So hoc VNMath.pdf

7/28/2019 So hoc VNMath.pdf

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Chuyn

S HCCh 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
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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 tbc 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 bi

b 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 D ng (duongld) , Nguyn Trn Huy (yeutoan11), NguynTrung Hiu (nguyentrunghieua), Phm Quang Ton (Phm Quang

Ton), Trn Nguyn Thit Qun (L Lawliet), Trn Trung Kin (Is-pectorgadget), Nguyn nh Tng (tungc3sp)... cng s gp sc

i
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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 nhtrao i cng cc bn nhiu dng bi tp hay v kh tcp 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 trong

chng 2. Chng 3 ni su hn v Cc bi ton chia ht. Phngtrnh nghim nguyn, Phng trnh ng dc phc ha trongcc chng 4 v 5. H thng dv nh l Thng dTrung 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
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Mc lc

i Li gii thiu

1Chng 1

c v Bi

1.1 c s, c s chung, c s chung ln nht 11.2 Bi s, bi s chung, bi s chung nh nht 41.3 Bi tp ngh 6

9Chng 2S Nguyn T

2.1 Mt s kin thc c bn v s nguyn t 92.2 Mt s bi ton c bn v s nguyn t 132.3 Bi tp 192.4 Ph lc: Bn nn bit 24

29Chng 3

Bi ton chia ht

3.1 L thuyt c bn 293.2 Phng php gii cc bi ton chia ht 31

57Chng 4

Phng trnh nghim nguyn

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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 955.6 ng dng nh l Euler gii phng trnh

ng d 965.7 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

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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 chkhia chia ht chod. Ta ni d chia ht a, k hiu d|a.Tp hp cc c ca a l: U(a) = {d 2 N : d|a}. 4

1
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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 snguyn 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:

U SC(a; b) = {d 2 N : (d|a) ^ (d|b)}= {d 2 N : (d 2 U(a)) ^ (d 2 U(b))}.

Tnh cht 1.2 NuU SC(a; b) = {1} tha v b nguyn tcng nhau.

nh ngha 1.3 Sd 2 N c gi lc schung ln nht ca a v b(a; b 2 Z) khi d l phn t ln nht trong tp U SC(a; b). K hiu cchung ln nht ca a v b l UCLN(a; b), (a; b) haygcd(a; b). 41.1.2 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, 8m 6= k, {m; k} 2 {1;2; . . . ; n} th ta ni cca1; a2; . . . ; an i mt nguyn t cng nhau.

c 2 U SC(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).

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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 dcui cng khc 0 trong thut ton Euclide.

1.1.4 Bi tp v d

V d 1.1. Tm (2k

1; 9k + 4), k2N.

4Li gii. Ta t d = (2k 1; 9k + 4). Theo tnh cht v c s chungta c d|2k 1 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 u

chia 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


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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 nhsau :Li gii. 987654321 = 123456789.8+9 th (987654321; 123456789) =

(123456789; 9).

123456789 = 9.1371421.

(123456789; 987654321) = 9.

V d 1.3. Ch ng minh r ng dy s An =1

2n(n + 1), n 2 N ch a

nh ng dy s v hn nh ng 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 2 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 bi

s, bi s chung, bi s chung nh nht v mt s bi tp v d minhha.



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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 khi

v ch

khi m chia h

t cho a haya l m

t

c s

c

a m. 4Nhn xt. Tp hp cc bi s ca a 6= 0 l: B(a) = {0; a; 2a; . . . ; ka}, k 2Z.

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 haya l mt c s ca m 4

nh ngha 1.6 Nu 2 tp B(a) vB(b) c phn tchung 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 d ng 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).



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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. Ch ng minh r ng [1; 2; . . . 2n] = [n + 1; n + 2; . . . ; 2n]. 4

Li 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 2 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

2N. Chng minh (A; B) = (a; b).

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



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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 2 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 2 N).

a. Chng minh rng (22n 1; 22n + 1) = 1.

b. Tm (2m 1; 2n 1).

Bi 7. Cho m, n 2 N vi (m, n) = 1. Tm (m2 + n2; m + n).

Bi 8. Cho A = 2n+3; B = 2n+1+3n+1(n2N); C = 2n+2+3n+2(n

2N). Tm (A; B) v (A; C).Bi 9. Cho su s nguyn dng a; b; a0; b0; d; d0 sao cho (a; b) = d; (a0; b0) =

d0. Chng minh rng (aa0; bb0; ab0; a0b) = dd0.

Bi 10. Chng minh rng dy s Bn =1

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

v hn nhng s nguyn t cng nhau.

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

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

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

+ 1(n 2 N) l dy snguyn t cng nhau.

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



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8 1.3. Bi tp ngh

Bi 15. Chng minh rng vi mi n 2 N, phn s 21n + 114n + 3

ti gin.

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

Bi 17. Cho a; b 2 N. Chng minh rng tn ti v s n 2 N sao cho(a + n; b + n) = 1.

Bi 18. Gi sm; n 2 N(m n) tha mn (199k1; m) = (19931; n).Chng minh rng tn ti t(t 2 N) sao cho m = 1993t.n.

Bi 19. Chng minh rng nu a; m 2 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.



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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 19

2.4 Ph lc: Bn nn bit 24

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

2.1 Mt s kin thc cbn v s nguyn t

2.1.1 nh ngha, nh l cbn

nh ngha 2.1 S nguyn t l nhng s t nhin ln hn 1, chc 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.

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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 d1 (*)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 th as nguyn t mt cch duy nht (khng k th t cc th a s).

Chng minh. * Mi s tnhin 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 l

tch cu cc tha s nguyn t.* Sphn 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 sc 2 cch phn tch n ra tha s nguyn t khc nhau:

n = p.q.r....

n = p0

.q0

.r0

....

Trong p, q, r..... v p0, q0, r0.... 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 nhtrn, 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 p0 ln lt l cc s

nguyn t nh nht trong phn tch thnht v thhai.V n l hp s nn n > p2 v n > p02. Do p 6= p ) n > p.p0



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

Xt m = n pp0 < n c phn tch ra tha s nguyn t mt cchduy nht ta thy:

p|n ) p|n pp0 hay p|mKhi phn tch ra tha s nguyn t ta c: m = n pp0 = p0p.P.Q... viP, Q 2 P ( P l tp cc s nguyn t).) pp0|n ) pp0|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 skhng 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 tnh 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 dth 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 thnht (nu trn), l da vo nh l c bn:

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

pA.

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



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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 t 2 n pA 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 2 P; xi 2 N; i = 1, n

Tnh cht 2.1 Sccc sca A tnh b ng 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 . 4

Tnh cht 2.2 Tng ccc mt sca A tnh b ng cng thc:

(A) =nY

i=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 khi

v ch khi chng c c chung ln nht (CLN) bng 1. 4Tnh cht 2.3 Hai st nhin lin tip lun nguyn tcng nhau.

Tnh cht 2.4 Hai snguyn tkhc nhau lun nguyn tcng nhau.

Tnh cht 2.5 Cc sa,b,c nguyn tcng nhau khi v chkhi(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



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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 2N

, a, b l 2 s

nguyn t

cng nhau).

Vic chng minh nh l ny kh phc tp, trmt 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 st 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 cbn v s nguyn t

2.2.1 C bao nhiu s nguyn t dng ax + b

V d 2.2. Ch ng minh r ng: c v s s nguyn t c dng 3x 1.4Li gii. Mi s tnhin 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.



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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 dkhc 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 th M 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. Ch ng minh r ng: 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 nhth N phi

c 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.



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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 2 N, (a, b) = 1.2.2.2 Chng minh s nguyn t

V d 2.4. Ch ng minh r ng: (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 ny

khng 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 2 P ) p nguyn t cngnhau vi mi tha s ca (p 1)! (pcm).

V d 2.5. Cho 2m

1 l s nguyn t. Ch ng minh r ng m cng l

s nguyn t. 4Li gii. Gi sm l hp s ) m = p.q (p, q2 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. Ch

ng minh r

ng: m

i

c nguyn t

c

a 1994! 1

u l

nhn 1994. 4Li gii. Gi p l c s nguyn t ca 1994! 1Gi sp 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. Ch ng minh r ng: n >2 th gi a n v n! c t nht 1 snguyn t (t suy ra c v s s nguyn t). 4



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Li gii. V n > 2 nn k = n! 1 > 1, do k c t nht mt c snguyn t p. Tng tbi 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 mn

bi. Vy p = 3.

V d 2.9. Tm k 2 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: t1 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 5

s 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 ccc snguyn tp : 2p+p2 cng l snguyn

t. 4Li gii. Xt 3 trng hp:



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p = 2 ) 2p +p2 = 22 + 22 = 8 62 P p = 3 ) 2p +p2 = 23 + 32 = 17 2 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 62 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 2 P : p|2p + 1 ) p > 2 ) (2;p) = 1

Theo nh l Fermat, ta c: p|2p1

1. Mp|2p + 1 ) p|2(2p1 1) + 3 ) p|3 ) p = 3

Vy: 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 l

s nguyn t th s cn li l s nguyn t hay hp s? 4Li gii. Nu p = 2 ) 8p + 1 = 17 2 P; 8p 1 = 15 62 P

Nu p = 3 ) 8p 1 = 23 2 P; 8p 1 = 25 62 P Nu p > 3, xt 3 s tnhin 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 2 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 2 P nn p c dng 3k + 1 hoc 3k + 2

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



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

l 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 8i = 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). Ch ng minh r ng 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 8i =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 snguyn tsao 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 = 5Khi : 5bc = 5(5 + b + c) , 5 + b + c = bc , (c 1)(b 1) = 6

Do vy:

2664

b 1 = 1c 1 = 6 ,

b = 2c = 7

chnb 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, q2 P sao cho p2 = 8q+ 1. 4



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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 2 NVy q > 2. V q2 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 snguyn tnhhn 100 c cha mt snguynt 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 s

nguyn tchn duy nht l 2 v l snguyn tnhnht. Vys nguyn t nh nht trong 3 s nguyn t l 2.



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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 snguyn 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.



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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 + 24f) 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 2 P. Chng minh rng: p + 8 l hp s.b) 2p + 1

2P. Chng minh rng: 4p + 1 l hp s.

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



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22 2.3. Bi tp

d) p + 8 2 P. Chng minh rng: p + 4 l hp s.e) 4p + 1 2 P. Chng minh rng: 2p + 1 l hp s.

f) 5p + 1 2P

. Chng minh rng: 10p + 1 l hp s.g) 8p + 1 2 P. Chng minh rng: 8p 1 l hp s.h) 8p 1 2 P. Chng minh rng: 8p + 1 l hp s.i) 8p2 1 2 P. Chng minh rng: 8p2 + 1 l hp s.

j) 8p2 + 1 2 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 2 N) l cc s nguyn t ln hn

3 th k...6.

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

b) Mt s nguyn t chia cho 30 c s dr. Tm s dr 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 stnhin.

Bi 7. Tm s t nhin c 4 ch s, ch s hng nghn bng ch s

hng n v, chs hng trm bng chs 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 2 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.



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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 2 N) l

cc 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. Ch

ng minh r

ng

iu ki

n c

n v

p v 8p

2

+ 1 l cc s

nguyn 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 tnhin 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)(p

2) 1...p.



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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)(p 1) + 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 chs:

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.

Tmc 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 cho

hiu 2 s

lin ti

p lun lun nh

nhau.

iu

l d

i v

i cc chuyngia nhng h mun i xa hn. Cng khng c vn g kh khn i

vi mt dy 7 s. H cn sh 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 chs, 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 th2....

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:



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2.4. Ph lc: Bn nn bit 25

Bi ton 2.1. Cho tr c s nguyn d ng n tu . Ch ng minh r ngtn ti n s t nhin lin tip m mi s trong chng u l hp s.4

Vy 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? 4

y 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.Tbi 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 tnhin m tng cc cs tnhin thc sca 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



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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 t

Merseme. 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 2132049 1 gm39751 chs ghi trong h thp phn. Gn y 2 sinh vin M tm

ra mt s nguyn t ln hn na l s 2216091 1 gm 65050 chs.

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 thxem 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 KtThng 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 tgic v hiu qu.

Mc ch ca ti ny l trang b nhng kin thc c bn c o

su c nng cao v rn luyn tduy ton hc cho hc sinh, to ra nntng tin cy cc em c vn kin thc nht nh lm hnh trang cho



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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 nn

trong khun kh ti ny phn chia dng ton, loi ton ch c tnhtng i. ng thi cng mi ch a ra li gii chcha 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 sc ngha cao p nhcu ngn ngPhp vit:

Ton hc l Vua ca cc khoa hc

S hc l N hong



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37/150

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 cbn

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 .4

Nhvy, khi a chia cho b th c th a ra cc s dr 2 {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
http://diendantoanhoc.net/forum/index.php?/user/89233http://diendantoanhoc.net/forum/index.php?/user/89233http://diendantoanhoc.net/forum/index.php?/user/89233http://diendantoanhoc.net/forum/index.php?/user/89233


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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 2 {0;2;4;6;8}5 | N , 5 | a0 , a0 2 {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 + + 2na03.2 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 nhnh 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 dkhi 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,



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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).Nhvy 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 snguyn 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 vb l hai st nhin. Ch ng minh r ng5a2+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 s7 | 3a + b. t 3a + b = 7c (c 2 Z. Khi b = 7c 3a. Nhvy

) 5a2

+ 15ab b2

= 5a

2

+ 15a(7c 3a) (7c 3a)2

= 49(c2 + 3ac a2)



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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.Ch ng minh r ng 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. Ch ng minh r ng 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). Nhvy

A 1 + 1 + + 1| {z }10 s 1+0 (mod 11)

10 (mod 11) ) 11 -A

V d 3.4. Cho p v q l hai s nguyn tphn bit. Ch ng minh r ngpq1 + 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.



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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 stnhin c p 1 chs v cc chs u bng 1 th n chiaht cho p.

Bi 4. Gi s n 2 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 2 {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 ri

l lun ra p s. Sau y l mt s v d

V d 3.5. Tm k 2 N tn ti n 2 N sao cho

4 | n2 k

vi k 2 {0;1;2;3}. 4

Li gii. Gi stn ti k 2 N tn ti n 2 N tha mn 4 | n2

k.Ta xt cc Trng hp: (m 2 N)



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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. Ch ng minh r ng vi mi n 2 N th 6 | n(2n +7)(7n + 1).4Li gii. Ta thy mt trong hai s n v 7n + 1 l s chn 8n 2 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) 8n 2N.

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)

Ch

ng minh r

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



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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 dth 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 tnhin n 7 | (2n 1).ii) Chng minh rng 7 -(2n + 1) 8n 2 N.

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

Bi 3. Tm n

13 | 3

2n

+ 3

n

+ 1.Bi 4. Chng minh rng vi mi a, b 2 N th ab(a2b2)(4a2b2) lun

lun 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 ht

cho 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 2 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 ?



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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 2 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 2 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 2 N.



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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 mn

a + b + c + d + e v a2

+ b

2

+ c

2

+ d

2

+ e

2 u chia h

t cho p.Chng minh rng s a5 + b5 + c5 + d5 + e5 5abcde cng chia

ht 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.Chng

minh 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 Choa, b l cc snguyn v n l snguyn d ng. Ta

ni, 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. 4Nhvy a n (mod n) () n | (a b). V d: 2012 2 (mod 5).

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

Tnh cht 3.10

a a (mod n),a b (mod n) , b a (mod n),a b (mod n), b c (mod n) ) a c (mod n).



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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), 8k 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 d ng ca a, b v m th a b(mod m) th

a

d b

d(mod

m

d).

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

a

c b

c(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. Ch ng minh r ng 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. Ch ng minh r ng 19 | 7 52n + 12 6n. vi mi s nguynd ng 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. Ch ng minh r ng 1998 | A. 4



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

A = 111 1000777 + 112 1000776 + + 888.

Do 1000k 1 (mod 999), 8k 2 N nn

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

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

V d 3.20. Ch ng minh r ng 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 dthc.

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,

8a,b,c

2Z. Chng minh rng 3 | a b c.

Bi 5. Chng minh rng 222333 + 333222 chia ht cho 13.



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Bi 6. Chng minh rng 9n + 1 khng chia ht cho 100, 8n 2 N.Bi 7. Chng minh rng vi mi s nguyn khng m n th 25n+3 +

5

n

3

n+1

chia ht cho 17.

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

Tm s dtrong php chia X cho 10001.

Bi 10. Chng minh rng 100 | 77777

777 .Bi 11. Cho

b

2

4acv

b

2

+ 4acl hai s chnh phng vi

a,b,c 2N.

Chng minh rng 30 | abc.

3.2.5 Quy np

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

Kim tra mnh ng vi n = p.

Gi smnh ng vi n = k. Ta i chng minh mnh cngng vi n = k + 1.

V d 3.21. Ch ng minh r ng A = 4n + 15 1 chia ht cho 9 vi min 2 N. 4Li gii. Vi n = 1 =

)A = 18 chia ht cho 9.

Gi sbi ton ng vi n = k. Khi 9 | 4k+15k1, hay 4k+15k1 =9q vi q2 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.



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V d 3.22. (HSG 9 TQ 1978)Ch ng minh r ng s c to bi3n ch

s ging nhau th chia ht cho 3n

vi 1 n, n 2 N. 4Li gii. Vi n = 1, bi ton hin nhin ng.Gi sbi ton ng vi n = k, tc 3k | aa a| {z }

3n s a

.

Vi n = k + 1 ta c:

aa a

| {z }3k+1= aa a

| {z }3kaa a

| {z }3kaa a

| {z }3k= aa a| {z }

3k

1 00 0| {z }3k1

00 0| {z }3k1

1

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

V d 3.23. Ch ng minh r ng vi mi n2N, 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,nnt k = 2m + 1 vi m 2 N, th khi (k + 1)(k 1) = 4k(k + 1) chiaht cho 23 = 8.Gi sbi ton ng vi n = p, tc 2p+2 | k2

p 1 hay k2p = q 2p+2 + 1vi q

2N.

Ta chng minh bi ton ng vi n = p + 1. Tht vy

A = k2p+1 1 = k22p 1 = k2p2 1

=

k2p 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.



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Bi tp ngh

Bi 1. Mt s bi ton cc phng php nu trn c th gii bng

ph

ng php quy n

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

Bi 3. Chng minh rng 64 | 32n+3 + 40n 27 vi n 2 N.

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

Bi 5. Chng minh rng 676 | 33n+3 16n 27 vi n 2 N, n 1.

Bi 6. Chng minh rng 700 | 292n 140n 1 vi n 2 N.

Bi 7. Chng minh rng 270 | 2002n 138n 1 vi n 2 N.

Bi 8. Chng minh rng 22 | 324n+1

+ 234n+1

+ 5 vi n 2 N.

Bi 9. Chng minh rng s 23n

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

2N.

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

chia ht cho 2n+5 vi n 2 N.

Bi 11. Chng minh rng vi mi s tnhin 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 v

c bn chs tn cng l 2008.

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

Bi 14. Tm phn d khi chia 32n

cho 2n+3, trong n l s nguyndng.

Bi 15. Cho n 2N

, n 2. t A = 77...

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



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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 ch a t nht n + 1 con th.

Chng minh. Gi skhng 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. Ch ng minh r ng lun tn ti s c dng

20112011 201100 0

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

2011; 20112011, , 20112011 2011| {z }2012 s 2011

.

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

| {z }m s 2011

v 20112011 2011

| {z }n s 2011

(m > n >

0).

=) 2012 | 20112011 2011| {z }m s 2011

20112011 2011| {z }n s 2011



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=) 2012 | 20112011 2011| {z }mn s 2011

00 00| {z }n s 2011

Vy tn ti s tha mn bi.

V d 3.25. Ch ng minh r ng 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 trong

101 s nguyn cho tn ti hai s c chs tn cng ging nhau.

V d 3.26 (Tuyn sinh 10 chuyn HSPHN, 1993). Cho 5 snguynphn bit ty a1, a2, a3, a4, a5. Ch ng minh r ng 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 tng

qut, 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 tbn s c thlp thnh su hiu khc nhau chia ht cho 2. Do 32 | P.



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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. Ch ng minhr ng tn ti mt schia 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 2012

s 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 2 N, 1 n < m 2012) th hiu

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

chia ht cho 2012.



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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. Ch ngminh r ng 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 mrng). (Tp ch Ton Tui Th s 115)Cho n l mt s chuyn d ng v n s nguyn d ng a1, a2, , an ctng bng 2n 1. Ch ng minh r ng 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 chs 0, 1, 2, 3.

Bi 2. (HSG 9 H Ni, 2006) Chng minh rng tn ti s t nhin

n 6= 0 tha mn 313579 | (13579n 1).Bi 3. Chng minh rng trong 52 s nguyn dng bt k lun lun

tm 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 2 {0, 1, 1}, (i = 1, 10) khng ng thi bng0 sao cho

A = c1a1 + c2a2 + + c10a10

chia ht cho 1032.

Bi 5. Chng minh rng tn ti s tnhin 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.



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3.2.7 Phn chng

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

Gi sngc li p | A(n).

Chng minh iu ngc li sai.

V d 3.28. Ch ng minh r ng vi mi snguynn thn2+n+1 khngchia ht cho 9. 4Li gii. Gi s9 | (n2 + n + 1). Khi n2 + n + 1 = (n +2)(n

1 ) + 3

chia 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 d3, 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 nguynd ng v k l s t nhin l. Gi thit x v y l cc s t nhin m

p |

x2t

+ y2t

. Ch ng minh r ng khi x v y ng thi chia ht cho

p. 4Li gii. Gi stri 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.2t 2 (mod p). (i)

Theo gi thit th x2t

+ y2t 0 (mod p).



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Do k l nn

xk.2t

+ yk.2t

= x2t

k

+ y2t

k ...x

2t + y2t

)

xk.2t

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

mi n 2 N.Bi 4. Chng minh 4n3 6n2 + 3n + 37 khng chia ht cho 125 vi

mi n

2N.

Bi 5. Chng minh n3 + 3n 38 khng chia ht cho 49 vi mi n 2 N.



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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) chkhng 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 ph ng trnh nghim nguyn

13x + 5y = 175 (4.1)

57
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58 4.1. Xt tnh chia ht

Li gii. Gi sx, 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 (t2Z). Thay vo phng trnh (4.1), ta c

13.5t + 5y = 175 , 13t + y = 35 , y = 35 13tDo , phng trnh (4.1) c v s nghim nguyn biu din di dng

(x; y) = (5t; 35 13t), (t 2 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 ph ng 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 nhsau:

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

Nhvy, n y ta c x v y nguyn nn 3x + 1 v y + 2 phi l cca 54. Nhng nu nhvy 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)



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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:2

664

3x + 1 = 1y + 2 = 54

,

x = 0y = 52

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

x = 1y = 56

V d 4.3. Gii ph ng 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 d2, vn dng iu ny tac th gim bt s trng hp cn xt.

V d 4.4. Gii ph ng 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)



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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 2 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 = 5b = 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 = 3

a b = 3 ,(a = 1

b = 2,(x y = 1

xy = 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 ph ng trnh:

x2 + 1 y2 + 1 + 2 (x y) (1 xy) = 4 (1 + xy) (4.13)



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

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

2

+ 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 2 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 tta 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 = 2

y 1 = 1 , x = 3

y = 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 ph ng trnh

x6 + 3x3 + 1 = y4 (4.14)



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

4x6 + 12x3 + 4 = 4y4

, (4x6

+ 12x

3

+ 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 2 Z nnta suy ra c cc trng hp sau:

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

,

x3 y2 = 1x3 + y2 = 1

,

x3 = 0y2 = 1

,

8>>>:

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

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

x3 y2 = 2x3 + y2 = 4 ,

x3 = 3y2 = 1 (loi)

2x3 2y2 + 3 = 5

2x3 + 2y2 + 3 = 1, x3 y2 = 1

x3 + y2 = 1 , x3 = 0

y2 = 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 ph ng trnh nghim nguyn d ng:1

x+

1

y=

1

p(4.15)

trong p l s nguyn t. 4



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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 ph ng 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 2 {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 nhvn 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 tcon s 18 xung ch cn 2!



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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 = 18x + 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.



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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 dtng 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 2 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.


7/28/2019 So

Documents

So hoc VNMath.pdf