các chuyên đề số học

8/3/2019 cc chuyn s hc

1/99

Nguyn Vn Tho Chuyn S Hc - Phn

1

Li ni uS hc l mt phn rt quan trng trong chng trnh Ton ph thng. Trong hu

ht cc thi hc sinh gii th bi S hc thng xuyn xut hin v lun l mt thchthc ln i vi hc sinh.

Hin nay, khng cn h chuyn cp Trung hc c s nn cc em hc sinh chuynTon cng khng c hc nhiu v phn ny nn thng gp rt nhiu kh khn khi giicc bi ton . V vy, ti bin son ti liu ny nhm gii quyt phn no nhng khkhn cho cc em hc sinh chuyn Ton.

Chuyn gm ba chng:

-Chng I. Cc bi ton chia ht

-Chng II. Cc bi ton ng d

-Chng III. Cc bi ton khc.

mi bi u c trnh by ba phn: H thng l thuyt; h thng cc v d vcui cng l h thng cc bi tp t gii. Cc v d v bi tp lun c sp xp vi kh tng dn - theo quan im ca tc gi.

Tuy nhin, do trnh c hn nn khng th trnh khi nhiu thiu st, rt mongc cc thy c ng gp hon thin hn. Xin chn thnh cm n!

NGUYN VN THO


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Chng I

CC BI TON V CHIA HT

I.1 Chia ht

I.1.1 L thuyt

I.1.1.1 nh ngha

Cho m v n l hai s nguyn , n 0. Ta ni rng m chia ht cho n (hay n chia htm) nu tn ti mt s nguyn ksao cho m = kn.

K hiu: m n, (c l m chia ht cho n) hay n | m, (c l n chia ht m).

I.1.1.2 Cc tnh cht c bn

Cho cc s nguynx, y, z. Ta c:

a)x x, x 0.

b) Nu x y vx 0 th |x| |y|.c) Nuxz, y zth ax + byzvi mi s nguyn a, b.

d) Nuxzvx yzthyz

e) Nuxy vyx th |x| = |y|.

f) Nuxy vyzthx z.

g) Nux |y vy 0 th |x

.

Chng minha) x = 1.x nnxx vi mix 0.

b)Nuxy , x 0 th tn ti kZsao chox = ky, k 0 |x| = |k||y| |y| do |k| 1.

Cc phn cn li cng kh n gin, vic chng minh xin nhng li cho bn c.

I.1.2 Cc v d

V d 1. Cho n l mt s t nhin ln hn 1. Chng minh rng

a)

2n

l tng ca hai s l lin tip.b) 3n l tng ca ba s t nhin lin tip.

Li gii

a) Ta c 2n = (2n-1 - 1) + (2n-1 +1) suy ra pcm.b) Ta c 3n = (3n-1 - 1) + (3n-1) + (3n-1 + 1) suy ra pcm.

V d 2. Chng minh rng:


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a) nu m n chia ht mp + nq th m n cng chia ht mq + np.b) nu m n chia ht mp th m n cng chia ht np.

Li gii

Nhn xt:Hai biu thc (mp + nq) v (mq + np) l hai biu thc c hnh thc ging nhi xng loi hai v vy khi xt cc biu thc loi ny thng ngi ta kim tra hiu

ca chng.

a) Ta c (mp + nq) (mq + np) = (m - n)(p - q) (m - n)

Nn nu (mp + nq) (m - n) th hin nhin (mq + np) (m - n).

b) Chng minh tng t.

V d 3. Chng minh rng nu a3 + b3 + c3 chia ht cho 9 th mt trong ba s a, b, cphi chia ht cho 3.

Li gii

Nhn xt:Vi nhng bi ton chng minh a chia ht cho mt s c th lun kh ngin! Ta c th xt ht cc trng hp xy ra ca s d khi a chia cho s . ( Cng vic chnh l xt v h thng d y - y l tp hu hn nn c th th trc tip)

Gi s khng c s no trong ba s a, b, c chia ht cho 3. Khi

a = 3m 1; b = 3n 1; c = 3p 1

Do a3 + b3 + c3 = (3m 1)3 + (3n 1)3 + (3p 1)3

=

9 3

9 1

9 3

9 1

A

a

a

A

+ +

khng th chia ht cho 9.

T suy ra pcm.

V d 4.

Chng minh rng nu a2 + b2 chia ht cho 3 th c a v b u chia ht cho 3.

Li gii

TH1: c 1 s khng chia ht cho 3, gi s l a

Khi a = 3k 1; b = 3q suy ra a2 + b2 = (3k 1)2 + (3q)2

= 3(3k2 2k+ 3q2) + 1 khng chia ht cho 3.

TH2: c hai s khng chia ht cho 3.

Khi a = 3k 1; b = 3q 1 suy ra a2 + b2 = 3A +2

Do c a v b phi chia ht cho 3.

V d 5. Chng minh rng vi mi s t nhin chn n v mi s t nhin l k th


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S = 1k+ 2k+ + nk lun chia ht cho n + 1.

Li gii

Ta c 2S= (1k+ nk) + (2k+ (n - 1)k) + n + 1

M n chn nn n + 1 l nn (2, n+ 1) = 1

Do Sn + 1.

V d 6. Chop l s nguyn t,p > 3 v 3

122 =

p

n .Chng minh rng

n22 n .

Li gii

Vp l s nguyn t v p>3 )3(mod12 1 p

Mt khc (2,p) = 1 nn theo nh l Fermat ta c

)(mod121

pp

Do pp 312 1

Ta c

n.22n12n123

12nVi

12122p1-nrasuy3

)12)(12(4

3

441

3

121

n1-n2p2p

2p1-n

112

=

+=

=

=

pppp

n

T suy ra iu phi chng minh.

V d 7. Chox, y l hai s nguyn khc -1 sao cho

1

1

1

1 33

++

+++

x

y

y

xl mt s nguyn

Chng minh rng 12004 x chia ht choy+1.

Li gii

Trc ht ta t d

c

xb

a

y

x

=+

+

=+

+

1

1y

;1

1 33

vi a, b, c, dnguyn v b > 0, d> 0, (a,b) = 1, (c,d) = 1.

Ta c

bd

bcad

d

c

b

a +=+ nguyn

Do


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bdbadbbcadbd ++ bcad v (a,b)=1 (1)

Mt khc

(2)dadacbdac

)1)(1(

1

1.

1

1. 22

33

++=

+

+

+

+= Zyyxx

x

y

y

x

d

c

b

a

V (c,d)=1 nn t (1) v (2) suy ra

ab suy ra b = 1 v (a,b) = 1

V

(3)1y1x)1(1x1

1 333

+++=+=++

yab

a

y

x

M 1x1-)(x1 366432004 += x

Kt hp vi (3) suy ra iu phi chng minh.

V d 8. Cho n 5 l s t nhin .Chng minh rng

n

n )!1( n-1 .

Li gii

a) Trng hp 1. n l s nguyn t

Theo nh l Winson (n-1)! -1(mod n) suy ra ((n-1)!+1 n

Ta c

11)!1(11)!1()!1(

+

=

+

=

n

n

nn

n

n

n(v 1

10


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Do p =2 n, n 5 suy rap 123p3 2 +> pp

12 2


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3a2 3.602 (mod 191) -87 (mod 191)

Vy vi mi m, ch cn chon b m - a3 (mod 191)

l cP(x) (x + a)3 + b (mod 191).

Ta c, vi mi i, j nguyn thP(i) P(j) (mod 191)

(i + a)3

(j + a)3

(mod 191) (i + a)3.63(j + a)2 (j + a)3.63 + 2 (mod 191)

(j + a) (mod 191)

(j + a)2 (i + a)189(j + a)3 (mod 191)

(i + a)192 (mod 191)

(i + a)2 (mod 191)

(i + a)3.63(j + a)2 (i + a)189.(i + a)2 (mod 191)

i + a (mod 191)T suy ra

P(i) P(j) (mod 191) i = j (mod 191)

T suy ra tp {P(1), P(2), ..., P(191)} c 191 s d khc nhau khi chia cho 191

Do phi tn ti mt s nguyn n {1, 2, ..., n} sao cho P(n) 191

Vy ta c iu phi chng minh.


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I.1.3 Bi tp

Bi 1. Chng minh rng vi mi s nguyn m, n ta c:

1) n3 + 11n 62) mn(m2 n2) 33) n(n + 1)(2n + 1) 6.4) n3 + (n + 1)3 + (n + 2)3 9.5) n2(n2 - 12) 126) mn(m4 n4) 307) n5 n 308) n4 + 6n3 + 11n2 + 6n 249) n4 4n3 4n2 + 16n 384 ( n chn v n > 4)10)n2 + 4n + 3 811)n3 + 3n2 n 3 4812)n12 n8 n4 + 1 51213)n8 n6 n4 + n2 1152.14)n3 4n 48 ( n chn)15)n2 3n + 5 khng chia ht cho 121.16)(n + 1)(n + 2)(2n) 2n17)n6 n4 n2 + 1 128 ( n l)

Bi 2. Chng minh rng tch ca n s nguyn lien tip lun chia ht cho n!

Bi 3. Chop l s nguyn t l. Chng minh rng vi mi kN, ta lun c

S = 12k+ 1 + 22k+ 1 + + (p - 1)2k+ 1 chia ht chop.

Bi 4. Chng minh rng nu a3 + b3 + c3 chia ht cho 9 th mt trong ba s a, b, c phichia ht cho 9.

Bi 5. Cho a, b nguyn. Chng minh rng nu an bn th a b.

Bi 6. Tm s nguyn dng n sao cho n chia ht cho mi s nguyn dng khng vtqu n .

Bi 7. Chng minh rng a2 + b2 + c2 khng th ng d vi 7 modulo 8.

Bi 8. Tng n s nguyn lin tip c chia ht cho n hay khng? ti sao?

Bi 9. Chng minh rng khng tn ti cp s nguyn ( x, y) no tha mn mt trongnhng ng thc sau:

a)x2 +1 = 3y


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b)x2 + 2 = 5y.

Bi 10. Chng minh rng vi n 1 th

(n + 1)(n + 2) ... (n + n)

chia ht cho 2n.

Bi 11. Tm ch s tn cng ca s FermatFn = 122

+

n

, n 2.Bi 12. Tm cc s nguyn dngp, q, rsao cho

pqr- 1 (p - 1)(q - 1)(r- 1).

Bi 13. Chng minh rng tn ti mt s t nhin c 1997 ch s gm ton ch s 1 v 2sao cho s chia ht cho 21997.

Bi 14. Cho a l mt s nguyn dng v a > 2. Chng minh rng tn ti v s snguyn dng n tha mn

an - 1 n.

Bi 15. Chng minh rng tn ti v s s nguyn dng n sao cho2n + 1 n.

Bi 16. Chng minh rng trong 12 s nguyn t phn bit bt k lun chon ra c 6 sa1, a2, ..., a6 sao cho

(a1 - a2)(a3 - a4)(a5 + a6) 1800.

Bi 17. Cho a, b, c, dnguyn bt k. Chng minh rng

(a - b)(a - c)(a - d)(b - c)(b - d)(c - d) 12.

Bi 18. Tm s t nhin n sao cho 2n - 1 chia ht cho 7. Chng minh rng vi mi s tnhin n th 2n + 1 khng th chia ht cho 7.Bi 19. Tm s t nhin n sao cho n5 - n chia ht cho 120.

Bai 20. Tm tt c cc cp s nguynx > 1,y > 1 sao cho

+

+

.13

13

xy

yx

Bi 21. Chox1,x2 l hai nghim ca phng trnhx2 - mx + 1 = 0 vi m l s nguyn ln

hn 3. Chng minh rng vi mi s nguyn dng n th Sn =nn xx 21 + l mt s nguyn v

khng chia ht cho m - 1.Bi 22. Tm tt c cc cp s nguyn dng a, b sao cho

2

22

+

ab

a

l mt s nguyn.

Bi 23. (30.4.2003) Tm ba s nguyn dng i mt phn bit sao cho tch ca hai sbt k u chia ht cho s th 3.


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B i 24. Chng minh rng vi mi s t nhin n th gia n2 v (n + 1)2 lun tn ti ba st nhin phn bit a, b, c sao cho a2 + b2 c2.

Bi 25. Cho s t nhinAn = 19981998...1998 (gm n s 1998 vit lin nhau)

a) Chng minh rng tn ti s nguyn dng n < 1998 sao choAn 1999.

b) Gi kl s nguyn dng nh nht sao cho Ak 1999. Chng minh rng

1998 2k.

Bi 26. Cho hai s nguyn dng m v n sao cho n + 2 m. Hy tnh s cc b ba snguyn dng (x, y, z) sao chox + y + z m trong mi sx, y, zu khng ln hn n.

Bi 27. (APMO 98) Tm s nguyn dng n ln nht sao cho n chia ht cho mi s

nguyn dng nh hn 3 n .

Bi 28. Tm tt c cc s nguyn dng m, n sao cho n3 + 1 chia ht cho mn - 1.

Bi 29. Tm tt c cc cp s nguyn dng a, b sao cho

12

2

+

abba

l mt s nguyn.

Bi 30. Tm tt c cc cp s nguyn dng sao cho

72

2

++++

bab

baba

l mt s nguyn.

Bi 31. Cho n l s nguyn dng ln hn 1.p l mt c nguyn t ca s Fermat Fn.

Chng minh rng p - 1 chia ht cho 2n+2.Bi 32. Chox, y , p l cc s nguyn vp > 1 sao chox2002 vy2002 u chia ht cho p.Chng minh rng 1 +x + y khng chia ht chop.

Bi 33. (USA - 98) Chng minh rng vi mi s nguyn dng n 2, tn ti mt tp hpn s nguyn sao cho vi hai s a, b bt k (ab) thuc tp th (a - b)2 chia ht ab.

Bi 34. Gi s tp S = {1, 2, 3, ..., 1998} c phn thnh cc cp ri nhau

{ai, bi| 1 i 1998} sao cho |ai - bi| bng 1 hoc bng 6. Chng minh rng

= 999

1 ||i ii ba = 10k+ 9.

Bi 35. Tm tt c cc cp s nguyn dng a, b sao cho

2

22

+

ab

a

l mt s nguyn.

Bi 36. Chng minh rng vi mi n N* lun tn ti s t nhin a sao cho


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64a2 + 21a + 7 2n.

Bi 37. (Nga - 1999) Cho tpA l tp con ca tp cc s t nhin n sao cho trong 1999 st nhin lin tip bt k lun c t nht mt s thucA.

Chng minh rng tn ti hai s m, n thucA sao cho m n.

Bi 38. Tmx, y, znguyn dng vx


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I.2 c s chung ln nht - Bi s chung nh nht

I.2.1. L thuyt

I.2.1.1. c s chung ln nht

I.2.1.1.1 nh ngha 1

Cho a, b l hai s nguyn. S nguyn dng dln nht chia ht c a v b c gil c chung ln nht ca a v b.

K hiu:d = (a, b) hoc d= gcd(a, b)

Nu d= 1 th ta ni a v b l hai s nguyn t cng nhau.

I.2.1.1.2. Cc tnh cht ca c chung ln nht

a) Nup l mt s nguyn t th (p,m) =p hoc (p, m) = 1.

b) Nu (a, b) = dth a = dm, b = dn v (m, n) = 1.

c) Nu (a, b) = d, a = dm, b = dn v (m, n) = 1 th d= d.d) Nu m l mt c chung ca a v b th m | (a,b).

e) Nupx || m v py || n thpmin(x,y) || (m, n).

f) Nu a = bq + rth (a, b) = (r, b).

g) Nu c | ab v (a,c) = 1 th c | b.

h) Nu (a, c) = 1 th (ab, c) = (b, c).

I.2.1.2 Bi s chung nh nht.

I.2.1.2.1 nh ngha.

Cho a, b l hai s nguyn. S nguyn dng nh nht chia ht cho c a v b cgi l bi s chung nh nht ca a v b.

K hiu: [a,b] hay lcm(a,b).

I.2.1.2.2. Cc tnh cht ca bi chung nh nht.

a) Nu [a, b] = m v m = a.a= b.b th (a, b) = 1.

b) Nu m = a.a = b.bv (a, b) = 1 th [a, b] = m.

c) Nu m = [a, b] v m l mt bi chung ca a v b th m | m.

d) Nu a | m v b | m th [a, b] | m.

e) Cho n l mt s nguyn dng, ta lun c n[a, b] = [na, nb].

f) Nu a = 1 2 1 21 2 1 2. ... ; . ...k kn mn n m m

k kp p p b p p p=


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Th [a, b] = min( , )

1

i i

kn m

ii

p=

.

I.2.1.3 nh l Bzout

Phng trnh mx + ny = (m,n) lun c v s nghim nguyn.

Nhn xt: Phng trnh ax + by = c c nghim nguyn khi v ch khi c lbi ca (a, b).

Phng trnh ax + by = 1 c nghim nguyn khi v ch khi

(a, b) = 1.

I.2.1.4 Mi quan h gia c s chung ln nht v bi s chung

nh nht

Cho a v b l cc s nguyn khc 0, ta c

[a,b]( , )

aba b

=

I.2.2 Cc v d

V d 1. Chng minh rng vi mi s nguyn a, b ta lun c

(3a + 5b, 8a + 13b) = (a, b).

Li gii

Ta c (3a + 5b, 8a + 13b) = (3a + 5b, 8a + 13b 2(3a + 5b))

= (3a + 5b, 2a + 3b) = (a + 2b, 2a + 3b)= (a + 2b, b) = (a, b).

pcm.

V d 2. Nu (a, b) = dth (a +b, a - b) c th nhn nhng gi tr no?

Li gii

Ta c m = (a + b, a - b) = (a + b, 2a) = (a + b, 2b).

Do m l c chung ca 2a v 2b v a + b.

Nu a + b l th (a + b, a - b) = d

Nu a + b chn th (a + b, a - b) = 2d.

V d 3. Chng minh rng phn s sau ti gin

21 4

14 3

n

n

++

Li gii

Ta c (21n + 4, 14n + 3) = (7n + 1, 14n + 3)


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= (7n + 1, 14n + 3 2(7n + 1))

= (7n +1, 1) = 1


V d 4. Cho a, b l cc s nguyn dng phn bit sao cho ab(a + b) chia ht cho a2 +ab + b2. Chng minh rng

3| |a b ab >

Li gi

tg= (a, b) a =xgv b = ygvi (x, y) = 1.

Khi

2 2 2 2

( ) ( )ab a b gxy x y

a ab b x xy y

+ +=

+ + + +

l mt s nguyn.

Ta c (x2 +xy +y2,x) = (y2,x) = 1

(x2 +xy +y2,y) = 1.

V (x + y, y) = 1 nn ta c

(x2 +xy +y2,x + y) = (y2,x + y) = 1

Do x2 +xy +y2 |g

Suy ragx2 +xy +y2

Mt khc |a - b|3 =g3|x - y|3

=g2|x - y|3gg2.1. (x2 +xy +y2) = ab.

T ta c iu phi chng minh.

V d 5. Cho n l mt s nguyn dng, d= (2n + 3, n + 7).

Tm gi tr ln nht ca d.

Li gii

Ta c

(2n + 3, n + 7) = (2(n + 7) 2n -3, n + 7)= (11, n + 7) 11.

Mt khc khi n = 11k+ 4 th n + 7 = 11(k+ 1)

(11, n + 4) = 11.

Do gi tr ln nht ca dl 11.


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V d 6. (India 1998) Tm tt c cc b (x, y, n) nguyn dng sao cho

(x, n+1) = 1 (1)

v xn + 1 =yn+1. (2)

Li gii

T (2) ta cxn =yn+1 1

= (y - 1)(yn +yn-1 + + 1) (*)

t m =yn +yn-1 + + 1

Suy raxnm

M (x, n+1) = 1 nn ta phi c (m, n +1) = 1

Ta li c

m =yn yn-1 + 2(yn-1 yn-2) + + n(y - 1) +n + 1

= (y 1)(yn-1 + 2yn-2 + + n) + n + 1

n + 1 (m, y - 1)

M (m,n + 1) = 1 (m, y - 1) = 1 (**)

T (*) v (**) suy ra m phi l lu tha n ca mt s nguyn dng.

Tc l m = qn vi q l mt s nguyn dng no

V y > 0 nn ta cyn 1 (v l)

Vy n = 1 x =y2 1

V (x, n +1) = (x, 2) = 1 nn x= 2k+ 1 y chn

Do (x, y, n ) = (4a2- 1, 2a, 1) vi a nguyn dng.

V d 7. Chng minh rng nu mt s nguyn dng c s c s l l th phi l schnh phng.

Li gii

Gi n l s t nhin nh vy.

Nhn thy, nu dl mt c s ca n th nd

cng l mt c s ca n.

Do vy nu vi mi dm dn

dth s c ca n phi l chn.

Nn tn ti dl c ca n sao cho d=n

dn = d2 (pcm).


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V d 8. (APMO - 1999) Tm s nguyn dng n ln nht sao cho n chia ht cho mi s

t nhin nh hn 3 n .

Li gii

Cu tr li l 420.

Tht vy, ta c[1, 2, 3, 4, 5, 6, 7] = 420. 7 < .84203 <

Gi s n > 420 v tha mn iu kin u bi 73 >n n 420.

Do n 2.420 = 480 93 n .

Ta c

[1, 2, ..., 9] = 2520 n 2520 133 >n .

Gi m l s nguyn dng ln nht nh hn 3 n n 13 v m3 < n (m +1)3.

Do n [1, 2, ..., m] n [m - 3, m - 2, m - 1, m]

Mt khc

6

)3)(2)(1(],1,2,3[

mmmmmmmm

nn

3)1(6

)3)(2)(1(+

mn

mmmm

suy ra

)3)(2)(1(

)1(6 3

+

mmm

mm

)3

41)(

2

31)(

1

21(6

+

+

+

mmmm

)3

41)(

2

31)(

1

21(6)(

+

+

+=

mmmmmf 0

M rng tp xc nh ca m trn tp s thc ta d dng chng minh c f(m) l hm sng bin trn tp [13; +)Do vi mi m 13 th f(m) f(13) > 0 (v li)

nn iu gi s l sai.T suy ra iu phi chng minh.


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I.2.3 Bi tp

Bi 1. Cho m, n l hai s nguyn dng phn bit v (m, n) = d.

Tnh (2006m + 1, 2006n + 1).

Bi 2. Chng minh rng nu cc s a, b, c i mt nguyn t cng nhau th(ab + bc + ca, abc) = 1.

Bi 3. Tm

a) (21n + 4, 14n + 3)

b) (m3 + 2m, m4 + 3m2 + 1)

c) [2n - 1, 2n + 1].

Bi 4. Chng minh rng (2p - 1, 2q - 1) = 2(p, q) - 1.

Bi 5. Cho a, m, n l cc s nguyn dng, a > 1 v (m, n) = 1. Chng minh rng(a - 1)(amn - 1) (am - 1)(an - 1).

Bi 6. Chng minh rng

[1, 2, ..., 2n] = (n +1, n + 2, ..., 2n)

Bi 6. Chng minh rng vi mi s nguyn dng m >n ta c

[ ] [ ]nm

mnnmnm

>+++

21,1, .

Bi 7. Chng minh rng dy 1, 11, 111, ... cha v hn cp (xn,xm) nguyn t cng nhau.

Bi 8. Cho n l mt s nguyn dng, a v b nguyn dng v nguyn t cng nhau.

Chng minh rng ),( baba

ba nn

bng 1 hoc n.

Bi 9. Cho m, n l cc s nguyn dng, a l mt s nguyn dng ln hn 1.

Chng minh rng

.1)1,1( ),( = nmnm aaa

Bi 10. (Hn Quc 1998) Tm tt c cc s nguyn dng l, m, n i mt nguyn t cng

nhau sao cho)

111)((

nmlnml ++++

l mt s nguyn.

Bi 11. (Canada - 97)Tm s cc cp s nguyn a, b (a b) tho mn


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18

[ ]

=

=

!50,

!5),(

ba

ba

Bi 12. (Hungari - 1998) Tm n nguyn dng sao cho tn ti cc cp s nguyn a, b thamn

(a, b) = 1998 v [a, b] = n!

Bi 13. (Nga - 2000) Cho 100 s nguyn dng nguyn t cng nhau xp trn mt vngtrn. Xt php bin i nh sau: Vi mi s nguyn trn vng trn ta c th cng thmc chung ln nht ca hai s k bn n. Chng minh rng sau mt s hu hn php bini , ta c th thu c cc s mi i mt nguyn t cng nhau.

Bi 14. (Hungari - 1997) Cho tp A gm 1997 s nguyn phn bit sao cho bt k 10 sno trongA cng c bi chung nh nht nh nhau. Tm s ln nht cc s i mt nguynt cng nhau c th c trongA.

Bi 15. Cho s v t l cc s nguyn dng khc 0. Vi cp (x, y) bt k, gi Tl php bini (x, y) thnh cp (x - t, y - s). Cp (x, y) c gi l Tt nu sau hu hn php bini T ta thu c cp mi nguyn t cng nhau.

a) Tm (s, t) sao cho (s, t) l mt cp Tt

b) Chng minh rng vi mis, tth lun tn ti cp (x, y) khng Tt.

Bi 16. Tn ti hay khng cc cp s nguyn dng a, b sao cho

(30a + b)(30b + a) = 42001?


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I.3. S nguyn T

I.3.1. L thuyt

I.3.1.1 nh nghaMt s nguyn dngp c gi l s nguyn t, nu n ch c hai c s dng

l 1 v chnh n.Nup khng phi s nguyn t thp c gi l hp s.

Nhn xt: 2 l s nguyn t chn duy nht.

I.3.1.2 nh l 1 ( nh l c bn ca s hc)

Mi s t nhin ln hn 1 u c th phn tch mt cch duy nht thnh tch cctha s nguyn t.

I.3.1.3. nh l 2

Tp hp cc s nguyn t l v hn.

I.3.1.3. nh l 3

Chop l mt s nguyn t. Nu p | ab thp | a hocp | b.

(Vic chng minh cc nh l trn kh n gin v ta c th tm c trong bt k mtquyn sch s hc no, v vy s khng c trnh by ti y)


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I.3.2 Cc v d

V d 1. Tm tt c cc s nguyn dng n sao cho cc s 3n 4, 4n 5, 5n 3 u l ccs nguyn t.

Li giiTa c (3n - 4) + (5n - 3) = 8n 7 l s l

Do trong hai s trn phi c mt s chn v mt s l.

Nu 3n 4 chn th 3n 4 = 2 n = 2 4n 5 = 3 v 5n 3 = 7 u l cc s nguynt.

Nu 5n 4 chn th 5n 3 = 2 n = 1 3n 4 = -1 (loi)

Vy n = 2.

V d 2. Tm s nguyn t p sao cho 8p2 + 1 v 8p2 1 cng l nhng s nguyn t.

Li giiNup = 2 th 8p2 + 1 = 33 3 nn khng tha mn.

Nup = 3 th 8p2 + 1 = 73 v 8p2 1 = 71 u l s nguyn t nn p = 3 tha mn

Nup > 3 vp nguyn t nnp khng chia ht cho 3.

Do p = 3k+ 1 hoc p = 3k 1

+)p = 3k+ 1 8p2 + 1 = 8(3k+ 1)2 + 1

= 72k2 + 48k+ 9 3

V hin nhin 8p2 + 1 > 3 nn 8p2 + 1 l hp s.+)p = 3k 1 8p2 + 1 3 v 8p2 1 > 3 nn khng tha mn.

Vyp = 3.

V d 3. Chop 5 tha mnp v 2p + 1l s nguyn t. Chng minh rng 4p + 1 l hps.

Li gii.

+)p = 3k+ 2 4p + 1 = 4(3k+ 2) + 1 = 12k+ 9 53 4p + 1 l hp s.

+p = 3k+ 1 2p + 1 = 2(3k+ 1) + 1 = 6k+ 3 3 (v l v 2p + 1 l s nguyn t ln hn11.)

V d 4. Tm s nguyn tp sao cho 2p + 1 l lp phng ca mt s t nhin.

Li gii

Ta c 2p + 1 = n3 2p = n3 1 = (n - 1)(n2 + n + 1) (*)

Do vi mi s t nhin n th n2 + n + 1 > n 1 v mi s nguyn tp thp 2


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21

Nn t (*) ta c2

1 2

1

n

n n p

=

+ + =

T tm cp = 13.

V d 5. Chng minh rng vi mi s nguyn dng a > 2, tn ti v s s nguyn dng

n sao cho: an

1 n.Li gii

Xt dy sx0 = 1,xn+1 = 1nxa

Ta s chng minh vi mi k N th 1kxa k (*)

+) n = 0, hin nhin (*)ng.

+) Gi s (*)ng ti n k, ta c

1 ( 1)

1

1 . 1 1 1

( ) 1 1

xkk x k

k k

x x mxak

x xm

k

a m x a a a

a a x

+

+

= = =

= =

T suy ra (*)c chng minh.

Do a > 2 nn (xn) l dy s tng suy ra (xn) l dy s v hn.

pcm.

V d 6. Chop, q l hai s nguyn t phn bit. Chng minh rng1 1 1q pp q pq + .

Li gii

Dop v q l hai s nguyn t phn bit nn (p, q) = 1Theo nh l Fermat nh ta cpq 1 1 qpq-1 + qp-1 1 q.

Tng t ta cng cpq-1 + qp-1 1 q

Do ta c iu phi chng minh.

V d 7. Bit rng 2n 1 l mt s nguyn t. Chng minh rng n l s nguyn t.

Li gii

Do 2n 1 l s nguyn t nn n > 1.

Gii s n l hp s

Khi n = pq trong p v q u ln hn 1 2n 1 = 2pq 1 2p 1 v 2q 1

M 2p 1 v 2q 1 u ln hn 1 nn 2n 1 khng phi s nguyn t (mu thun)


V d 8. Chng minh rng an + 1 (a v n nguyn dng) l s nguyn t th n =2k.

Li gii


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Gi s n 2k th n = 2m.q trong q l mt s nguyn dng l .

Khi an + 1 = 2 2 2 2 ( 1)1 ( ) 1 ( 1)( ... 1)m m m mq q qa a a a + = + = + + +

Suy ra an + 1 khng th l s nguyn t (Mu thun)


V d 9. Tm tt c cc s nguyn tp v qsao chop + q = (p - q)3.Li gii

V (p - q)3 =p + q 0 nnp v q phn bit v (p, q) = 1.

Mt khc ta li cp q 2p (modp + q)

Suy ra 8p3 chia ht chop + q

Li c 1 = (p, q) = (p + q, p) do p3 vp + q cng nguyn t cng nhau

Nn 8 p+q

2

48

, 8.

p q

p qp q

p q

+ =

+ = + = n > 0.Gi p l mt c nguyn t ca 2 2 2 22 2 (mod )

n n n n

a a p+ (*)

Bnh phng hai v ca (*) m n ln ta c:

2 22 (mod )m m

a p

Do 2 2 22 2.2 (mod )m m m

a p+ nnp khng th l c ca 2 22m m

a + .


V d 11. (Nga - 2001) Tm s nguyn dng l n > 1 sao cho a v b l hai c nguyn

t cng nhau bt k ca n th a + b 1 cng l c ca n.Li gii

TH1: n =pk trong p l mt s nguyn t th n tha mn yu cu.

TH2: n khng l lu tha ca mt s nguyn t.

Gip l mt c nguyn t nh nht ca n. Khi n =pk.s v (p,s) = 1.

p + s 1 | n


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Gi q l mt c nguyn t cas th q > p.

D thys


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24

I.3.3 Bi tp

Bi 1. Cho a N*. Chng minh rng nu am + 1 l s nguyn t th m = 2n. iu ngcli c ng khng?

Bi 2. Gi s phng trnhx2 + ax + b + 1 = 0 (a, b Z) c nghim nguyn. Chng minhrng a2 + b2 l hp s.

Bi 3. Cho a, b, c l cc s nguyn khc 0 v ac tha mn

22

22

bc

ba

c

a

++

=

Chng minh rng a2 + b2 + c2 khng phi l s nguyn t.

Bi 4. Tm n sao cho n4 + 4n l mt s nguyn t.

Bi 5. Chop l s nguyn t. Chng minh rng s

ppp

9...99...2...221...11 - 123456789

chia ht chop.

Bi 6. Tm s t nhin n sao cho

A = n2005 + n2006 + n2 + n + 2

l mt s nguyn t.

Bi 7. Tm n nguyn dng mi s sau y l s nguyn t:

a) n4 + 4

b) n4 + n2 +1.

Bi 8. Chng minh rng nup l mt s nguyn t ln hn 3 th

p2 - 1 24.

Bi 9. Cho 2m - 1 l mt s nguyn t. Chng minh rng m l mt s nguyn t.

Bi 10. Tm s nguyn tp sao cho 2p + 1 = a3, vi a nguyn dng.

Bi 11. Tm s nguyn t p sao chop + 4 vp + 8 cng l s nguyn t.

Bi 12. Tm s nguyn tp sao cho 8p2 + 1 v 8p2 - 1 l nhng s nguyn t.

Bi 13. Chop l mt s nguyn t vp = 30k+ r. Chng minh rngr= 1 hoc rl mt s nguyn t.

Bi 14. Cho aN*, a > 1. Chng minh rng an + 1 l mt s nguyn t th n = 2k.

Bi 15. (Iran 1998) Cho a, b, x l cc s nguyn dng tham mn

xa + b = abb.

Chng minh rngx = a v b =xx.


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Bi 16. Chng minh rng tn ti mt dy v hn {pn} cc s nguyn t phn bit sao chopn 1 (mod 1999

n) vi mi n = 1, 2, ...

Bi 17. Tm tt c cc s nguyn tp sao cho

f(p) = (2 + 3) - (22 + 32) + ... - (2p-1 + 3p - 1) + (2p + 3p) chia ht cho 5.

Bi 18. (Trung Quc 2001) Cho cc s nguyn dng a, b, c sao cho a, b, c, a + b - c,c + a - b, b + c - a v a + b + c l by s nguyn t phn bit. dl s cc s nguyn phn

bit nm gia s b nht v ln nht trong by s . Gi s rng s 800 l mt phn tca tp {a + b, b + c, c + a}. Tm gi tr ln nht ca d.

Bi 19. Chng minh rng vi mi s nguyn a th lun tn ti mt dy v hn {ak} saocho dy {ak+ a} cha hu hn s nguyn t.

Bi 20. Chng minh rng mi s t nhin u biu din c di dng hiu ca hai st nhin c cng s c nguyn t.

Bi 21. Chop l s nguyn t l v a1, a2, ..., ap - 2 l dy cc s nguyn dng sao cho pkhng l c ca akv ak

k- 1 vi mi k= 1, 2, ..., p - 2.

Chng minh rng tn ti mt s phn t trong dy trn c tich khi chia chop d 2.

Bi 22. Chng minh rng nu c s nguyn t nh nht p ca s nguyn dng n khng

vt qu 3 n thp

nl s nguyn t.

Bi 23. (Balan 2000) Cho dy cc s nguyn t p1, p2, ... tha mn tnh cht: pn l cnguyn t ln nht capn - 1 +pn - 2 + 2000. Chng minh rng dy s trn b chn.

Bi 24. Cho a1, a2, ..., an l cc s t nhin i mt khc nhau v c c nguyn t khngln hn 3. Chng minh rng

.31

...11

21


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26

Chng II

CC BI TON V NG D

II.1. nh ngha v cc tnh cht c bn ca ng d

II.1.1. L thuyt

II.1.1.1 nh ngha

Cho ba s nguyn a, b, m (m 0). Ta ni a ng d vi b theo modulo n nu a bchia ht cho m.

K hiu: a b (mod m).

II.1.1.2 Cc tnh cht

a) a b (mod m) a bm

b) Nu aibi (mod m) vi mi i = 1, 2, , n th1 1

n n

i ii i

a b= =

(mod m).

c) Nu ab (mod m) v c d(mod m)

th a c b d(mod m).

d) a b (mod m) v b c (mod m) ac(mod m).

e) Nu aibi (mod m) vi mi i = 1, 2, , n th1 1

n n

i ii i

a b=

(mod m)

f) Nu ab (mod m) th anbn (mod m).

g) ChoP(x) l a thc tu vi h s nguyn

nu ab (mod m) th P(a) P(b) (mod m)

Chng minh

a) Hin nhin

b) Do aibi (mod m) nn ta c ai bim

M1 1 1

( )n n n

i i i ii i i

a b a b= = =

=

m

T suy ra pcm.

c) Do ab (mod m) v cd(mod m)

nn a b v c du chia ht cho m

M (a - c) (b - d) = (a - b) (c - d) o (mod m)

Suy ra pcm.


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d) Gi s a = mq + r

V ab (mod m) nn b = mp + r

V bc (mod m) nn c = ml + r

a b (mod m).

(Cc phn cn li xin nhng bn c)


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II.1.2. Cc v d

V d 1.A vB l hai s c 7 ch s khc nhau t 1 n 7 vA > B.

Chng minh rngA khng chia ht cho B.

Li gii

K hiu S(n) l tng cc ch s ca n.Khi S(A) = S(B) = 1 + 2 + + 7 = 28.

MA S(A) (mod 9); BS(B) (mod 9).

T suy raA B 1 (mod 9).

Gi sA B hay A =pB,p nguyn dng.

DoA >B nnp > 1.

VA vB c 7 ch s khc nhau t 1 n 7 nn

1111111


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29

V d 3. Cho 11 s nguyn dng a1, a2, , a11. Chng minh rng lun tn ti cc sxi{-1, 0, 1} , i = 1, 2, , 11 khng ng thi bng 0 sao cho:

x1a1 + x2a2 + + x11a11 chia ht cho 2047.

Li gii

t S= {b1a1 + b2a2 + + b11a11| bi {0, 1}}.

Khi |S| = 2048

Do khi chia cc s ca Scho 2047 phi c 2048 s d.

V vy phi tn ti hai s trong Sm khi chia cho 2047 c cng s d

Gi s hai s l11 11

1 1

,m i i n i ii i

b a A c a= =

= = .

Ta cAm An 2047.

M11 11

1 1

( )m n i i i i ii i

A b c a x a= =

= =

V bi, ci {0, 1} nnxi {-1, 0, 1}

Li c (b1, b2, , b11) (c1, c2, , c11) nnxi khng ng thi bng 0.

pcm.

V d 4. Xt 100 s t nhin lin tip 1, 2, , 100.

GiA l s thu c bng cch xp mt cch tu 100 s thnh mt dy, B l s thu

c bng cch t mt cch tu cc du cng vo gia cc ch s caA. Chng minhrng cA vB u khng chia ht cho 2010.

Li gii

K hiu S(n) l tng cc ch s ca s t nhin n.

Ta thy t 1 n 100 xut hin 21 ch s 1, 20 ch s 2, 3, 4, 5, 6, 7, 8, 9.

Do S(A) = 21.1 + 20(2 + 3 + + 9) = 901

M AS(A) 1 (mod 3)

NnA khng chia ht cho 3 do A khng chia ht cho 2010.

Gi s sau khi t cc du cng vo gia cc ch s caA ta c

B = b1 + b2 + + bn.

Khi

B = b1 + b2 + + bnS(b1) + S(b2) + + S(bn) (mod 3).

M S(b1) + S(b2) + + S(bn) = S(A) 1 (mod 3)

Suy raB 1 (mod 3). Do B khng chia ht cho 2010.


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V d 5. Chng minh rng 11n+2 + 122n+1 chia ht cho 133 vi mi s t nhin n.

Li gii

Ta c 11n+2 + 122n+1 = 121.11n + 12.144n

= 133.11n + 12(144n 11n)

M 144n

11n

144-11 = 133Nn ta c pcm.

V d 6. Chng minh rng 19.8n + 17 l hp s vi mi s t nhin n.

Li gii

Ta c

19.8n + 17 19.(-1)n + 17 (mod 3)

Nu n chn th

19.(-1)n + 17 19 + 17 0 (mod 3)

M 19.8n + 17 > 3 nn 19.8n + 17 l hp s khi n chn.

Nu n = 4k+ 1

Ta c

19.8n + 17 = 19.84k + 1 + 17

48.642k + 17 (mod 13)

48.(-1)2k + 17 (mod 12)

0 (mod 13)

Do 19.8n + 17 l hp s.

Nu n = 4k+ 3 th

19.84k+3 + 17 0 (mod 5)

Tm li vi mi n th 19.8n + 17 l hp s.

V d 7.(Nga 2000) Tm cc s nguyn tp v q sao chop + q = (p - q)3.

Li gii

Ta c

(p - q)3 =p + q 0

Nnp v q l hai s nguyn t phn bit.

Ta c

p q 2p (modp + q)

(p - q)3 8p3 (modp + q)

8p3 0 (modp + q). (*)


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Do (p, q) = 1 (p, p + q) = 1

(p3,p + q) = 1

T (*) 8 p + q

p + q 8

Mp, q 2 nn 2 q < p 5Do (p, q) = (5, 3).

V d 8. Cho a l mt s nguyn dng l. Chng minh rng

2 2 2 22 , 2n n m m

a a+ +

nguyn t cng nhau vi moi s nguyn dng m n.

Li gii

Gi s m > n

Gip l mt c nguyn t ca 2 22n na + Ta c

2 22 (mod )n n

a p (*)

Bnh phng hai v ca (*) m n ln ta c:

2 22 (mod )m m

a p

Do 2 2 22 2.2 (mod )m m m

a p+ nn p khng th l c ca 2 22m m

a + .



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II.1.3. Bi tp

Bi 1. Chng minh rng vi mi n nguyn dng ta c:

a) 124 22 ++nn

0 (mod 7)

b) )9(mod011522 + nn .

Bi 2. Tm cc s t nhinx, y, ztha mn

2x.3y + 1 = 17z.

Bi 3. Tm tt c cc s nguyn dng n sao cho 5n - 1 (mod 72000).

Bi 4. Chng minh rng vi mi n l th

1n + 2n + ... + nn 1 + 2 +... + n.

Bi 5. Cho p l s nguyn t. a, b l hai s nguyn bt k. Chng minh rng

(a + b)pap + bp (mod p).Bi 6. Cho a, n nguyn dng,p nguyn t tha mn

2p + 3p = an

Chng minh rng n = 1.

Bi 7. Tmx nguyn sao cho |x| 1997 sao cho x2 + (x + 1)2 1997.

Bi 8. Tm s nguyn dng n = 2p3q sao cho n + 25 l mt s chnh phng.

Bi 9. Cho cc s nguyn khng m a1 , a2 < ... < a101 < 5050. Chng minh rng tn ti

bn s nguyn phn bit sao cho(ak+ al - am - an) 5050.

Bi 10. Chng minh rng tn ta s nguyn dng a sao cho mi s nguyn kth cc sm = k2 + k- 3a v n = k2 + k+ 1 - 3a u khng chia ht cho 20002001.

Bi 11. Chng minh rng tn ti mt s t nhin 2001 ch s gm ton ch s 1, 2 vchia ht cho 22001.

Bi 12. Chng minh rng t 11 s t nhin tu lun chon c ra hai s sao cho hiubnh phng ca chng chia ht cho 20.


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II.2. nh l Fermat, nh l euler v nh l Wilson

II.2. 1. H thng d y v h thng d thu gn.

II.2.1.1 nh ngha 1

Nu x y (mod m) ta niy l mt thng d cax modulo m.

Tp S= {x1,x2, ,xm} c gi l mt h thng d y modulo m nu mi snguyny tu u tn ti duy nht mt sxi sao choy xi (mod m).

II.2.1.2 Cc tnh cht c bn

+) Tp {1, 2, , m - 1} l mt h thng d y modulo m.

+) Mi h thng d y modulo m u c ng m phn t.

+) Mt h gm m phn t l h thng d y modulo m khi v ch khi hai phnt khc nhau bt k ca n khng ng d vi nhau modulo m.

+) Mi s nguyn m lun c v s h thng d y .

+) Vi mi s nguyn a, m > 0. Tp tt c cc s x nguyn tho mnx a (mod m)lp thnh mt cp s cng. Tp hp ny pc gi l mt lp thng d modulo m.

+) Vi mi s nguyn dng m th lun c m lp thng d modulo m.

II.2.1.3 nh l 1

Cho a, b, m l cc s nguyn. Khi ab (mod m) th (a, m) = (b, m).

Chng minh

Ta c a b (mod m) a b 0 (mod m) a = b + mq

(a, m) = (b + mq, m) = (b, m).pcm.

II.2.1.4 nh ngha 2

Tp S= {x1, x2, , xn} vi cc s xi phn bit gi l mt h thng d thu gnmodulo m nu (xi, m) = 1 vi mi i = 1, 2, , n v mi s nguyny nguyn t cng nhauvi m u tn ti sxi sao choy xi (mod m).

Nhn xt:

+) Ta c th thu c mt h thng d thu gn bng cch loi ra khi h thng dy nhng s khng nguyn t cng nhau vi m.

+) Mi h thng d y u c cng s phn t, s phn t ca mt h thng dthu gn k hiu l (m). (m) gi l phi hm Euler.

+) Nup l s nguyn t th (p) =p -1.

+) (m) bng s cc s nguyn khng vt qu m v nguyn t cng nhau vi m.


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II.2.1.5 nh l 2

Cho (a, m) = 1. Nu S = {x1,x2, ,xn l} l mt h thng d thu gn (hoc y )modulo m th aS= {ax1, ax2, , axn } cng l mt h thng d thu gn ( hoc tngng y ) modulo m.

Chng minh

Ta c (a, m) = 1 Nu axiaxj (mod m) thxixj (mod m)

Do / vi ij thxi khng ng d vi xjaxi cng khng ng d vi axj (mod m).

Suy ra cc phn t ca ca aSi mt phn bit theo modulo m.

M Sv aSc cng s phn t do nu Sl h thng d y th aScng l h thngd y .

Nu S l h thng d thu gn th ta ch cn chng minh cc phn t ca aSunguyn t cng nhau vi m.

Tht vy, v (a, m) = 1 v (xi, m) = 1 vi mi i (axi, m) = 1 aSl h thng dthu gn modulo m.

II.2.1.6 nh l Eurler

Cho a, m l cc s nguyn tho mn (a, m) = 1. Khi ( ) 1(mod )ma m

Chng minh

Giy1,y2, ,y(m) l mt h thng d thu gn modulo m.

V (a, m) = 1 nn ay1, ay2, , ay(m) cng l mt h thng d thu gn modulo m.

Do , vi mi i {1, 2, , (m)} u tn ti duy nhtj {1, 2, , (m)} sao choyiayj (mod m).

T ta c( ) ( ) ( )

( )

1 1 1

( ) (mod )m m m

mi i i

i i i

ay a y m

= = =

=

M (yi, m) = 1 vi mi i = 1, 2, , (m)Suy ra a (m) 1 (mod m).

II.2.1.7 nh l FermatChop l mt s nguyn t, a l mt s nguyn bt k khng chia ht chop. Khi

ap-1 1 (modp).

Chng minh


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Do a khng chia ht chop nn (a, p) = 1. Do , theo nh l Euler ta c

a (p) 1 (modp)

Mp l s nguyn t nn (p) =p 1 ap 1 1 (modp)

pcm.

+) Nhn xtT nh l Fermat suy ra vi mi s nguyn a v s nguyn tp th

apa (modp).

II.2.1.7 nh l Wilson

Chop l mt s nguyn t. Khi

(p - 1)! -1 (modp)

(Phn chng minh ca nh l kh n gin, xin nhng cho bn c )


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36

II.2.2. Cc v d

V d 1. Cho n l mt s t nhin bt k. Chng minh rng

n7 n 0 (mod 42).

Li gii

Do 7 l mt s nguyn t nn, theo nh l Fermat, ta cn7 n 0 (mod 7). (1)

Ta li c

n7 n = n(n6 - 1) = (n - 1)n(n + 1)(n4 + n2 + 1) 0 (mod 6) (2)

M (6, 7) = 1 nn t (1) v (2) suy ra pcm.

V d 2. Chop l mt s nguyn t ln hn 7. Chng minh rng

(3

p

2

p

- 1) 42p.Li gii

Ta c

3p 2p 1 = (3p - 3) (2p - 2) 0 (mod p). (1)

Mt khc

3p 2p 1 = (3p - 1) 2p 2 (2)

Mp > 7 p l

Do

3p 2p 1 - (-1)p 1 0 (mod 3) (3)

By gi ta cn chng minh 3p 2p 1 7

Ta c

3p 2p - 1 = 3.3p-1 2p 1

=1 1

2 23.9 2 1 3.2 2 1p p

p p

(mod 7)

=1 1

2 22 2 2 1p p

p+

+

Do (p, 3) = 1 nnp = 3k+ 1 hocp = 3k+ 2

Nup = 3k+ 1 th1 1

2 22 2 2 1p p

p+

+ =1 1

2 28 1 2 2 2 2 (mod 7)p p

k p p+ +

+

1 1 12 2 22 (2 1) 2 (8 1) 0(mod 7)

p p pk

+ +

= =


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Nup = 3k+ 2 ta chng minh tng t.

T suy ra pcm.

V d 3. Chox l mt s nguyn t. Khi , phng trnh x2 -1 (modp) (1) c nghimkhi v ch khi p = 2 hocp 1 ( mod 4).

Li gii

Nup = 2, phng trnh c nghimx = 1.

Nup 1 (mod 4) 1

2

p l s chn

Khi ta s chng minhx =1

( )!2

p l mt nghim ca (1).

Tht vy, ta c

1 ( 1)(mod )

2 ( 2)(mod )...

1 1( )(mod )

2 2

p p

p p

p pp

+

Do 1

21 1

( )! ( 1) ( 1)( 1)...2 2

pp pp p

+ (mod p)

1

2 2

12

1 1 1( !) ( 1) .1.2...( ). ...( 1)(mod )2 2 2

( 1) ( 1)!

p

p

p p p p p

p

+

=

M theo nh l Wilson ta c (p - 1)! -1 (modp)

p 1 (mod 4) nn1

2

p l s chn

T suy ra1

2( 1) ( 1)! 1(mod )p

p p

Do x = 1!2

p l mt nghim ca (1).

Ngc li, nup l s nguyn t l v (1) c nghim

Khi , gi a l mt nghim ca (1)

Ta c


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1 11 2 2 2( ) ( 1) (mod )

p ppa a p

=

M theo nh l Fermat, ta c ap-1 1 (modp) nn1

21 ( 1) (mod )p

p

suy ra 12

p chn hayp 1 (mod 4)

pcm.

V d 4. Cho a, b l hai s nguyn tha mn 24a2 + 1 = b2. Chng minh rng c mt vch mt trong hai s chia ht cho 5.

Li gii

Ta c 24a2 b2 = 1 khng chia ht cho 5 nn a v b khng th cng chia ht cho 5.

Gi s a v b cng khng chia ht cho 5

Theo nh l Fermat ta c

a4 1 0 (mod 5)

b4 1 0 (mod 5)

Do

(a2 b2)(a2 + b2) = a4 b4 0 (mod 5)

Nu a2 + b2 0 (mod 5) th 25a2 + 1 = a2 + b2 0 (mod 5) (v l)

Vy a2 b2 0 (mod 5)

23a2 + 1 = b2 a2 0 (mod 5) 23a2 + 1 0 (mod 5)

V (a, 5) = 1 nn a 1 (mod 5) hoc a 2 (mod 5)

Nu a 1 (mod 5) th

0 23a2 + 1 23( 1)2 + 1 -1 (mod 5) (v l)

Nu a 2 (mod 5) th

0 23a2 + 1 23( 2)2 + 1 3 (mod 5) (v l).

Vy iu gi s l sai

T ta c pcm.

V d 5. Chop l mt s nguyn t, a v b l hai s nguyn dng.Chng minh rng:

abp bap 0 (modp).

Li gii

Ta c

abp bap = ab(bp-1 ap-1 )


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Nu ab p th hin nhin abp bapp.

Nu ab khng chia ht chop (a, p) = (b, p) = 1

ap-1 1 0 (modp) v bp-1 1 0 (modp)

ap-1 bp-1 0 (modp)

abp bap 0 (modp)pcm.

V d 6. Cho a l mt s nguyn. Chng minh rng a2 + 1 khng c c nguyn t dng4k+ 3. T suy ra cc phng trnh sau khng c nghim nguyn dng.

a) 4xy x y =z2b)x2 y3 = 7.

Li gii

Gi s a2 + 1 c c nguyn tp = 4k+ 3 (a, p) = 1.

Khi

ap-1 + 1 = a4k+ 2 + 1 = (a2)2k + 1 + 1 a2 + 1 (1)

Mt khc, theo nh l Fermat, ta c

ap 1 1 p (2)

T (1) v (2) suy ra 2 pp = 2 (v l v 2 khng c dng 4k+ 3)

Vy a2 + 1 khng c c nguyn t dng 4k+ 3.

p dng

a) Ta c

4xy x y =z2

(4x - 1)(4y - 1) = 4z2 + 1

(4x - 1)(4y - 1) = (2z)2 + 1

Do 4x 1 3 vi mi x nguyn dng v c dng 4k+ 3 nn n c t nht mt cnguyn t dng 4k+ 3

M (2z)2

+ 1 khng c c nguyn t dng 4k+ 3 nn phng trnh trn v nghim.b) Ta c

x2 y3 = 7 x2 + 1 =y3 + 8

x2 + 1 = (y + 2)(y2 2y + 4)

Nu y chn th

(y + 2)(y2 2y + 4) 4


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x2 + 1 4 x2 -1 (mod 4) v l

Do y l y = 4k+ 1 hocy = 4k+ 3

Nuy = 4k+ 1

y + 2 = 4k+ 3 nn c c t nht mt c nguyn t dng 4k+ 3.

M x2

+ 1 khng c c nguyn t dng 4k+ 3 phng trnh trn khng c nghimtrong trng hp ny.

Nu y = 4k+ 3 y2 2y + 4 (-1)2 2(-1) (mod 4) 3 (mod 4)

nn phng trnh trn cng v nghim trong trng hp ny.

T ta c pcm.

V d 7. Cho a, b l cc s nguyn. p l mt s nguyn t c dng 4k + 3. Chng minhrng nux2 +y2p thxp vyp. T suy ra phng trnh sau v nghim nguyn:

x2 + 2y + 4y2 = 37Li gii

Gi sp = 4k+ 3 l s nguyn t tha mnx2 +y2p nhngx khng chia ht chopx2khng chia ht chopy khng chia ht chop.

Theo nh l Fermat ta c

xp 1 1 p (x2)2k + 1 1 p

Tng t, ta cung c

(y2

)2k + 1

1 psuy ra

(x2)2k + 1 + (y2)2k + 1 2 (modp)

M (x2)2k + 1 + (y2)2k + 1x2 +y2p 2 pp = 2 (v l)

Vyxp vy p.

p dng

Ta c

x

2

+ 2x + 4y

2

= 37 (x +1)2 + (2y)2 = 38 19 = 4.4 + 3

Do x + 1 19 v 2y 19.

Vx + 1 v 2y khng th cng bng 0 nn |x + 1| 19 hoc |2y| 19

Khi (x + 1)2 + (2y)2 192 > 38 nn phng trnh trn v nghim.

V d 8. Chop 7,p nguyn t. Chng minh rng s S = 111 (p 1 ch s 1) chia htchop.


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41

Li gii

Ta c

S=110 1

9

p

Vp 7 nn (10,p) = 1Do

10p 1 1 p

10p 1 1 (10 - ) = 9

M (p, 9) = 1 nn 10p 1 1 9p

T suy ra S p.

pcm.

V d 9. [IMO - 2005] Cho dy s (an) xc nh nh sau

an = 2n + 3n + 6n 1 vi n = 1, 2,

Tm s t nhin nguyn t cng nhau vi mi s hng ca dy trn.

Li gii

Ta s chng minh rng vi mi s nguyn tp u tn ti mt s hng an chia ht chop.

Tht vy, ta c a2 = 22 + 32 + 62 1 = 48 chia ht cho 2 v 3.

Xtp 5

Ta c

(2,p) = 1; (3,p) = 1; (6,p) = 1

Do , t nh l Fermat suy ra

2p 1 3p 1 6p 1 1 (modp)

T d dng chng minh c 6ap 2p

M (p, 6) = 1 nn ap 2p

Do ch c s 1 l s t nhin duy nht nguyn t cng nhau vi mi s hng ca dy(an).

V d 10. [IMO - 2003] Tm s nguyn dng knh nht sao cho tn ti cc sx1,x2, ,xksao cho:

x13 +x2

3 + +xk3 = 20022002.

Li gii

Ta c 2002 4 (mod 9) 20023 43 1 (mod 9)

20022002 = (20023)667.2002 2002 (mod 9) 4 (mod 9)


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42

Mt khc, vi mi s nguyn a ta c

a3 1 (mod 9) hoc a3 0 (mod 9)

Do

x13 ;x1

3 +x23;x1

3 +x23 +x3

3 khng th ng d vi 4 modulo 9 c.

Tc l vi k 3 th phng trnh trn khng c nghim nguyn.Ta s chng minh k= 4 l gi tr cn tm.

Tht vy, ta c 2002 = 103 + 103 + 13 + 13

M 2002 = 3. 667 + 1

20022002 = 2002. (2002667)3

= (103 + 103 + 13 + 13 ) (2002667)3

= (10.2002667)3 + (10.2002667)3 + (2002667)3 + (2002667)3

Vy vi k= 4 th phng trnh trn c nghim.

KL: k= 4 l gi tr cn tm.

V d 11. (Balan - 98) Cho dy s (an) c xc nh nh sau:

a1 = 1, 1[ ]2

n n na a a= + vi mi n 2.

Chng minh rng dy (an) c v s s hng chia ht cho 7.

Li gii

Gi s trong dy trn c hu hn s hng chia ht cho 7.

Khi gi ak l s hng cui cng ca dy chia ht cho 7.iu ny l tn ti v c a5 = 7 7.

Ta c

a2k = a2k-1 + ak; a2k + 1 = a2k+ ak.

Do

a2k + 1a2ka2k 1x (mod 7).

Vix 7.

Mt khca4k 2

= a4k 3 + a2k 1a4k 3 +x (mod 7)

a4k 1 = a4k 2 + a2k 1a4k 2 +x (mod 7) a4k 3 + 2x (mod 7)

a4k = a4k 1 + a2ka4k 3 + 3x (mod 7)

a4k + 1 = a4k+ a2ka4k 3 + 4x (mod 7)

a4k + 2 = a4k + 1 + a2k + 1a4k 3 + 5x (mod 7).


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43

a4k + 3= a4k + 2 + a2k + 1a4k 3 + 6x (mod 7)

Ta c a4k 3 7 v x 7 nn trong cc s a4k 3 + ix (i = 1, 2, , 6.) phi c s chia htcho 7 nn trong cc s a4k 3 + i , i = 1, 2, , 6 phi c s chia ht cho 7.

Vy iu gi s l sai

T ta c iu phi chng minh.V d 12. Chng minh rng nup l mt s nguyn t th (p - 2)! 1 p.

Nup > 5 th (p - 2)! 1 khng l lu tha cap.

Li gii

Theo nh l Wilson ta c

(p - 1)! - 1 (mod p) p 1 (modp)

(p - 1)(p - 2)! p - 1 (mod p) (*)

Do (p, p

- 1) = 1 nn(*) (p - 2)! 1 (modp)

Vi p > 5, gi s (p - 2)! 1 =pn

Ta c

(p - 2)! p 1 pn + 1 p 1

M

pn + 1 = (pn 1) + 2 2 (modp - 1)

2 0 (mod p 1)

V l v p > 5

Vy c iu phi chng minh.

V d 13.

a) Cho a l mt s nguyn dng. Chng minh rng mi c nguyn tp ca a2 +1 vip > 2 thp u c dng 4k+ 1.

b) Chng minh rng c v s s nguyn t c dng 4k+ 1.

Li gii1) Gi sp l s nguyn t dng 4k+ 3 vp | a2 + 1

Khi

a2 -1 (modp) 1

2 2 2 12( ) ( )p

ka a

+= -1 (modp)

Mt khc theo nh l Fermat, ta c


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44

ap 1 1 (modp)

Do

2 0 (modp)

V l v p > 2

Vy c iu phi chng minhb) Theo phn a) ta c mi c nguyn t ca (n!)2 + 1 u c dng 4k+ 1

Gi s c hu hn c nguyn t dng 4k+ 1 khi gi p l s nguyn t ln nht cdng 4k+ 1.

Ta c mi c nguyn t ln hn 2 ca (p!)2 + 1 u c dng 4k+ 1 v u khng lnhnp (dop ln nht dng 4k+ 1)

((p!)2 + 1, q) = 1 vi mi q p d (p!)2 + 1 khng c c nguyn t nh hn hocbngp (v l)


V d 14. Chng minh rng vi mi s nguyn tp, tn ti v hn cc s nguyn dngn sao cho

2n np (*)

Li gii

Nup = 2 th mi n chn u tha mn (*)

Nup > 2, theo nh l Fermat, ta c

2p 1 1 (modp)

2m(p - 1) 1 modp

Chn m = kp -1, n = m(p - 1) = (kp - 1)(p - 1) 1 (modp)

Khi

2n - n = 2m(p - 1) n 1 1 0 (modp)

T suy ra pcm.

V d15. (Bulgarian 95) Tm s cc s t nhin n > 1 sao cho

a25 a 0 (mod n)Vi mi s t nhin a.

Li gii

Vi mi s nguyn t p th

(p2,p25 -p) =pp25 p p


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Do n p2 vi mi p nguyn t.

suy ra n l tch ca cc s nguyn t phn bit

(S t nhin nh vy c gi l s squarefree)

Mt khc

225 2 = 2.32.5.7.13.17.241Nhng n khng th chia ht cho 17 v 241 v

325 3 -6 (mod 17)

v 325 3 29 (mod 241)

By gi ta xtp = 2, 3, 5, 7, 13.

+)p = 2, ta c vi mi a nguyn th a25 a 2.

+)p = 3. Nu a 3 th a25 a 3

Nu a 3 a2 1 (mod 3) a25 = (a2)12. aa (mod 3)

a25 a 3.

Tng t chop = 5, 7, 13 th a25 ap vi mi a nguyn.

Do n chnh l tch ca k(1 k 5) s trong 5 s trn

T suy ra c 25 1 = 31 s t nhin nh vy

V d 16. Cho n 5 l s t nhin .Chng minh rng

n

n )!1( n-1 .

Li gii

a) Trng hp 1. n l s nguyn t

Theo nh l Winson (n-1)! -1(mod n) suy ra ((n-1)!+1 n

Ta c

11)!1(11)!1()!1(

+

=

+

=

n

n

nn

n

n

n(v 1

10


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)(n)k(nn

)!(n11

1=

.

+) n =p 2 vip l mt s nguyn t.

Do p =2 n, n 5 suy rap 1233 2 +> ppp

12 2


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II.2.3. Bi tp

Bi 1.

a) Cho a l s nguyn sao cho (a, 7) = 1. Chng minh rng

a12 - 1 7.

b) a l s nguyn dng sao cho (a, 240) = 1. Chng minh rnga4 - 1 240.

Bi 2. Cho a1 + a2 + ... + an 30, v a1, a2, ..., an nguyn. Chng minh rng

.30... 55251 naaa +++

Bi 3. Chng minh rng

n7 - n 42

vi mi s nguyn n.

Bi 4. Cho n nguyn dng. Chng minh rnga) 32

143 ++n

11.

b) 1921102 +

+n

23.

c)2622

+n

16 (mod 37).

Bi 5. Cho p l s nguyn t ln hn 17. Chng minh rng

P16 1 (mod 16320).

Bi 6. Chop l s nguyn t l. Chng minh rng

2(p-3)! -1 (modp).Bi 7. Cho n l hp s, n 4. Chng minh rng (n - 1)! 0 (mod n).

Bi 8. Chop v q l hai s nguyn t phn bit. Chng minh rng

qp - 1 +pq - 1 1 (modpq).

Bi 9. Cho a, b l cc s nguyn,p l s nguyn t. Chng minh rng

(a + b)pap + bp (modp).

Bi 10. Chng minh rng n v n + 2 l cp s nguyn t sinh i khi v ch khi

4[(n - 1)! + 1] + n 0 (mod n(n + 2)).

Bi 11. Chop l mt s nguyn t l. Chng minh rng s m =8

19 pl mt hp s l,

khng chia ht cho 3 v 3m - 1 1 (mod m).

Bi 12. Chng minh rng vi mi s nguyn t p, tn ti v hn s nguyn dng n thamn

2n - n p.


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Bi 13. Tm tt c cc s nguyn tp sao cho

p)(p mod1522

Bi 14. Cho a, b l hai s nguyn dng sao cho 2a - 1, 2b - 1 v a + b u l cc snguyn t. Chng minh rng aa + bb v ab + ba u khng chia ht cho a + b.

Bi 15. Tm s nguyn dng n sao cho n = a

2

+ b

2

vi a, b l hai s nguyn dngnguyn t cng nhau v ab chia ht cho mi s nguyn t nh hn hoc bng n .

Bi 16. Cho p l mt s nguyn t l. Chng minh rng khng tn ti x, y nguyn thamn h thc

xp +yp =p[(p - 1)!]p.

Bi 17. Tm ba s nguyn tp, q, rsao chop2 + q2 + r2 cng l s nguyn t.


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II.3. S chnh phng mod p

II.3.1. L thuyt

II.3.1.1 nh ngha 1Cho s nguyn tp. S nguyn a c gi l s chnh phng (mod p) nu tn ti

s nguynx sao chox2

a ( modp).Nhn xt:

+) Mi s chnh phng u l s chnh phng (modp)

+) a 0 (modp) th a2a (modp) nn mi a 0 (modp) u l s chnh phng(modp). Do , t y v sau, ta ch xt s nguyn a sao cho (a, p) = 1.

+) Mi s nguyn l u l s chnh phng (mod 2)

II.3.1.2 K hiu Legendre

Cho p l s nguyn t l.

1a

p

=

nu a l s chnh phng modp

a

p

= -1 nu a khng l s chnh phng modp.

K hiu trn gi l k hiu Legendre.

II.3.1.3 nh l 1

Chop l mt s nguyn t l. Khi 1

21 1(mod )pa

a pp

=

(1)

1

21 1(mod )pa

a pp

=

(2)

Chng minh

Gi s a l s chnh phng modp, khi , tn ti s t nhinx sao cho

x2a (modp)

Do (a, p) = 1 nn (x, p) = 1.

Theo nh l Fermat ta c1

1 21 (mod )p

px a p

Ngc li, nu c (1) th vi mi k {1, 2, ,p - 1} c duy nht mt s k {1, 2, ,p- 1} sao cho k.ka (modp)


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Nu tn ti k = kth k.k= k2a (modp) a l s chnh phng (modp).

Tri li, tp {1, 2, ,p - 1} c chia thnh1

2

ktp con {k, k} ri nhau sao cho k.k

a (modp). (p - 1)! 1

2 1(mod )p

a p

Mt khc, theo nh l Wilson, ta c(p - 1)! -1 (modp)

T suy ra

1 -1 (modp) p = 2 (v l vp l)

Vy (1) c chng minh xong.

Theo nh l Fermat, ta c ap 1 1 (modp)

1

2

1

2

1(mod )

1(mod )

p

p

a p

a p

Do , (2) c chng minh.

II.3.1.4 nh l 2(B Gauss)

Chop l s nguyn t l, a l s nguyn, (a, p) = 1. Xt tp

{ka| k= 1, 2, ,1

2

p }.

Gi rkka (modp), 1 rkp.

Gi n l s cc s rk thuc khong ( ;2

pp ). Khi

12 ( 1) (mod )

pna p

.

(n chnh l s bi s ca a trong khong ( ;2

pp ))

Chng minh

Ta c

kark(modp)

Cho kchy t 1 n (p - 1)/2 ri nhn cc ng thc li ta c:1

1 22

1

1! (mod )

2

pp

kk

pa r p

=


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GiA = |2k kp

r r >

; B = |2k kp

r r


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52

v2 1

82 ( 1)p

p

=

Chng minh

Gi sp = 4k+ 1, tp cc s chn trong khong (0;p) l {2i | 1 i 2k}

Khi , s cc s chn trong khong ( ;2p p ) l n = k.

Tng t nh vy, nup = 4k 1 th ta cng tnh c s cc s chn trong khong

( ;2

pp ) l n = k.

Do , theo h qu 1, ta c1

22 ( 1) (mod )p

n p

(-1)k(mod p)

2 l s chnh phng modp1

21 2 ( 1) (mod )p

k p

kchn p 1 (mod p).

2 khng l s chnh phng modpkl p 3 (modp)

pcm.

H qu 4. Cho p l s nguyn t l. Gi n l s cc bi ca 3 trong khong ( ;2

pp ).

Khi

123 ( 1) (mod )p n p

Chng minh

c suy ra hin nhin t nh l, trong trng hp a = 3.

H qu 5. Cho p l s nguyn t c dng 6k 1. Khi , 3 l s chnh phngmodp khi v ch khip 1 (mod 12).

Chng minh

Gi sp = 6k+ 1. Tp hp cc s l bi ca 3 trong khong (0;p) l{3i | 1 i 2k}

M

3i >1

12 6 6

p pi k i k > = + +


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53

Do , s cc s l bi ca 3 trong khong ( ;2

pp ) l n = k.

Tng t khip = 6k 1, ta cng chng minh c n = k.

Vy trong c hai trng hp ta u chng minh c s cc s l bi ca 3 trong khong

( ;2

p

p ) l n = k.

V vy, theo h qu 4, ta c1

23 ( 1) (mod )p

n p

(-1)k(mod p)

3

( ) 1p

= (-1)k 1 (modp)

kchn

p 1 (mod 12).pcm.

II.3.1.5 nh l 2

Chop l s nguyn t. a b (modp), (a, p) = (b, p) = 1. Khi

( ) ( )a b

p p= .

Chng minh

Do ab (modp)1 1

2 2 (mod )p p

a b p

( ) ( )a b

p p=

pcm.

II.3.1.5 nh l 4

Cho p l mt s nguyn t. a1, a2, , an l cc s nguyn khng chia ht cho p.Khi

1 2 1 2...( ) ( )( )...( )n na a a aa a

p p p p= Chng minh

(Phn chng minh kh n gin, xin nhng cho bn c)

II.3.1.6 nh l 5

Chop l s nguyn t l, a l s nguyn khng chia ht chop. Khi


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54

12 ( 1) (mod )

psa p

Vi s =

1

2

1

p

k

ka

p

=

Chng minh

Gi n l s cc bi ca a trong khong ( ;2

pp ), theo nh l 2, ta c

12 ( 1) (mod )

pna p

V vy, ta ch cn chng minhs n l s chn.

(Phn ny kh n gin, xin nhng li cho bn c).

II.3.1.6 nh l 7(Lut tng h Gauss)

Chop, q l hai s nguyn t l phn bit.

Khi :

a) Nu c t nht mt trong hai s c dng 4k+ 1 th

( ) ( )p q

q p= .

b) Nu c hai s c dng 4k+ 3 th

( ) ( ).p q

q p

=

Chng minh

a) t

s1 =

1 1

2 2

21 1

;

p q

i i

iq ips

p q

= =

=

Khi

s1

+ s2

=1 1

( )( )2 2

p q

(tnh cht ny c chng minh trong ti liu Mt s bi ton v phn nguyn cacng tc gi).

Dop hoc q c dng 4k+ 1 nn1 1

( )( )2 2

p q l s chn.

Nns1 vs2 c cng tnh chn l.


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55

Theo nh l 5, ta c

1

12 ( 1) (mod )

psq p

22 ( 1) (mod )q

sp q

T d dng suy ra iu phi chng minh.b) Chng minh tng t.

II.3.2. Cc v d

V d 1. Chop l s nguyn t c dng 8k+ 5 hoc 8k+ 7.x, y l hai s nguyn tha mnx2 + 2y2 chia ht chop.

Chng minh rngxp vyp.

Li gii

Gi s ngc lix p 2y2 py p.

M

x2 + 2y2px2 -2y2 (modp)

2 22

( ) ( )y

p p

=

1 =22 2 1 2

( )( ) ( ) ( )( )y

p p p p p

= = . (1)

Nup = 8k+ 5 2

12

18

1( ) ( 1) 1

2( ) ( 1) 1

p

p

p

p

= =

= =

Thay vo (1) c 1 = -1 (v l).

Nup = 8k+ 1 2

12

18

1( ) ( 1) 1

2( ) ( 1) 1

p

pp

p

= =

= =

cng khng tho mn (1)

Vy iu gi s l sai

T suy ra pcm.


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V d 2. Cho 22 1n

k= + vi n nguyn dng. Chng minh rng kl s nguyn t khi v

ch khi kl c ca1

23 1k

+ .

Li gii

Nu kl c ca

12

3 1

k

+ th ta c 123

k

-1 (mod k) (1)

3k -1 1 (mod k) (2)

Gi dl bc ca 3 modulo k

T (1) v (2) ta c d| k 1 nhng d li khng chia ht1

2

k

d = k 1 kl s nguyn t.

Ngc li, kl s nguyn tTa c kl s nguyn t dng 4l+ 1 nn theo lut tng h Gauss ta c

3( ) ( )

3

k

k=

M k 2 (mod 3) nn2

( ) ( ) 13 3

k= = (do 2 khng phi s chnh phng mod 3).

T suy ra1 1

2 23 1(mod ) 3 1 0(mod )

k k

k k

+ .pcm.

V d 3. Chng minh rng vi mi n nguyn dng, s 2n + 1 khng c c nguyn tdng 8k+ 7.

Li gii

Gi s tn ti s nguyn tp = 8k+ 7 sao chop | 2n + 1

Nu n chn, ta c

2n -1 (modp)

Suy ra -1 l s chnh phng modp

Do 1

4 321

1 ( ) ( 1) ( 1) 1p

k

p

+= = = = (v l)

Nu n l, ta c


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2n -1 (mod p) 2n + 1 -2 (mod p)

Suy ra -2 l s chnh phng mod p

Do , ta c

2 1 21 ( ) ( )( ) 1

p p p

= = = .1 = -1. (v l)


V d 4. Tmx, n nguyn dng sao cho

x3 + 2x + 1 = 2n. (1)

Li gii

Dox nguyn dng nnx3 + 2x + 1 4 n 2.

Nu n = 2, ta d dng tm cx = 1.

Xt n 3

(1) x(x2 + 2) = 2n 1 l s l nn x l

Do ,x2 + 2 3 2n (1 mod 3) (-1)n 1 (mod 3)

T suy ra n chn.

Mt khc

x3 + 2x + 1 = 2n

x3 + 2x + 3 = 2n + 2

(x + 1)(x2 x + 3) = 2n + 2.

Gi s p l mt c nguyn t cax2 x + 3 p l

Khi

2n + 2 0 (modp) 2n -2 (mod p)

M n chn suy ra2

( ) 1p

=

Ta c

1 =

2 1182

2 1 2( ) ( )( ) ( 1) ( 1)

pp

p p p

= =

T suy rap = 8k+ 1 hoc p = 8k+ 3.

M n 3 2n = 8.2n 3 0 (mod 8)

Do ta suy ra

x3 + 2x + 1 0 (mod 8).


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Mx l x = 8k 1,x = 8k 3

Nux = 8k 1 x3 + 2x + 1 ( 1)3 + 2( 1) + 1 0 (mod 8) (v l)

Nux = 8k+ 3 thx3 + 2x + 1 cng khng th chia ht cho 8.

Nux = 8k+ 5 th tha mn.

Khi x2 x + 3 52 5 + 3 7 (mod 8)

Hay

x2 x + 3 = 8l+ 7

M s dng 8l+ 7 th khng th c c nguyn t dng 8k+ 1 hoc 8k+ 3

Do (1) v nghim khi n 3.

Vyx = 1, n = 2.

V d 5. Cho p l s nguyn t dng 12k 1. Tmx, y, n nguyn khng m sao cho

3a2 + b2pn (1)

Vi a v b khng chia ht chop.

Li gii

Nu n = 0 th mi a, b nguyn khng m v a p, b p u tha mn (1)

Nu n 0 a> 0 3a2 + b2 p

Do b2 -3a2 (modp)

1 1 1

2 22 2 2( ) ( 3) ( ) (mod )p p p

b a p

bp 11

2( 3)p

ap 1 (modp)

M

(a, p) = (b, p) = 1 ap 1bp 1 1 (modp)

Suy ra1

2( 3) 1(mod )p

p

Mt khc1

6 123 1 3 3 3 3

1 ( ) ( )( ) ( 1) ( ) ( 1) ( ) ( )p

k

p p p p p p

= = = = = (*)


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Li cp = 12k 1 nn 3 l s chnh phng mod p hay3

( ) 1p

=

Thay vo (*) ta c 1 = -1 (v l)

Vy khi n > 0 th khng tn ti a, b, n tha mn.

V d 6. (Serbi - 2008) Gii phng trnh12x +y4 = 2008z

Vix, y, zl cc s nguyn khng m.

Li gii

Nuz= 0 x = y = 0.

Nu z> 0 y > 0 v 12x 251 m 2008 = 251.8

TH1:x chn

Ta c 12x +y4 = 2 2 2 2 22(12 ) ( )x

y a b+ = +

Vi a = 12x/2, b =y2.

V 2008 = 251.8 a2 + b2 251

M 251 l s nguyn t v 251 = 4k+ 3 nn c a v b u phi chia ht cho 251

V l v 12x/2 251.

TH2.x l lm tng t.

Vy phng trnh c nghim duy nhtx = y = z= 0.


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II.3.3. Bi tp

Bi 1. Chng minh rng mi c nguyn t ca n4 - n2 + 1 u c dng 12k+ 1.

Bi 2. (Ba Lan 2007) Chng minh rng phng trnh x2 + 5 = y3 khng c nghimnguyn.

Bi 3. Chng minh rng vi mi s nguyn t p th tn ti cc s nguyn a, b sao cho a2 +b2 + 1 chia ht chop.

Bi 4. Cho s nguyn tp. Chng minh rng 3p + 7p - 4 khng l bnh phng ca mt snguyn.

Bi 5. (IMO 2006) Tm tt c cc nghim nguyn ca phng trnh

1 + 2x + 22x + 1 =y2.

Bi 6. Cho m, n l cc s nguyn sao cho A =m

m n

3

1)3( ++l mt s nguyn. Chng

minh rngA l s l.

Bi 7. ( ngh IMO 2004)Chng minh rng 2n + 1 khng c c nguyn t dng 8k+ 7.

Chng minh rng 123 +n

c t nht n c nguyn t dng 8k+ 3.

Bi 8. Tm tt c cc s nguyn dng n sao cho 3n - 1 2n - 1.


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61

II. 4. nh l thng d Trung Hoa

II.4.1 L thuyt

II.4.1.1. nh l 1(nh l thng d Trung Hoa)Cho h phng trnh

1 1

2 2

(mod )(mod )

...

(mod )n n

r mr m

r m

(1)

Trong m1, m2, , mn l cc s nguyn i mt nguyn t cng nhau. Khi hphng trnh trn lun c nghim. Nux1 v x0 l hai nghim ca (1) th x1 x0 (modm1m2mn).

Chng minh

t si = 1

n

kk

i

m

m=

.

Do m1, m2, , mn i mt nguyn t cng nhau nn (si, mi) = 1.

Nn vi mi s nguyn dngsi lun tn ti mt s nguyn hi sao cho

sihi 1 (mod mi).

tx0 =1

n

i i ii

h r=

Vsimj vi miji nn

x0sihiri (mod mi) ri (mod mi)

Do x0 chnh l mt nghim ca (1).

Gi sx1 cng l mt nghim ca (1)

Ta c

x1x0 (mod mi) x1 x0mi.vi mi iM m1, m2, , mn i mt nguyn t cng nhau nn

x1 xom1m2mn.



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62

II.4.1.2. nh l 2

Cho h phng trnh

1 1

2 2

(mod )

(mod )

...(mod )n n

r m

r m

r m

(1)

iu kin cn v h trn c nghim l

rirj (mod (mi, mj)).

Khi h trn c nghim duy nht theo modulo [m1, m2, , mn].

Chng minh

(xin nhng cho bn c)

II.4.1.3. Phng php gii h (1) trong trng hp cc s mi imt nguyn t cng nhau.

Bc 1. t m = m1m2mn = Mimi vi i = 1, 2, ,n.

Bc 2. Tm cc sNi nghim ng phng trnh

Mix 1 (mod m)

Bc 3. Tm c mt nghim ca h l

x0 = 1

n

i i ii N r= l mt nghim ca h

Bc 4. Kt lun x =x0 + mt.


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63

II.4.2 Cc v d

V d 1. Gii h

4(mod11)

3(mod17)

x

x

Li giiXt phng trnh

17y 1(mod 11)

6y 1 (mod 11)

Ta d dng tm c mt nghim lN1 = 2.

Xt phng trnh

11y 1 (mod 17)

Ta cng d dng tm c mt nghim lN2 = 14.Suy ra mt nghim lx0 = 17.2 .4 +11.14.3

T tm c nghim ca h l x = 17.2.4 + 11.14. 3 + 11.17t.

V d 2. Gii h

1(mod2)

2(mod3)

3(mod5)

x

x

x

Li giiXt phng trnh

15y 1 (mod 2)

Phng trnh ny c mt nghim lN1 = 1

Xt phng trnh

10y 1 (mod 3)

Cng d dng tm c mt nghim lN2 = 1.

Xt phng trnh

6y 1 (mod 5)

Cng tm cN3 = 1 l mt nghim.

Vy x0 = 15.1.1 + 10.1.2 + 6.1.3 = 53 l mt nghim ca h

Do h phng trnh cho c nghim l

x 53 (mod 30) 23 (mod 30)


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64

V d 3. Gii h

0(mod2)

0(mod3)

1(mod5)

6(mod7)

x

x

x

x

Li gii

Xt phng trnh

42y 1 (mod 5)

2y 1 (mod 5)

Tm cN3 = 3.

Xt phng trnh

30y 1 (mod 7)

2y 1 (mod 7)

Tm cN3 = 4

T suy rax0 = 42.3.1 + 30.4.6 l mt nghim ca h

Vy h c nghim l x

V d 4. Chng minh rng vi mi s nguyn dng ktu lun tn ti ks nguyn lintip ton hp s.

Li giiGip1


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65

(2n + 1; 5n + 2) = (2n + 1, n) = 1

Do m | n(2n + 1)(5n +2) khi v ch khi xy ra mt trong cc trng hp sau:

1) m | n

2) m | 2n + 1

3) m | 5n + 24) 34096 | n v 2232008 | 2n +1

5) 34096 | n v 2332008 | 5n + 2

6) 34096 | 5n + 2 v 2232008 | 2n + 1

7) 34096 | 5n + 2 v 2332008 | n

8) 34096 | 2n + 1 v 2232008 | 5n + 2

9) 34096 | 2n + 1 v 2232008 | n

Trong mi trng hp y, theo nh l thng d Trung hoa, c duy nht mt s t nhin n

(modulo m) tha mn.Nn c tt c 9 s t nhin tha mn bi.

V d 6. ( ngh IMO 2002) Trong li im nguyn ca mt phng ta Oxy, mtim c ta l cc s nguyn A(x, y) Z2 c gi l nhn thy c t O nu trnon OA khng c im no thuc Z2, tr O vA. Chng minh rng vi mi s t nhin ntu , lun tn ti hnh vung n x n c cc nh nguyn v mi im nguyn bn trong vtrn bin ca hnh vung u khng nhn thy c t O.

Li gii

Ta c nu (x, y) = dth im );(

d

y

d

xM l im nguyn thuc on OA viA(x; y).

Do ,A(x, y) l im nhn thy c t O khi v ch khi (x, y) = 1.

Gi jip , l cc s nguyn t i mt khc nhau vi 0 i, j n

(c (n +1)2 s nguyn t nh vy)

Xt hai h sau:

)...(mod

...

)...(mod1

)...(mod0

,1,0,

,11,10,1

,01,00,0

nnnn

n

n

pppnx

pppx

pppx


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66

v

)...(mod

...

)...(mod1

)...(mod0

,,1,0

1,1,11,0

0,0,10,0

nnnn

n

n

pppnx

pppx

pppx

Theo nh l thng d Trung Hoa, tn ti cc s t nhin y nh vy.Mx + i v y + j u chia ht chopi,j

Do , mi im trong hnh vung n x n vi (n + 1)2 im nguyn Ai,j(x + i, y + j) trnu khng nhn thy c t O.

V d 8. (Nordic - 98) Tn ti hay khng mt dy c hn cc s t nhin

{x1,x2, ...,xn, ...} = {1, 2, ..., n, ...}

Sao choxi xj vi mi i j v

x1 +x2 + ... +xk k vi mi k= 1, 2, ...

Li gii

Ta xy dng mt dy tha mn bi nh sau:

Chnx1 = 1,x2 = 3,x3 = 2.

Gi sx1,x2, ...,xn l dy s tha mn

x1 +x2 + ... +xk k vi mi k= 1, 2, ..., n.

Gi m l s nguyn dng b nht khng nm trong dyx1,x2, ...,xn.

Do (n + 1, n + 2) = 1 nn, theo nh l thng d Trung Hoa, tn ti s nguyn x ln hn

max {x1,x2, ...,xn} v tha mn

+

+

)2(mod

)1(mod

nsmx

nsxvis =x1 +x2 + ... + xn.

Khi , t xn+1 = x,xn + 2 = m

T ta c dyx1, x2, ...,xn+1,xn + 2 tho mn

x1 + x2 + ... + xn + xn+1 = s + x n +1.

x1 + x2 + ... + xn+1 + xn + 2 = s + x + n n + 2.

Do x1 + x2 + ... + xk kvi mi k= 1, 2, ..., n + 2

C tip tc nh vy ta thu c dy s v hn tha mn yu cu bi ton.


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67

II.4.3. Bi tp

Bi 1. Gii cc h phng trnh sau:

a)

)7(mod3

)11(mod1

x

x

b)

)7(mod5

)5(mod3

)3(mod1

x

x

x

c)

)11(mod37

)8(mod25

)12(mod15

x

x

x

Bi 2. (IMO 1989) Chng minh rng vi mi s t nhin n, lun tn ti n s t nhin lintip m c n s u khng phi lu tha ca mt s nguyn t.

Bi 3. (Hn Quc 1999) Tm tt c cc s t nhin n sao cho 2n - 1 chia ht cho 3 v tn

ti m Z sao cho 4m2 + 1 chia ht cho3

12 n.

Bi 4. Chng minh rng vi mi s nguyn dng k ln tu u tn ti ks nguynlin tip gm ton hp s.

Bi 5. Cho S= {a1, a2, ..., an} Z. Chng minh rng tn ti mt s bZsao cho tpb.S= {ba1 , ba2 , ..., ban} m mi phn t ca n u l lu tha ln hn 1 ca mt snguyn.Bi 6. Chof(x) l mt a thc vi h s nguyn. Gi s c mt tp hu hn cc s nguynt {p1, p2 ,..., pn} m vi mi n th u tn ti pi | f(n). Chng minh rng tn ti mt snguyn tp sao cho vi mi n thf(n) p.


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68

Chng III

MT S VN KHC

III.1. Lu tha ca mt s nguyn

III.1.1 S chnh phng

III.1.1.1 nh ngha

S t nhin n c gi l mt s chnh phng nu tn ti m nguyn sao cho n = m2.

Nhn xt: Nu phn tch tiu chun n = kkppp ...21 21 th:

+) n l s chnh phng khi v ch khi i chn vi mi i = 1, 2, ..., k.

+) n l lu thas ca mt s nguyn khi v ch khi i s vi mi i = 1, 2, ..., k.

III.1.1.2 Cc v d

V d 1. Tm tt c cc s chnh phng c dngA = ab1985Li gii

Ta c

4452 = 198025 0

Do

b > k 20a + b + k> 20a + b - k> 20a

T suy ra

nma

kbakbaabc .

4

)20)(20(=

+++=


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69

M 20a + b + kv 20a + b - ku ln hn 4a nn m v n u ln hn 1

Nn abc l hp s, tri vi gi thit


V d 3. Chng minh rng mt s chnh phng lun c s c s nguyn dng l l vngc li mt s t nhin c s c nguyn dng l l th s phi l s chnh phng.

Li gii

a) NuA l mt s chnh phng th phn tch tiu chun caA l

nkn

kk pppA 22221 ...

21=

T suy ra s c s nguyn dng caA l

(2k1 + 1)(2k2 + 1)...(2kn + 1) l s l.

b) Ngc li: NuA c s c s l l

Xt phn tch tiu chun caA

nkn

kk pppA ...21 21=

T suy ra s c s caA l

(k1 + 1)(k2 + 1)...(kn + 1) l s l

Do , tt c cc nhn t trn u l

Suy ra ki chn vi mi i = 1, 2, ..., n

T suy raA l s chnh phng.

V d 4. (Romani - 2004) Tm tt c cc s khng m n sao cho tn ti hai s nguyn av b sao cho

n2 = a + b v n3 = a2 + b2.

Li gii

Ta c

2(a2 + b2) (a + b)2 2n3 n4 0 n 2.

Nu n = 0, chn a = b = 0

Nu n = 1, chn c a = 1, b = 0.

Nu n = 2, chn c a = b = 2.Vy n = 0, n = 1, n = 2 l cc gi tr cn tm.

V d 5. Chng minh rng nu 2n (n N*) l tng ca hai s chnh phng th n cng ltng ca hai s chnh phng.

Li gii

Gi n l s t nhin tha mn


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70

2n = a2 + b2 vi a, b N. (1)

T (1) suy ra a v b cng tnh chn l

Do a +b v a - b l s chn

t

==+

ybaxba

22

Khi a = x + y v b = x - y

Thay vo (1) ta c

2n = (x + y)2 + (x - y)2

n =x2 +y2.


V d 6. Cho a, b l hai s nguyn dng sao cho a

2

+ b

2

ab + 1.Chng minh rng

1

22

++

ab

bal s chnh phng.

Li gii

t

1

22

++

=ab

bak (1)

Khng mt tnh tng qut ta gi s a b.

Khi

(1) a2 + b2 - kab - k= 0

a2 - kb.a + b2 - k= 0.(*)

Gi s rng kkhng phi s chnh phng. Xt phng trnh

x2 - 2bx + b2 - k= 0 (3)

Ta c al mt nghim ca (3), gi a1 l mt nghim cn li

Khi

a + a1 = kbDo a, k, b nguyn nn suy ra a1 cng l s nguyn.

+) Nu a1 = 0 th k= b2, tri vi iu gi s.

+) Nu a1 < 0, ta c

a12 - kba1 + b

2 = k

M a1 nguyn m nn a12 -kba1 + b

2 > -kba1 = (-a1b)k> k(v l)


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71

+) Do o a1 > 0 th

a.a1 = b2 - ka1 = a

a

kb a1 + b1. ta c

a12

- kb1a1 + b12

- k= 0 (4)Li xut pht t (4) suy ra phng trnh (3) c nghim t nhin mi (a2, b2) m

a2+ b2 < a1 + b1.

T suy ra phng trnh hai bin (*) c v hn nghim t nhin tha mn

(a1, b1), (a2, b2), ... tha mn a1 + b1 > a2 + b2 > ...(v l)

Nn iu gi s l sai.


Vd 7. Tm n nguyn dng sao cho

A = 28 + 211 + 2n

l mt s chnh phng.

Li gii

NuA l s chnh phng th

28 + 211 + 2n = x2 2n = (x - 48)(x + 48)

Vix l mt s nguyn dng no .

Khi

x - 48 = 2s vx + 48 = 2n - s, n > 2sT suy ra

2n - s - 2s = 96 2s(2n - 2s - 1) = 3.25.

T suy ra

=

=

=

= 12

5

312

52 n

sssn

KL: n = 12.

V d 8. Chng minh rngA = 1k+ 9k+ 19k+ 2013k

khng phi s chnh phng vi mi knguyn dng l.

Li gii

Vi mi k nguyn dng l, ta c

1k 1 (mod 4)


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72

9k 1 (mod 4)

19k -1 (mod 4)

2013k 1 (mod 4)

Nn A 2 (mod 4)

VyA khng th l s chnh phng.V d 9. Tm s t nhin n sao cho n - 50 v n + 50 u l s chnh phng.

Li gii

Ta c

=+

=2

2

50

50

bn

an

vi a, b nguyn dng v a > b.

Suy ra b2

- a2

= 100 (b - a)(b + a) = 22

.52

.Do b - a < b + a v chng c cng tnh chn l nn a + b v b - a phi l cc s chn. Do

=

=

=+

=

26

24

50

2

b

a

ab

ab

T tm c n = 626.

V d 10. Chng minh rng vi mi s nguyn dng n th s

24)21217()21217(

nn

+

l mt s nguyn v khng phi s chnh phng.

Li gii

Ta c 4)12(21217 +=+ v 4)12(21217 =

Do

24

)21217()21217(nn +

=

22

)12()12(.

2

)12()12( 2222 nnnn +++

t

2

)12()12( 22 nnA

++= v

22

)12()12( 22 nnB

+=

S dng nh thc Niutn ta suy ra

( 2 + 1)2n = x + y 2 v ( 2 - 1)2n = x - y 2


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73

Vix, y l cc s nguyn dng.

T suy ra

xAnn

=++

=2

)12()12( 22v yB

nn

=+

=22

)12()12( 22

NnA.B l mt s nguyn dng.Mt khc, ta li c

A2 - 2B2 = (A - B 2 )(A + B 2 ) = ( 2 +1)2n( 2 - 1)2n = 1

Do A vB nguyn t cng nhau.

Vy chng minhAB khng phi s chnh phng ta ch cn chng minh mt trong hais khng phi l s chnh phng.

Ta c

2 )12()12(

22 nn

A ++= = [ ] 12 )12()12(2

++

nn

2

)12()12( 22 nnA

++= =

[ ]1

2

)12()12(2

++ nn

T d dng suy ra cA khng phi s chnh phng

Vy suy ra iu phi chng minh.

V d 11. (Polish - 2001) Cho a, b l cc s nguyn sao cho vi mi n th 2na + b u ls chnh phng. Chng minh rng a = 0.

Li giiGi s a 0.

Nu a 0, b = 0 th 21a + b v 22a + b khng th ng thi l s chnh phng.

Do a v b u phi khc 0.

T d dng suy ra c a v b u dng.

Xt hai dy s (xn) v (yn) nh sau

baybax nnn

n +=+=+2222

D thy (xn) v (yn) l hai dy s nguyn dng, n iu tng v hn.Ta c

(xn +yn)(xn -yn) = 3b

Suy ra 3b xn +yn vi mi n 3b xn + yn vi mi n (v l)

Vy iu gi s l sai



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V d 12. Tm n nguyn dng sao cho

n2 + 3n

l s chnh phng.

Li gii

Gi sn2 + 3n = m2

vi m l mt s nguyn dng no .

Ta c

(m - n)(m + n) = 3n

T suy ra

=+

=kn

k

nm

nm

3

3(*)

M m +n > m - n nn n - 2k 1.

Nu n - 2k= 1

T (*) suy ra 2n = 3n - k- 3k= 3k(3n - 2k- 1) = 2.3k

3k= n = 2k+ 1 k= 0, k= 1.

T tm c n = 1, n = 3.

Nu n- 2k 2 k n - k- 2

T (*) suy ra 2n = 3n - k- 3k 3n - k- 3n - k - 2 = 8.3n - k - 2

Theo bt ng thc Bernoulli ta c

8.3n - k - 2 = 8.(1 + 2)n - k - 2 8[1 + 2(n - k- 2)]

= 16n - 16k- 24.

Do

2n 16n - 16k- 24 8k+ 12 7n .

M n - 2k 2 nn n 2k+ 2 7n 14k+ 14

T suy ra 8k+ 12 14k+ 14 v l v k 0.

KL: n = 1, n = 3.


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III.1.2 Lp phng ca mt s nguyn

V d 1. Chng minh rng nu n l lp phng ca mt s nguyn (n -1) th

n2 + 3n + 3 khng th l lp phng ca mt s nguyn.

Li gii

n = 0 th n2

+ 3n + 3 = 3 m3

.Gi s n2 + 3n + 3 = k3 vi k l mt s nguyn no

V n l lp phng ca mt s nguyn nn

n(n2 + 3n + 3) = l3

(n + 1)3 - 1 = l3

Vi n 0, n 1, n nguyn th 0 < 3n2 + 3n < 3n2 + 3n + 1

n3 < n3 + 3n2 + 3n < n3 + 3n2 + 3n + 1

n3

< n3

+ 3n2

+ 3n < (n + 1)3

Do n3 + 3n2 + 3n khng th l lp phng ca mt s nguyn.

Vy iu gi s l sai

Do ta c iu phi chng minh.

V d 2. (Iran - 98)Chng minh rng khng c s t nhin no c dng abab trong h cs 10 l lp phng ca mt s nguyn. Hy tm c s b nh nht sach trong h c s b,

c t nht mt s c dng abab l lp phng ca mt s nguyn.

Li gii

Ta c abab = 101 ab l lp phng ca mt s nguyn th

101 | ab (v l)

Vy abab khng th l lp phng ca mt s nguyn.

Xt trong h c s b

Ta c

))(1()1( 2)(2)( bannabnabab nn ++=+=

vi a, b < n v n2 + 1 > an+ b.

Nu n2 + 1 khng chia ht cho mt s chnh phng no th n2 + 1 =p1p2...pk

Khi an + b phi chia ht cho (p1p2...pk)2 v l

Vy n2 +1 phi chia ht cho mt s chnh phng.

Th trc tip thy n = 7 l s nh nht nh vy (n2 + 1 = 50)

Mt khc thy 10002626 )7( = = 103.


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76

Do n = 7 chnh l s cn tm.

Vd 3. Chox, y l cc s t nhin tha mnx2 +y2 + 6 chia ht choxy.

Chng minh rngxy

yx 622 ++l lp phng ca mt s t nhin.

Li giiVx2 +y2 + 6 xy nnx2+y2 + 6 =pxy (1)

Gi (x0,y0) l nghim ca (1) v tha mnx0 +y0 nh nht.

Khng mt tnh tng qut, gi sx0 y0

Xt phng trnh

y2 -px0y +x02 + 6 = 0 (2)

D thyy0 l mt nghim ca (2). Giy1 l nghim cn li

Khi , theo Viet ta c

+=

=+

62010

010

xyy

pxyy(3)

D thyy1 > 0 nn (x0,y1) cng l mt nghim nguyn dng ca (1)

Do

x0 + y0 x0 +y1y0 y1

Nux0 =y0 th t (1) ta c

20

020

20 6

2

62

xxx

x

p +=

+

= Z

x0 = 1 p = 8 = 23

Nux0


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Do tm cy0 = 1,y1 = 7 (khng tha mn iu kinx0


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78

III.1.3 Cc bi ton biu din mt s nguyn thnh tng cc lutha

V d 1. Chng minh rng nu3

12 nl tch ca hai s t nhin lin tip th n l tng

bnh phng ca hai s nguyn lin tip.

Li gii

Gi s n l s t nhin m

3

12 n= a(a + 1) (1)

vi a l mt s t nhin no .

Ta c

(1) n2= 3a2 + 3a + 1

4n2

= 12a2

+ 12a + 4 (2n - 1)(2n + 1) = 3(2a + 1)2.

Do 2n - 1 v 2n + 1 l hai s l lin tip v (2n - 1, 2n + 1) = 1 nn

=

=+

=+

=

(**)12

312

(*)12

312

2

2

2

2

pn

mn

pn

mn

T (*) suy rap2

= 3m2

+2 (v l v s chnh phng chia 3 ch d 0 v 1)Do , ch c (**) xy ra.

T (**) suy ra m vp u l

tp = 2k+ 1 ta c

2n =p2 + 1 = (2k+1)2 + 1 = 4k2 + 4k+ 2

n = (k+1)2 + k2.

l iu phi chng minh.

V d 2. (Nga - 1996) Chox, y, p, n, kl cc s t nhin tha mnxn +yn =pk. (1)

Chng minh rng nu n > 1, n l vp l mt s nguyn t l th n l mt lu tha cap.

Li gii

t m = (x, y) thx = ma,y = mb vi (a, b) = 1.

Khi


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79

(1) mn(an + bn) =pk. (2)

Do p nguyn t nn t (2) suy ra m =pq

Do

(2) an + bn =pk- nq.

M n l nnan + bn = (a + b)(an - 1 - an - 2b + ... - abn - 2 + bn - 1) (3)

= (a + b)A

ViA = an - 1 - an - 2b + ... - abn - 2 + bn - 1.

Vp l nnp > 2 suy ra hai sx vy khc tnh chn l nn t nht mt trong hai s a v bphi ln hn 1.

Do an + bn > a+ bA > 1

M

A(a + b) =pk- nq > 1

Suy raA v a + b u chia ht chop

Hay a + b =pr

T suy ra

A = an - 1 - an - 2(pr- a) + ... - a(pr- a)n - 2 + (pr- a)n - 1 = nan - 1 +Bp.

MA pnan - 1 p

a + b p, (a, b) = 1 nn a khng chia ht chop

Do n pn= mpThay vo (1) ta d dang suy ra c n = pl

V d 3. Chop l s nguyn t v a, n l cc s nguyn dng. Chng minh rng nu

2p + 3p = an

th n = 1.

Li gii

Nup = 2 th 2p + 3p = 13 n = 1

Nup > 2 p lTa c

2p + 3p 2p + (-2)p (mod 5) 0 (mod 5)

Do a 5

Gi s n > 1.


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80

Khi an 25 v32

32

++ pp

5 2p - 1 - 2p - 2.3 + ... + 3p - 1 5

M

2p - 1 - 2p - 2.3 + ... + 3p - 1 p.2p - 1 (mod 5)

suy ra p.2

p - 1

5 p = 5 (dop l s nguyn t)M 25 + 35 = 753 an, ( vi n > 1)

Vy iu gi s l sai.


III.1.4. Bi tp

Bi 1. Tm cc hm sf: NNtho mn ng thi cc iu kin sau

i)f(2n) =f(n) + n

ii)f(n) l s chnh phng th n l s chnh phng

iii)fl hm tng nghim ngt.

Bi 2. Tm tt cc cc s nguyn dng n sao cho 2n + 1 v 3n + 1 u l s chnhphng. Khi chng minh rng n chia ht cho 40.

Bi 3. Chng minh rng tng ca 3, 4, 5, hoc 6 s nguyn lin tip khng th l s chnhphng.

Bi 4. (IMO 86) Cho d l mt s nguyn khc 2, 5, 13. Chng minh rng lun tm chai s nguyn a, b phn bit trong tp {2, 5, 13, d} sao cho ab - 1 khng phi s chnh

phng.Bi 5. Tm tt c cc s nguyn dng n sao cho lu tha 4 ca s c ca n chnh bngn.

Bi 6. Chng minh rng bnh phng ca mt s nguyn l lun c dng 8k+ 1.

Bi 7. (Hungari 1998) Chox, y, zl cc s nguyn vz> 1. Chng minh rng

(x + 1)2 + (x + 2)2 + ... (x + 99)2yz.

Bi 8. Chng minh rng vi mi s t nhin n th gia n2 v (n + 1)2 lun tn ti ba s tnhin phn bit a, b, c sao cho a2 + b2 c2.

Bi 9. Cho n l s t nhin c s c s t nhin l s. Chng minh rng tch tt c cc

c s bng n sn .

Bi 10. Tm tt c cc cp s nguyn dng x, y sao cho

(x + 1)y - 1 =x!

Bi 11. Tm cc s nguyn tp sao cho tn ti cc s nguyn dng n, x, y tha mn

x3 +y3 =pn.


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81

Bi 12. Cho n l s nguyn dng. Chng minh rng nu 2n+1 v 3n + 1 l cc s chnhphng th 5n + 3 khng phi s nguyn t.

Bi 13. Cho a v b l hai s nguyn dng. S (36a + b)(a + 36b) c th l lu tha ca 2c hay khng?

Bi 14. Cho tpA gm cc s nguyn dng v |A| > 3. Bit rng tch ba phn t bt k

ca A u l s chnh phng. Chng minh rng mi phn t ca A u l s chnhphng.

Bi 15. Cho n l mt s nguyn dng sao cho 1282 2 +n l mt s nguyn. Chng

minh rng 2 + 1282 2 +n l mt s chnh phng.

Bi 16. Cho a, b, c l cc s nguyn dng sao cho

0 < a2 + b2 - abc c

Chng minh rng a2 + b2 - abc l mt s chnh phng.

Bi 17. Cho a v b l cc s nguyn dng sao cho a2 + b2 chia ht cho ab + 1. Chngminh rng

1

22

++

ab

ba

l mt s chnh phng.

Bi 18. Chox, y, zl cc s nguyn dng.

Chng minh rng (xy + 1)(yz+1)(zx + 1) l cc s nguyn dng khi v ch khixy + 1,yz+1,zx + 1 u l s chnh phng.

Bi 19. Cho a v b l cc s nguyn sao cho vi mi s khng m n th 2na = b u l schnh phng. Chng minh rng a = 0.

Bi 20. Cho p l mt s nguyn dng l. Chng minh rng tng cc lu tha bcp caps nguyn lin tip chia ht chop2.

Bi 21. Tm cc s nguyn dng n sao cho n.2n + 3n chia ht cho 5.

Bi 22. Tm cc s nguyn dng n sao cho n.2n + 3n chia ht cho 25.

Bi 23. Tm cc s t nhin n sao cho nn + 1 + (n + 1)n chia ht cho 5.

Bi 24. Tm cc s nguyn dng m, n, kln hn 1 sao cho

1! + 2! +... + n! = m

k

.Bi 25. Tm cc s nguyn t c dng 1722002 +

n

(n l s t nhin) biu din di dnghiu lp phng ca hai s t nhin.


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82

III. 2. p dng t hp vo cc bi ton s hc

III.2.1. Mt s nh l c bn

III.2.1.1. Nh thc Newton

(a + b)n =

=

n

k

kknkn baC

0

vi a, b l cc s thc tu , n nguyn dng.)!(!

!

knk

nCkn

=

III.2.1.2. nh l

Chop l mt s nguyn t. Khi k

pC p vi mip = 1, 2, ...,p - 1.

Chng minh

Ta c

!

)1)...(1(

)!(!

!

k

kppp

kpk

pCpn

+=

=

Do p nguyn t v k {1, 2, ...,p -1} nn (p, k!) = 1

M Cpk nguyn nn (p-1)(p-2) ...(p - k+ 1) k!

Hay

Zak

kppp=

+!

)1)...(2)(1(


III.2.1.3 Mt s h thc c bn

1) knnkn CC

=

2) 111 +=

kn

kn

kn CCC (H thc Pascal)

3)

+

=


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83

S= }C,...,C,{C 21-n

n2n

1n

cha l cc s l.

Li gii

t

Sn = 21

21 ...

+++n

nnn CCC

Khi

2Sn = 2...10 +++ nnnn CCC = 2

n - 2.

Sn = 2n - 1 - 1 l s l

Vy tp Sphi cha l cc s l.

V d 2. (Trung Quc - 1998) Tm s t nhin n 3 sao cho

22000

1 + .321nnn CCC ++

Li gii

Theo bi ta c

1 + knnn CCC 2321 =++ (0 k 2000, knguyn)

knnn

26

)6)(1( 2=

++

(n + 1)(n2 - n + 6) = 3.2k + 1

t m = n + 1 (m 4)Khi ta c

m(m2 - 3m + 8) = 3.2k + 1.

Do , ch c th xy ra mt trong hai trng hp sau:

+) Trng hp 1: m = 2s

Do m 4 nns 2

m2 - 3m + 8 = 22s - 3.2s + 8 = 3.2k + 1 - s.

Nu s 4 th 2

2s

- 3.2

s

+ 8 8 (mod 16). 8 3.2k + l - s (mod 16) 2k + 1 - s = 8 m2 - 3m + 8 = 24

(khng c nghim nguyn)

Nus = 2 m = 4 n = 3 (tha mn)

Nus = 3 m = 8 n = 7 (tha mn)

+) Trng hp 2. m = 3.2s


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Lm tng t nh trn, ta tm c n = 23.

Vy n = 3, n = 7, n = 23 l nhng gi tr cn tm.

V d 3. Chop l mt s nguyn t. Chng minh rng

)(mod2 22 pCpp

Li giiTheo h thc Vandermon, ta c

01102 ... p

pp

ppp

ppp

pp CCCCCCC +++=

M kpC p vi mi p = 1, 2, ..., p - 1.

Do ip

pi

pCC p2 vi mi i = 1, 2, ..., p - 1


V d 3. (Hungari - 2001) Cho m, n l cc s nguyn dng v 1 m n. Chng minhrng

1

1

)1(m

kn

kCn m.

Li gii

Ta c

)()1(...)()(

)1...(1

1

212

1

1

1

1

1

0

1

0

1

11210

+++++=

+m

n

m

n

m

nnnnn

mn

mnnn

CCCCCCC

CCCC

= (-1)m - 11

1

mnC

Do

1

1

)1(m

kn

kCn = (-1)m - 1n 11

mnC = (-1)

m-1.m mnC m.

T c iu phi chng minh.

V d 4. (VMO - 2011) Cho dy s nguyn (an) xc nh bi

a0 = 1, a1 = - 1an= 6an - 1 + 5an - 2 , vi n = 2, 3, ...

Chng minh rng a2012 - 2010 chia ht cho 2011.

Li gii

D dng tm c s hng tng qut ca dy s l


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85

14

)143)(1427()143)(1427( nnna

+++=

Do

14

)143)(1427()143)(1427( 201220122012

+++=a

Mt khc, theo khai trin Newton ta c

14)14.(3)143(2012

0

20122012

2012 BACk

k

kk +==+ =

14)14.(3)1()143(2012

0

20122012

2012 BACk

k

kkk == =

.

Trong

.14...14.33 10062012201220102

201220120

2012 CCCA +++=

.14.3....14.33 10052011201220093

2012201

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