Upload
nguyen-nhut-truong
View
235
Download
0
Embed Size (px)
Citation preview
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
8/3/2019 cc chuyn s hc
2/99
Nguyn Vn Tho Chuyn S Hc - Phn
2
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:
8/3/2019 cc chuyn s hc
3/99
Nguyn Vn Tho Chuyn S Hc - Phn
3
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
8/3/2019 cc chuyn s hc
4/99
Nguyn Vn Tho Chuyn S Hc - Phn
4
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
8/3/2019 cc chuyn s hc
5/99
Nguyn Vn Tho Chuyn S Hc - Phn
5
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
8/3/2019 cc chuyn s hc
6/99
Nguyn Vn Tho Chuyn S Hc - Phn
6
Do p =2 n, n 5 suy rap 123p3 2 +> pp
12 2
8/3/2019 cc chuyn s hc
7/99
Nguyn Vn Tho Chuyn S Hc - Phn
7
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.
8/3/2019 cc chuyn s hc
8/99
Nguyn Vn Tho Chuyn S Hc - Phn
8
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
8/3/2019 cc chuyn s hc
9/99
Nguyn Vn Tho Chuyn S Hc - Phn
9
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.
8/3/2019 cc chuyn s hc
10/99
Nguyn Vn Tho Chuyn S Hc - Phn
10
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
8/3/2019 cc chuyn s hc
11/99
Nguyn Vn Tho Chuyn S Hc - Phn
11
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
8/3/2019 cc chuyn s hc
12/99
Nguyn Vn Tho Chuyn S Hc - Phn
12
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=
8/3/2019 cc chuyn s hc
13/99
Nguyn Vn Tho Chuyn S Hc - Phn
13
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)
8/3/2019 cc chuyn s hc
14/99
Nguyn Vn Tho Chuyn S Hc - Phn
14
= (7n + 1, 14n + 3 2(7n + 1))
= (7n +1, 1) = 1
T suy ra iu phi chng minh.
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.
8/3/2019 cc chuyn s hc
15/99
Nguyn Vn Tho Chuyn S Hc - Phn
15
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).
8/3/2019 cc chuyn s hc
16/99
Nguyn Vn Tho Chuyn S Hc - Phn
16
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.
8/3/2019 cc chuyn s hc
17/99
Nguyn Vn Tho Chuyn S Hc - Phn
17
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
8/3/2019 cc chuyn s hc
18/99
Nguyn Vn Tho Chuyn S Hc - Phn
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?
8/3/2019 cc chuyn s hc
19/99
Nguyn Vn Tho Chuyn S Hc - Phn
19
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)
8/3/2019 cc chuyn s hc
20/99
Nguyn Vn Tho Chuyn S Hc - Phn
20
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
8/3/2019 cc chuyn s hc
21/99
Nguyn Vn Tho Chuyn S Hc - Phn
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)
Vy ta c iu phi chng minh.
V d 8. Chng minh rng an + 1 (a v n nguyn dng) l s nguyn t th n =2k.
Li gii
8/3/2019 cc chuyn s hc
22/99
Nguyn Vn Tho Chuyn S Hc - Phn
22
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)
Vy ta c iu phi chng minh.
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 + .
T suy ra iu phi chng minh.
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
8/3/2019 cc chuyn s hc
23/99
Nguyn Vn Tho Chuyn S Hc - Phn
23
Gi q l mt c nguyn t cas th q > p.
D thys
8/3/2019 cc chuyn s hc
24/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
25/99
Nguyn Vn Tho Chuyn S Hc - Phn
25
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
8/3/2019 cc chuyn s hc
26/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
27/99
Nguyn Vn Tho Chuyn S Hc - Phn
27
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)
8/3/2019 cc chuyn s hc
28/99
Nguyn Vn Tho Chuyn S Hc - Phn
28
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
8/3/2019 cc chuyn s hc
29/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
30/99
Nguyn Vn Tho Chuyn S Hc - Phn
30
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). (*)
8/3/2019 cc chuyn s hc
31/99
Nguyn Vn Tho Chuyn S Hc - Phn
31
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 + .
T suy ra iu phi chng minh.
8/3/2019 cc chuyn s hc
32/99
Nguyn Vn Tho Chuyn S Hc - Phn
32
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.
8/3/2019 cc chuyn s hc
33/99
Nguyn Vn Tho Chuyn S Hc - Phn
33
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.
8/3/2019 cc chuyn s hc
34/99
Nguyn Vn Tho Chuyn S Hc - Phn
34
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
8/3/2019 cc chuyn s hc
35/99
Nguyn Vn Tho Chuyn S Hc - Phn
35
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 )
8/3/2019 cc chuyn s hc
36/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
+ +
= =
8/3/2019 cc chuyn s hc
37/99
Nguyn Vn Tho Chuyn S Hc - Phn
37
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
8/3/2019 cc chuyn s hc
38/99
Nguyn Vn Tho Chuyn S Hc - Phn
38
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 )
8/3/2019 cc chuyn s hc
39/99
Nguyn Vn Tho Chuyn S Hc - Phn
39
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
8/3/2019 cc chuyn s hc
40/99
Nguyn Vn Tho Chuyn S Hc - Phn
40
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.
8/3/2019 cc chuyn s hc
41/99
Nguyn Vn Tho Chuyn S Hc - Phn
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)
8/3/2019 cc chuyn s hc
42/99
Nguyn Vn Tho Chuyn S Hc - Phn
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).
8/3/2019 cc chuyn s hc
43/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
44/99
Nguyn Vn Tho Chuyn S Hc - Phn
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)
Vy ta c iu phi chng minh.
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
8/3/2019 cc chuyn s hc
45/99
Nguyn Vn Tho Chuyn S Hc - Phn
45
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
8/3/2019 cc chuyn s hc
46/99
Nguyn Vn Tho Chuyn S Hc - Phn
46
)(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
8/3/2019 cc chuyn s hc
47/99
Nguyn Vn Tho Chuyn S Hc - Phn
47
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.
8/3/2019 cc chuyn s hc
48/99
Nguyn Vn Tho Chuyn S Hc - Phn
48
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.
8/3/2019 cc chuyn s hc
49/99
Nguyn Vn Tho Chuyn S Hc - Phn
49
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)
8/3/2019 cc chuyn s hc
50/99
Nguyn Vn Tho Chuyn S Hc - Phn
50
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
=
8/3/2019 cc chuyn s hc
51/99
Nguyn Vn Tho Chuyn S Hc - Phn
51
GiA = |2k kp
r r >
; B = |2k kp
r r
8/3/2019 cc chuyn s hc
52/99
Nguyn Vn Tho Chuyn S Hc - Phn
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 > = + +
8/3/2019 cc chuyn s hc
53/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
54/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
55/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
56/99
Nguyn Vn Tho Chuyn S Hc - Phn
56
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
8/3/2019 cc chuyn s hc
57/99
Nguyn Vn Tho Chuyn S Hc - Phn
57
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)
Vy ta c iu phi chng minh.
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).
8/3/2019 cc chuyn s hc
58/99
Nguyn Vn Tho Chuyn S Hc - Phn
58
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
= = = = = (*)
8/3/2019 cc chuyn s hc
59/99
Nguyn Vn Tho Chuyn S Hc - Phn
59
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.
8/3/2019 cc chuyn s hc
60/99
Nguyn Vn Tho Chuyn S Hc - Phn
60
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.
8/3/2019 cc chuyn s hc
61/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
T suy ra iu phi chng minh.
8/3/2019 cc chuyn s hc
62/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
63/99
Nguyn Vn Tho Chuyn S Hc - Phn
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)
8/3/2019 cc chuyn s hc
64/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
65/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
66/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
67/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
68/99
Nguyn Vn Tho Chuyn S Hc - Phn
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(=
+++=
8/3/2019 cc chuyn s hc
69/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
T ta c iu phi chng minh.
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
8/3/2019 cc chuyn s hc
70/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
T ta c iu phi chng minh.
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)
8/3/2019 cc chuyn s hc
71/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
Vy ta c iu phi chng minh.
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)
8/3/2019 cc chuyn s hc
72/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
73/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
T suy ra iu phi chng minh.
8/3/2019 cc chuyn s hc
74/99
Nguyn Vn Tho Chuyn S Hc - Phn
74
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.
8/3/2019 cc chuyn s hc
75/99
Nguyn Vn Tho Chuyn S Hc - Phn
75
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.
8/3/2019 cc chuyn s hc
76/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
77/99
Nguyn Vn Tho Chuyn S Hc - Phn
77
Do tm cy0 = 1,y1 = 7 (khng tha mn iu kinx0
8/3/2019 cc chuyn s hc
78/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
79/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
80/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
T ta c iu phi chng minh.
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.
8/3/2019 cc chuyn s hc
81/99
Nguyn Vn Tho Chuyn S Hc - Phn
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.
8/3/2019 cc chuyn s hc
82/99
Nguyn Vn Tho Chuyn S Hc - Phn
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(
T suy ra iu phi chng minh.
III.2.1.3 Mt s h thc c bn
1) knnkn CC
=
2) 111 +=
kn
kn
kn CCC (H thc Pascal)
3)
+
=
8/3/2019 cc chuyn s hc
83/99
Nguyn Vn Tho Chuyn S Hc - Phn
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
8/3/2019 cc chuyn s hc
84/99
Nguyn Vn Tho Chuyn S Hc - Phn
84
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
T suy ra iu phi chng minh.
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
8/3/2019 cc chuyn s hc
85/99
Nguyn Vn Tho Chuyn S Hc - Phn
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