 v m = 2 th phng trnhkhng c nghim nguyn dng.

    b, Gii phng trnh nghim nguyn dng khi m = 3.

    Bi 9. Gii phng trnh nghim nguyn dng (x3+y3)+4(x2+y2)+4(x+ y) = 16xy.

    Bi 10. Gii phng trnh nghim nguyn dng

    3(x4 + y4 + x2 + y2 + 2) = 2(x2 x+ 1)(y2 y + 1)

    Bi 11. Gii phng trnh nghim nguyn dng vi x, y, z l cc s imt khc nhau

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

    Chuyn S hc Din n Ton hc

  • 80 4.2. S dng bt ng thc

    4.2.3 Ch ra nghim

    Nhn xt. Phng php ny dnh cho nhng bi ton gii phngtrnh nghim nguyn khi m ta tm c chnh xc nghim nguynv mun chng minh phng trnh ch c duy nht cc nghim nguyn m thi.

    V d 4.21. Tm nghim nguyn dng ca phng trnh

    2x + 3x = 5x (4.26)

    Li gii. Chia 2 v ca phng trnh (4.26) cho s dng 5x, ta c:

    (4.26)(

    2

    5

    )x+

    (3

    5

    )x= 1

    Vi x = 1 th ta c2

    5+

    3

    5= 1:ng nn x = 1 l 1 nghim ca

    (4.26).Vi x > 1 th (

    2

    5

    )x+

    (3

    5

    )x>

    2

    5+

    3

    5= 1

    Do mi gi tr x > 1 u khng l nghim ca (4.26). Vy nghimnguyn dng ca (4.26) l x = 1.

    Nhn xt. v d trn, ta d nhn thy x = 1 l nghim duy nhtca phng trnh nn ch cn chng minh vi x > 1 th phng trnhv nghim. Ngoi ra, t bi ton trn ta c th m rng thnh hai biton mi:

    Bi ton 4.1. Tm nghim nguyn dng ca phng trnh

    (

    3)x + (

    4)x = (

    5)x

    Bng cch gii tng t ta cng tm c nghim duy nht ca phngtrnh trn l x = 4.

    Din n Ton hc Chuyn S hc

  • 4.2. S dng bt ng thc 81

    Bi ton 4.2. Tm nghim nguyn dng ca phng trnh

    3x + 4y = 5z

    Bi ton 4.2 r rng c nng cao ln r rt, nhng li gii ca biton ny l s dng phng php xt s d hc. Sau y l li giirt p ca khanh3570883 hin l iu hnh vin THPT ca VMF:Li gii. Xt theo module 3 ta c:

    5z (1)z (mod 3) 4y (1)z (mod 3) z = 2h (h N) (5h 2y)(5h + 2y) = 3x

    Do 5h 2y v 5h + 2y khng ng thi chia ht cho 3 nn 5h + 2y = 3xv 5h 2y = 1.Ta c 5h+2y (1)h+(1)y = 0 (mod 3) v 5h2y (1)h(1)y =1 (mod 3) h l v y chn.Nu y > 2 th 5h + 2y 1 (mod 4) 3x 1 (mod 4) 3x 1(mod 8).Mt khc 5 5h + 2y (mod 8) 5 3x (mod 8) 5 1 (mod 8):v l.Do y = 2. Suy ra x = y = z = 2.

    Phng php ny thng hay s dng cho cc phng trnh c n sm v cc phng trnh c nghim nh.

    4.2.4 S dng ca phng trnh bc 2

    Nhn xt. Vit phng trnh di dng phng trnh bc hai i vimt n, dng iu kin. 0 hoc l s chnh phng. Ta s tytrng hp chn mt trong hai cch xt vo vic gii ton.

    V d 4.22. Gii phng trnh nghim nguyn

    3x2 + (3y 1)x+ 3y2 8y = 0 (4.27)

    Chuyn S hc Din n Ton hc

  • 82 4.2. S dng bt ng thc

    Li gii. Coi (4.27) l phng trnh bc 2 n x. Xt x = 27y2 +9y + 1. (4.27) c nghim x th

    x 0 27y2 + 9y + 1 0 0, 01 y 3, 3 y {0; 1; 2; 3}

    Nu y = 0 3x2 x = 0 x = 0 v x Z.Nu y = 1 3x2 + 2x 5 = 0 x = 1 v x Z.Nu y = 2 hoc y = 3 th khng tm c x nguyn nn loi.Vy (4.27) c nghim nguyn (x; y) = (0; 0); (1; 1).

    V d 4.23. Gii phng trnh nghim nguyn

    3x2 y2 2xy 2x 2y + 8 = 0 (4.28)

    Li gii. Ta c

    (4.28) y2 + 2(x+ 1)y (3x2 2x+ 8) = 0y = (x+ 1)2 + 3x2 2x+ 8 = 4x2 + 9

    (4.28) c nghim th y = 4x2+9 l s chnh phng. t 4x2+9 =k2 vi k N, ta a v phng trnh c s v tm c x {2; 0;2}. Vi x = 2 ta c y2 + 6y 16 = 0 nn y {8; 2}. Vi x = 0 th y2 + 2y 8 = 0 nn y {4; 2}. Vi x = 2 th y2 2y 24 = 0 nn y {6; 4}.

    Kt lun. Vy phng trnh (4.28) c nghim (x; y) l (2;8), (2; 2),(0;4), (0; 2), (2; 6), (2;4). Nhn xt. Hai bi ton trn u c th s dng phng php a vphng trnh c s gii.

    Bi tp ngh

    Bi 1. Tm cc phng php trc (nht l phng php a vphng trnh c s) cc bi ton gii bng phng phpny.

    Din n Ton hc Chuyn S hc

  • 4.2. S dng bt ng thc 83

    Bi 2. Tm nghim nguyn ca phng trnh x+ xy + y = x2 + y2.

    Bi 3. Gii phng trnh nghim nguyn 10x2 + 5y2 + 38 12xy +16y 36x = 0.

    Bi 4. Tm nghim nguyn phng trnh 9x2 +x2 + 4y2 + 34 12xy+20y 36x = 0.

    Bi 5. Tm nghim nguyn dng ca x+ 2y2 + 3xy + 3x+ 5y = 14.

    Bi 6. Tm nghim nguyn phng trnh x2xy6y2+2x6y10 = 0.

    Bi 7. Tm nghim nguyn ca phng trnh x2+2y62+3xy+3x+5y =15.

    Bi 8. Tm nghim nguyn ca phng trnh 2x2+6y2+7xyxy =25.

    Bi 9. Tm nghim nguyn ca phng trnh 9x2 10y2 9xy+ 3x5y = 9.

    Bi 10. Tm nghim nguyn ca phng trnh 12x2+6xy+3y2 = 28(x+y).

    (Thi vo lp 10 chuyn, HKHTN-HQGHN nm 1994)

    Bi 11. Tm nghim nguyn ca phng trnh 3(x2 +xy+y2) = x+8y.

    Bi 12. Tm nghim nguyn ca phng trnh 7(x2 +xy+y2) = 39(x+y).

    Bi 13. Tm nghim nguyn ca phng trnh 2x2 + y2 + 3xy + 3x +2y + 2 = 0.

    Bi 14. Tm nghim nguyn ca phng trnh x2+2y2+3xyxy+3 =0.

    Bi 15. Tm nghim nguyn ca phng trnh 3x2+4y2+12x+3y+5 =0.

    Chuyn S hc Din n Ton hc

  • 84 4.2. S dng bt ng thc

    4.2.5 Phng php kp

    Nhn xt. S dng tnh cht ly tha cng bc ca s nguyn lintip hoc tch cc s nguyn lin tip ... a phng trnh nghimnguyn cn gii v dng phng trnh khc t n hn v quen thuchn. Phng php ny cn c cch gi khc l phng php kh n.Ta thng vn dng cc nhn xt sau:

    1. Xn Y n (X + a)n (a N) th Y n = (X + a i)n vii = 0; 1; 2; ; a.V d vi n = 2 th:

    Khng tn ti x Z a2 < x2 < (a+ 1)2 vi a Z. Nu a2 < x2 < (a+ 2)2 th x2 = (a+ 1)2

    2. X(X + 1) (X + n) Y (Y + 1) (Y + n) (X + a)(X + a+1) (X + a + n) th Y (Y + 1) (Y + n) = (X + i)(X + 1 +i) (X + a+ i) vi i = 0; 1; 2; ; a.V d:

    Khng tn ti b Z a(a+ 1) < b(b+ 1) < (a+ 1)(a+ 2)vi a Z. Vi a(a + 1) < b(b + 1) < (b + 2)(b + 3) th b(b + 1) =

    (b+ 2)(b+ 3).

    V d 4.24. Tm cc s nguyn dng x biu thc sau l s chnhphng

    A = x4 + 2x3 + 2x2 + x+ 3 (4.29)

    Li gii. V A l s chnh phng nn ta c th t

    A = x4 + 2x3 + 2x2 + x+ 3 = y2(y N)Ta thy

    y2 = (x4 + 2x3 + x2) + x2 + x+ 3

    = (x2 + x)2 +

    (x+

    1

    2

    )2+

    11

    4> (x2 + x)2

    y2 > (x2 + x)2, (i)

    Din n Ton hc Chuyn S hc

  • 4.2. S dng bt ng thc 85

    Nu x = 1 A = 9: l s chnh phng nn tha .Nu x > 1 th xt hiu

    (x2+x+1)2y2 = x2+x2 = (x+2)(x1) > 0 y2 < (x2+x+1)2, (ii)

    T (i) v (ii), ta c

    (x2 + x)2 < y2 < (x2 + x+ 1)2

    Suy ra, khng tn ti y N y2 = A khi x > 1.Vy x = 1 l gi tr cn tm.

    V d 4.25. Gii phng trnh nghim nguyn

    x4 + x2 + 4 = y2 y (4.30)

    Li gii. Ta c nh gi sau

    x2(x2 + 1) < x4 + x2 + 4 < (x2 + 2)(x2 + 3) (4.31)

    T (4.30) v (4.31) suy ra

    x2(x2 + 1) < y(y 1) < (x2 + 2)(x2 + 3). (4.32)

    V x, y, z nguyn nn t (4.32) suy ra

    y(y 1) = (x2 + 1)(x2 + 2) (4.33)

    T (4.30) v (4.33) th

    x4 + x2 + 4 = (x2 + 1)(x2 + 2) x2 = 1 x = 1

    T y d tm c y = 1 hoc y = 3.Vy pt cho c bn cp nghim

    (x, y) = {(1,2), (1, 3), (1,2), (1, 3)}

    Bi tp nghTm nghim nguyn ca cc phng trnh sau:

    Chuyn S hc Din n Ton hc

  • 86 4.3. Nguyn tc cc hn, li v hn

    Bi 1. x4 + x2 + 1 = y2

    Bi 2. 3(x4 + y4 + x2 + y2 + 2) = 2(x2 x+ 1)(y2 y + 1)Bi 3. 2x4 + 3x2 + 1 y2 = 0Bi 4. x2 + (x+ y)2 = (x+ 9)2

    Bi 5. y3 x3 = 2x+ 1Bi 6. x4 y4 + z4 + 2x2z2 + 3x2 + 4z2 + 1 = 0Bi 7. x3 y3 2y2 3y 1 = 0Bi 8. x4 + (x+ 1)4 = y2 + (y + 1)2

    Bi 9. 9x 3x = y4 + 2y3 + y2 + 2yBi 10. x4 + x2 y2 + y + 10 = 0Bi 11. x6 4y3 4y4 = 2 + 3y + 6y2

    Bi 12. (x 2)4 x4 = y3

    Bi 13. x3 + 8x2 6x+ 8 = y3

    4.3 Nguyn tc cc hn, li v hn

    4.3.1 Li v hn

    V d 4.26 (Korea 1996). Gii phng trnh nghim nguyn sau:

    x2 + y2 + z2 = 2xyz (4.34)

    Li gii. Gi s (x0; y0; z0) l b nghim nguyn ca (4.34) th ta c

    x20 + y20 + z

    20 = 2x0y0z0

    R rng VT (4.34) chn do VP (4.34) chn nn c 2 trng hp xyra:

    Din n Ton hc Chuyn S hc

  • 4.3. Nguyn tc cc hn, li v hn 87

    Trng hp 1. Trong x0; y0; z0, c 2 s l, 1 s chn. Khng mt tnhtng qut, gi s x0; y0 l cn z0 chn. Xt theo module 4 th

    V T (4.34) 2 (mod 4), V P (4.34) 0 (mod 4) : v l!Vy trng hp ny khng xy ra. Trng hp 2. x0; y0; z0 u chn. t x0 = 2x1; y0 = 2y1; z0 = 2z1vi x1; y1; z1 Z. Thay vo (4.34) v rt gn, ta thu c

    x21 + y21 + z

    21 = 4x1y1z1

    Lp lun nh trn, ta li c x1; y1; z1 u chn.

    Qu trnh din ra tip tc nn x0; y0; z0...2k vi k t nhin ty .

    iu ch xy ra khi v ch khi x0 = y0 = z0 = 0.

    4.3.2 Nguyn tc cc hn

    nh ngha 4.1 Nguyn tc cc hn hay cn gi l nguyn l khi ucc tr. V mt hnh thc th phng php ny khc vi phng phpli v hn nhng cch s dng u nh nhau u chng minh phngtrnh ch c nghim tm thng (nghim tm thng l nghim bng0). Phng php gii nh sau:

    Gi