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Din n MATHSCOPE
CHUYN S HC
Ch bin: Phm Tin Kha
Ph trch Latex: Phan c Minh, Nguyn Anh Huy
Thng 10/2013
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Mc lc
Li ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1 Bc nhy Viete 7
M u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Li gii nguyn thy ca bi ton v cc vn lin quan . . . . . . . . . . . . . . 8
Gi cho mt s bi ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Cc bi ton th sc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2 Vn dng phng php LTE vo gii cc bi ton s hc 17
Mt s khi nim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Hai b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Lifting The Exponent Lemma (LTE) . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Mt s v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Bi tp vn dng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Cc bi ton s hc hon v vng quanh 29
Phng php i xng ha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Phng php dng bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Bi tp t luyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4 Dy s s hc 39
Dy s nguyn v tnh cht s hc . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Dy s nguyn v tnh chnh phng . . . . . . . . . . . . . . . . . . . . . . . . . . 47
5 Mt s hm s hc v ng dng 61
Hm tng cc c s v s cc c s . . . . . . . . . . . . . . . . . . . . . . . . . . 61
Kin thc cn nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
V d p dng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
Bi tp c hng dn, gi . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Bi tp t gii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
Mt s hm s khc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Hm phn nguyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Hm tng cc ch s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3
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4Hm Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
6 Thng d bnh phng 73
Tnh cht c bn ca thng d bnh phng v k hiu Legendre . . . . . . . . . . . 73
Bi tp v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
K hiu Jakobil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
Bi tp v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Khai thc mt b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
Bi tp ngh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
7 Cp v cn nguyn thy 85
Cp ca mt s nguyn dng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Cn nguyn thy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
8 Hm phn nguyn v phn l 101
nh ngha, tnh cht v bi tp c bn . . . . . . . . . . . . . . . . . . . . . . . . 101
nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Cc tnh cht quen thuc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Bi tp c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
ng dng nh l Hermite v nh l Legendre . . . . . . . . . . . . . . . . . . . . 105
Hm c cha phn nguyn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
Hm phn nguyn trong vic tnh tng cc ch s . . . . . . . . . . . . . . . . . . 115
nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Tnh cht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Bi tp v d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
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Li ni u
S hc l b hong ca Ton hc
C th ni S hc l lnh vc xut hin sm nht trong lch s Ton hc. Khi con ngi bt
u lm vic vi nhng con s th khi y, S hc ra i.Tri qua hng nghn nm pht trin,
S hc vn gi c v p thun khit ca n. V p y c th hin qua cch pht biu
n gin ca bi ton, n ni mt hc sinh lp 6 cng c th hiu c. Th nhng, v p
y thng tim n nhng th thch su thm bn trong thch thc tr tu loi ngi. . .
Hy ni v nh l Fermat ln, mt nh l qu ni ting trong th gii Ton hc. Trong
mt tuyn tp vn hc vi ta Tho c vi Qu c truyn ngn Con Qu v Simon Flagg
ca Arthur Poges. Trong truyn ngn ny con Qu c ngh Simon Flagg t cho n mt cu
hi. Nu con Qu tr li c trong vng 24 gi, n s ly i linh hn ca Simon, cn nu n
u hng n s tr cho Simon 100.000 la. Simon t cho con Qu cu hi: nh l cui
cng ca Fermat c ng khng? Nghe xong, con Qu bin mt v bay vt i khp v tr
tip thu tt c tri thc ton hc tng c sng to ra. Ngy hm sau con Qu quay tr li
v th nhn: Simon, ngi thng, con Qu bun ru ni v nhn Si mon vi con mt y
thn phc. Ngay c ta, ta cng khng c kin thc ton hc trong mt thi gian ngn
nh th c th gii p c mt bi ton kh nh vy. Cng nghin cu su n cng rc ri
hn. . .Ch! Ngi c bit-Con Qu tm s- ngay c nhng nh ton hc gii nht trn cc
hnh tinh khc, h cn uyn bc hn nhng nh ton hc ca cc ngi nhiu, cng khng gii
ni cu khng? Th y, mt g trn sao Th nhn ging nh mt cy nm trn c kheo,
g c th gii nhm cc phng trnh vi phn o hm ring, m cng phi u hng thi.
Chnh v c cch pht biu n gin nhng cn nhng suy lun su sc v tinh t nn nhng
bi ton S hc trong cc k thi Olympic thng c dng phn loi hc sinh. Tuy rng
trn th trng c rt nhiu cun sch vit v S hc, nhng nhu cu sch v lnh vc ny
cha bao gi vi i. c bit, ngy cng nhiu cng phng php mi xut hin, cn mt u
sch khng nhng pht trin nhng phng php c m cn c th gii thiu nhng phng
php mi hoc nhng ci nhn mi v nhng vn c.
Chuyn S hc ca Din n Mathscope ra i chnh l p ng nhu cu ca ng
o hc sinh, sinh vin v gio vin trn khp c nc. Chuyn c thc hin bi cc thnh
vin ca Din n Mathscope, bao gm cc ch : Cp v cn nguyn thu, Cc bi ton s
hc hon v vng quanh, Dy s s hc, Hm s hc, B nng s m LTE, Phn nguyn,
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6Thng d chnh phng, Phng php bc nhy Viete. Hi vng y s l mt ti liu hu ch
cho cc bn c gn xa trong vic n luyn cho cc k thi Olympic.
hon thnh chuyn ny, ban bin tp xin gi li cm n su sc n tc gi ca cc
bi vit, cc thnh vin tham gia tho lun, ng gp trn Din n. c bit, xin gi li
cm n su sc n Ban qun tr ca Din n Mathscope lun to iu kin tt nht cho
ban bin tp hon thnh chuyn ny. Cun sch ny chnh l thnh qu qu trnh lao
ng nghim tc ca cc thnh vin trong ban bin tp, nhng hn ht y vn l mt sn
phm ng qu ca cng ng Mathscope ni ring v cng ng Ton hc Vit Nam ni chung.
Tuy c kim tra k cng nhng chuyn khng trnh khi nhng sai st. Mi gp v
chuyn xin c gi ln Din n Mathscope (mathscope.org) hoc gi v hp mail
[email protected]. Xin chn thnh cm n.
Thnh ph H Ch Minh, ngy Ph n Vit Nam 20-10-2013
Thay mt Ban bin tp
Phm Tin Kha
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Chuyn 1:Bc nhy Viete
Phm Huy Hong 1
M u
Trong cc k thi hc sinh gii, cc bi ton v phng trnh Diophante bc hai khng cn
xa l. Phng trnh Pell l mt trong cc v d ni bt nht v phng trnh Diophante bc
hai, tuy nhin do lng bi ton v phng trnh Pell kh nhiu, nn trong k thi IMO 1988
xut hin mt dng bi phng trnh Diophante bc hai rt mi m thi by gi:
Cho a, b l hai s nguyn dng tha mn ab+ 1|a2 + b2. Chng minh rnga2 + b2
ab+ 1
l s chnh phng.
Bi ton ny c coi l kh nht trong cc k thi IMO trc nm 1988 v cng l bi ton
kh nht trong k thi ny. Tc gi Authur Engel tng bnh lun v bi ton ny (nguyn
vn):
" Nobody of the six members of the Australian problem committee could solve
it. Two of the members were George Szekeres and his wife, both famous problem
solvers and problem creators. Since it was a number theoretic problem it was sent
to the four most renowned Australian number theorists. They were asked to work
on it for six hours. None of them could solve it in this time. The problem committee
submitted it to the jury of the XXIX IMO marked with a double asterisk, which
meant a superhard problem, possibly too hard to pose. After a long discussion, the
jury finally had the courage to choose it as the last problem of the competition.
Eleven students gave perfect solutions."
Dch ra ting Vit nm na l:
"Khng c ai trong s su thnh vin ca hi ng gim kho ca c gii c
bi ton ny. Hai thnh vin ni bt trong , u l nhng ngi ni ting gii v
sng to cc bi ton, l George Szekeres v v ng. y l bi ton s hc nn n
c gi cho bn nh s hc ln nht ca c by gi. H c yu cu gii bi
ton trong su gi v khng c ai gii c sau . Hi ng thm nh np cho
1i hc Khoa hc T Nhin
7
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8ban gim kho IMO XXIX bi ton ny vi hai du hoa th, ni ln l n rt
kh, hoc l qu kh ra trong k thi. Sau mt hi bn bc, hi ng IMO XXIX
quyt nh chn bi ton ny lm bi cui ca k thi. C mi mt hc sinh cho li
gii hon chnh ca bi ton."
Trong s 11 hc sinh c Gio s Ng Bo Chu ca chng ta.
Li gii nguyn thy ca bi ton v cc vn lin quan
Chng ta bt u vi bi ton gc:
Bi ton 1. Cho a, b l cc s nguyn dng tha mn ab+ 1 | a2 + b2.Chng minh rng
a2 + b2
ab+ 1
l s chnh phng
Li gii. t k = a2+b2
ab+1. C nh k v xt tt c cc cp (a, b) nguyn dng tha mn phng
trnh
k =a2 + b2
ab+ 1,
c ngha l xt tp
S =
{(a, b) N N|k = a
2 + b2
ab+ 1
}V S l tp cc cp s nguyn dng nn lun tn ti mt cp (a0, b0) trong S m a0 + b0 t
gi tr nh nht v a0 > b0.Xt phng trnh
x2 + b20xb0 + 1
= k x2 kx.b0 + b20 k = 0
l mt phng trnh bc hai n x. Ta bit rng phng trnh trn c mt nghim l a0. Nh
vy theo nh l Viete th tn ti nghim a1 tha mn phng trnh bc hai vi n x trn v
a1 = kb0 a0 = b20 ka0
.
T y ta c a1 cng l s nguyn. Ta chng minh a1 khng m. Tht vy, nu a1 < 0 th
a21 kb0a1 + b20 k > a21 + k + b20 k > 0,mu thun. Do ta c a1 > 0. T y ta c:Nu a1 > 0 th (a1, b0) l mt cp thuc S. Theo nh ngha ca (a0, b0) ta c:
a0 + b0 6 a1 + b0 a0 6 a1.Do cng theo Viete th:
a20 6 a0a1 = b20 k < b20 a0 < b0,tri vi gi thit ban u.
Do a1 = 0, v vy suy ra k = b20 l mt s chnh phng, ta c iu cn chng minh. r
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9T bi ton ny ta c th thy c cc bc gii bi ton dng phng php ny nh sau:
1. Nhn dng bi ton thuc lp phng trnh Diophante bc hai (tr ln).
2. C nh mt gi tr nguyn m bi cho, ri gi s tn ti mt cp nghim tha mn
mt vi iu kin m khng lm mt tnh tng qut ca bi ton.
3. Da vo nh l Viete tm cc mi quan h v s mu thun, t tm c kt lun
ca bi ton.
im mu cht ca cc bi ton ny l nguyn l cc hn: Trong tp hp cc s nguyn dng
th lun tn ti s nguyn dng nh nht. Mnh trn khng nhng hu dng trong cc lp
bi ton ny m cn trong nhiu bi ton t hp, t hp s hc v s hc.
T nhng bi ton tip theo, ti s trnh by vn tt cc bc lm v cch lm thay v trnh
by y nh bi ton trn, cc bn c th t pht huy tnh t lm vic ca mnh. Phn
gi s c cui bi vit.
Bi ton 2. Tm tt c cc s nguyn dng n sao cho phng trnh sau c nghim nguyn
dng :
x2 + y2 = n(x+ 1)(y + 1).
Chng minh. Chng ta s lm theo cc bc nh bi ton trn:
1. C nh n, gi s tn ti cp (x0, y0) m tng x0 + y0 min v x0 > y0.
2. Xt phng trnh bc 2 n X nh sau:
X2 X.n(y + 1) + y20 ny0 n = 0.
Phng trnh c nghim l x0 nn c nghim x1.
3. p dng nh l Viete:
x0 + x1 = n(y0 + 1), x0x1 = y20 ny0 n.
4. Tng t bi trc, cc bn chng minh x1 > 0 v t s chng minh x1 = 0 bngcch chng minh x1 > 0 th s dn n mu thun.
5. T i n kt lun bi ton: x1 = 0 v y20 = n(y0 + 1) suy ra y0 + 1 | y20, l iu khng
th xy ra khi y0 nguyn dng.
Do khng tn ti s nguyn dng n tha mn phng trnh u tin. r
T bi ton ny, ta dn n c bi ton th v sau:
Bi ton 3. Gi s a, b nguyn dng tha mn:
b+ 1 | a2 + 1, a+ 1 | b2 + 1.
Chng minh rng a, b u l cc s l.
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Chng minh. Nhn vo bi ton trn, t gi thit ta khng nhn thy mi tng quan gia a, b
v tnh chn l ca hai s . V vy, nh bn nng v kinh nghim, cch lm tt nht thm
d kin l s dng phng php phn chng.
Gi s a, b u l cc s chn. T gi s ny, bn c hy chng minh hai mu cht sau:
1. a+ 1 v b+ 1 nguyn t cng nhau.
2. a+ 1 | a2 + b2, b+ 1 | a2 + b2.T suy ra tn ti n nguyn dng sao cho a2 + b2 = n(a + 1)(b + 1) v theo bi ton trn
ta c iu mu thun. V vy a, b l. r
Bi ton trn cng l mt b quan trng ca mt bi ton trong IMO Shortlist 2009.
Bi ton 4. Tm tt c cc s nguyn dng n sao cho tn ti dy s nguyn dng a1, a2, ..., an
tha mn
ak+1 =a2k + 1
ak1 + 1 1
vi mi k tha mn 2 6 k 6 n 1.(Phn gi s c cui bi vit).
Khi gii c hai bi ton trn th a s cc bi ton vi hai bin x, y s thnh chuyn "n
gin". Xin mi bn c th sc vi bi ton sau, v iu n sau mi l iu th v:
Bi ton 5. Tm tt c cc s n nguyn dng sao cho phng trnh sau c nghim nguyn
dng:
(x+ y)2 = n(4xy + 1).
Chng minh. Chng ta tun t theo cc bc trn. p n l vi n l s chnh phng th
phng trnh lun c nghim nguyn dng. r
ng ch l bi ton n gin nh vy m li l mt b cc k quan trng cho mt bi
ton kh sau:
Bi ton 6 (Taiwan MO 1998). Cho m,n l hai s l vi m > n > 1 tha mn
m2 n2 + 1 | n2 1.Chng minh rng m2 n2 + 1 l s chnh phng.Chng minh. Nhn vo bi ton ny v bi ton trn, chng ta khng th thy ngay s lin h.
Gi cho bi ton ny l lm th no chuyn v bi ton trc.
T gi thit ta c m2 n2 + 1 | n2 1, cho tin v gn hn, ta c m2 n2 + 1 | m2. T yta c th t m2 = k.(m2n2 + 1) vi k nguyn dng. n y th chc khng kh nhn rami lin h: T gi thit m,n l, tn ti hai s nguyn dng a, b sao cho m = a+ b, n = a b.Do phng trnh trn tr thnh:
(a+ b)2 = k(4ab+ 1),
chng ta quay v bi ton trn. Vy k l s chnh phng, do 4ab + 1 cng l s chnh
phng hay m2 n2 + 1 l s chnh phng. r
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Nhn bi ton trn, nh mt ngi lm ton, chng ta khng khi thc mc l: liu c tn ti
hai s m,n nh vy khng tha mn m2n2 + 1|n2 1 ri suy ra m2n2 + 1 l s chnhphng? Bng li suy ngh chng ta nn tm th mt nghim ca bi ton:
Bi ton 7. Tm mt cp nghim (m,n) l nguyn dng tha mn iu kin bi ton trn.
Chng minh. Th mt vi gi tr, bi ton khng h d nh cc bn tng: Chng ta khng
th "m" nghim ri suy ra c. Chng ta nn bt u vi cch lm t nhin nht: t
m2 = k(m2 n2 + 1). V bi ton ch yu cu mt nghim, ta bt u vi k = 1 l s chnhphng u tin:
m2 = m2 n2 + 1 n = 1,khng tha mn v m > n > 1.
Tip tc vi k = 4 l s chnh phng tip theo:
m2 = 4(m2 n2 + 1) 4n2 3m2 = 4.
T phng trnh trn suy ra 2 | m hay t m = 2t. T phng trnh tng ng vi
n2 3t2 = 1,
tr v phng trnh Pell quen thuc v phng trnh ny chc chn c nghim v 3 khng
phi l s chnh phng. Tm nghim ca phng trnh ny khng h kh, cc bn c th t
tm bng cch s dng cng thc nghim tng qut ca phng trnh Pell hoc h phng
trnh. r
Vy tt c cc bi ton trn u c th tm c nghim tha mn. Bi ton va xong ch l
mt nghim n gin. Cu hi l c th tm c nghim tng qut khng? Cu tr li l c.
Trong k thi chn hc sinh gii Ton quc gia nm 2012, bi ton cng s dng phng php
bc nhy Viete gii c bi ton. Xin trch dn bi, bi gii trong ti liu "Nhn xt
v nh gi thi VMO 2012" ca Thy Trn Nam Dng:
Bi ton 8. Xt cc s t nhin l a, b m a l c s ca b2 + 2 v b l c s ca a2 + 2.
Chng minh rng a v b l cc s hng ca dy s t nhin (vn) xc nh bi
v1 = v2 = 1; vn = 4vn1 vn2 ,n > 2.
Chng minh. Gi s (a, b) l cp s t nhin l m a l c s ca b2 + 2 v b l c s ca
a2 + 2. Trc ht ta chng minh (a, b) = 1. Tht vy, t d = (a, b) th do d | a | b2 + 2 nnd | 2. M a, b l nn d l, suy ra d = 1.Xt s N = a2 + b2 + 2 th do a2 + 2 chia ht cho b nn N chia ht cho b. Tng t, N chia ht
cho a. V (a, b) = 1 nn t y suy ra N chia ht cho ab. Vy tn ti s nguyn dng k sao
cho
a2 + b2 + 2 = kab (1)
Tip theo, ta chng minh k = 4. Tht vy, t A = {a+ b|(a, b) NN, a2 + b2 + 2 = kab}.Theo gi s trn th A 6= . Do tnh sp th t tt ca N, A c phn t nh nht. Gi s
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a0, b0 l cp s tha mn iu kin (1) vi a0 + b0 nh nht.
Khng mt tnh tng qut, c th gi s a0 > b0. Xt phng trnh a2 kb0a + b20 + 2 = 0 cnghim a0. Theo nh l Viete th phng trnh trn cn c 1 nghim na l a1 = kb0a0 = b
20+2
2.
Theo cng thc nghim th r rng a1 nguyn dng. Nh vy (a1, b0) cng l mt nghim ca
(1). Do tnh nh nht ca a0 + b0, ta c a0 + b0 6 a1 + b0, tc l a0 6 kb0 a0 suy ra a0b0 6 k2 .Ta c a20 + b
20 + 2 = ka0b0 suy ra
a0b0
+b0a0
+2
a0b0= k (2)
Do a0b06 k
2nn t (2) ta c k 6 k
2+ 2 + 1 hay k 6 6.
Mt khc, p dng bt ng thc AM-GM ta c a20 + b20 > 2a0b0 nn k > 2.
Nu k 6= 4 th (a0, b0) 6= (1, 1), do a0b0 > 2. Dng (2) nh gi ta c k 6 k2 + 1 + 1 nnk 6 4. Vy cc gi tr k = 5, 6 b loi. Nu k = 3 th do a20+b20+2 = 3a0b0 nn suy ra a20+b20+2chia ht cho 3, suy ra mt trong hai s a0, b0 chia ht cho 3, s cn li khng chia ht cho 3.
Nu b0 = 1 th a0 chia ht cho 3, khi v tri khng chia ht cho 9 cn v phi chia ht cho
9, mu thun. Vy b0 > 1. T suy ra a0b0 > 6. Li s dng (2) nh gi, ta suy ra
k 6 k2
+ 1 +2
6 k < 8
3.
M k N nn k 6 2, mu thun.Nh vy ta chng minh c nu a, b l cc s t nhin l tha mn iu kin bi th
a2 + b2 + 2 = 4ab (3)
Ta s chng minh trong trng hp nh vy th tn ti s nguyn dng n sao cho (a, b) =
(vn, vn+1) vi vn l dy s c nh ngha bi.
Trc ht, ta c nhn xt : Nu a, b l nghim ca (3) th (4a b, a) v (4b a, b) cng lnghim ca (3). T , do (v1, v2) l nghim ca (3) nn (4v2 v1, v2) cng l nghim ca (3),tc l (v2, v3) cng l nghim ca (3). T y bng quy np suy ra (vn, vn+1) l nghim ca
(3).
Gi s tn ti cp s (a, b) tha mn (3) nhng khng tn ti n sao cho (a, b) = (vn, vn+1).
Trong cc cp s nh th, chn (a, b) c tng a + b nh nht. Khng mt tnh tng qut, gi
s a > b (ch a khng th bng b v nu a = b suy ra a = b = 1, khi (a, b) = (v1, v2).
Theo nhn xt trn th 4b a, b cng l nghim ca (3). Nhng do 4b a = b2+2a
< a (V a > b
nn ab b2 = (a + b)(a b) > 3)), nn 4b a + b < a + b. Theo nh ngha ca (a, b) trn,phi tn ti n sao cho (4b a, b) = (vn, vn+1). S dng ng thc 4b a = b2+2a v b > 1, tasuy ra 4b a 6 b. Nh vy 4b a = vn, b = vn+1. Nhng t y a = 4vn+1 vn = vn+2, tcl (a, b) = (vn+1, vn+2) mu thun. Vy iu gi s l sai, tc l phi tn ti s t nhin n sao
cho (a, b) = (vn, vn+1) v nh vy a, b l s hng ca dy (vn). Bi ton c gii quyt hon
ton. r
Nhn xt:
Bi ton ny c 2 chnh:
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13
1. Chng minh rng nu k l s nguyn dng sao cho tn ti a, b nguyn dng tha
mn iu kin a2 + b2 + 2 = kab th k = 4. Phn ny kh quen thuc vi cc bn
bit v phng php phng trnh Markov hay bc nhy Viete.
2. M t tt c cc nghim ca phng trnh a2 + b2 + 2 = 4ab thng qua cp s hng
lin tip ca dy vn, ci ny gi l phng php gien.
Phng trnh (3) cn c th gii thng qua phng trnh Pell z2 3b2 = 2. Tuy nhincch gii ny s kh cng knh v y khng phi l phng trnh Pell loi 1.
Nh vy qua nhn xt ca bi ton trn, thy Trn Nam Dng tng kt li cc bc lm
chnh: l s dng bc nhy Viete tm c k v s dng phng php Gien, phng
php li v hn tm c im c bit ca nghim, l nu (a, b) l nghim th (4ba, b)cng l nghim, dn n tm c nghim tng qut ca phng trnh trn.
Ngoi ra thy cng cp ti phng trnh Markov, l mt phng trnh rt ni ting, c
th ni l "thy t" ca phng php bc nhy Viete:
Bi ton 9 (Phng trnh Markov). Tm tt c cc s nguyn dng k sao cho phng trnh
sau c nghim nguyn dng:
x2 + y2 + z2 = kxyz.
Chng minh. Chng ta va gii quyt xong cc lp bi ton vi hai n, vy bi ton ba n th
li gii c khc khng? V khc nh th no? Liu ta c th tm c cch gii tng qut cho
bi ton vi ba n nh vi hai n khng?
Cch gii bi ton ny ph thuc vo cch gii bi ton i vi hai s, trc ht ta cn b
sau:
B 1. k = 3 l s nguyn dng duy nht phng trnh sau lun c nghim nguyn
dng (x, y):
x2 + y2 + 1 = kxy.
B trn l mt bi ton s dng bc nhy Viete quen thuc, mi bn c t gii.
Tr li bi ton:
1. Thy k = 1 th phng trnh c nghim (3, 3, 3) v k = 3 th phng trnh c nghim
(1, 1, 1).
2. Xt k 6= 1, 3. Gi s phng trnh c nghim (x0, y0, z0). Khng mt tnh tng qut, gis x0 6 y0 6 z0 v x0 + y0 + z0 nh nht trong tt c cc tng x + y + z vi x, y, z lnghim ca phng trnh.
Nu y0 < z0, xt phng trnh bc hai:Z2 k.x0y0.Z + x20 + y20 = 0.
Phng trnh trn c mt nghim l z0. Theo nh l Viete th c nghim th hai
z1 tha mn:
z1 = kx0y0 z0 = x20 + y
20
z0.
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14
T suy ra z1 nguyn dng v (x0, y0, z1) l nghim th hai. Do theo gi thit
cc hn ta c:
x0 + y0 + z0 6 x0 + y0 + z1 z0 6 z1,dn n
x20 + y20 kx0y0 = z1z0 z1 z0 = (z1 1)(z0 1) 1 > y20 1,
v suy ra
1 > x0(ky0 x0) > x0(kx0 x0) > x0,m x0 nguyn dng suy ra x0 = 1. Ta tr v bi ton y
2 + z2 + 1 = kyz, chnh l
b suy ra k = 3, tri gi thit.
Nu y0 = z0. Ta c2y20 py20 + x20 = 0 x20 = y20(px0 2) > x20(px0 2),
v t dn n 3 > px0, m px0 > 2 nn px0 = 3 suy ra p {1, 3}, tri gi thit
Vy k {1, 3} th phng trnh lun c nghim nguyn dng. rTrn y l mt s bi ton m ti mun gii thiu vi bn c, l cc bi ton kh quen
thuc v c lp li trong nhiu k thi. Ti hi vng qua bi vit ny bn c c th nm bt
phng php gii mt lp cc bi ton v phng trnh bc hai Diophante nhiu n.
Sau y l gi cho cc bi ton v mt s cc bi ton th sc.
Gi cho mt s bi ton
2. Chng minh x1 > 0:
x21 x1n(y0 + 1) n(y0 + 1) + y20 = 0 x1 =x20 + y
20
n(y0 + 1) 1 > 1,
m x1 nguyn nn x1 > 0.
3. Vi a, b chn: c b+ 1 | b2 1, b+ 1 | a2 + 1 nn b+ 1 | a2 + b2. Tng t a+ 1 | a2 + b2.Gi d = (a+ 1, b+ 1, hy chng minh d | 2 m d l do a+ 1, b+ 1 l nn d = 1, t suyra a2 + b2 = k(a+ 1)(b+ 1).
4. n = 1, 2, 3, 4. D dng ch ra dy vi n = 1, 2, 3. Dy c di 4: 4,33,27,1384.
Phn chng tn ti dy di 5: a1, a2, a3, a4, a5. T y chng minh hai mnh sau:
(a) a2, a3 chn (s dng phn chng)
(b) a2 + 1 | a23 + 1, a3 + 1 | a22 + 1.
5. Chng minh x1 > 0: t phng trnh ta c
4x1y0 =(x1 + y0)
2
n 1 > 1 x1 > 1
4y0
m x1 nguyn nn x1 > 0.
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15
Cc bi ton th sc
Bi ton 10. Chng minh rng nu a, b l cc s nguyn dng sao cho k = a2+b2+6ab
l s
nguyn th k = 8.
Bi ton 11. 1. Tm n sao cho phng trnh sau c nghm nguyn dng:
(x+ y + z)2 = nxyz
2. Tm n sao cho phng trnh sau c nghm nguyn dng:
(x+ y + z + t)2 = nxyzt
Bi ton 12 (M rng phng trnh Markov). Cho a, b, c l ba s nguyn tha mn
a2 + b2 + c2 = kabc.
Chng minh rng hoc (a, b, c) = 1 hoc (a, b, c) = 3.
Bi ton 13. Chng minh rng phng trnh
x2 + y2 + z2 = n(xyz + 1)
c nghim nguyn dng khi v ch khi n c biu din di dng tng ca hai s chnh
phng.
Bi ton 14 (Adapted from Vietnam TST 1992). Tm tt c cc cp s nguyn dng (a, b)
tha mn
a2 + b2 = k(ab 1).
Bi ton 15 (Turkey TST 1994). Tm tt c cc cp (a, b) m ab | a2 + b2 + 3.
Bi ton 16. Cho a, b, c l ba s nguyn dng tha mn
0 < a2 + b2 abc 6 c,
chng minh rng a2 + b2 abc l s chnh phng.
Bi ton 17. Chng minh rng tn ti v hn cc cp (m,n) nguyn dng tha mn
m+ 1
n+n+ 1
m= 4.
Bi ton 18 (IMO Shortlist 2003). Tm tt c cc cp a, b tha mn
a2
2ab2 b3 + 1Bi ton 19. Chng minh rng tt c nghim nguyn dng ca phng trnh x2+y2+1 = 3xy
l (x, y) = (F2k1, F2k+1) vi Fn l s Fibonacci.
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Bi ton 20 (IMO 2007, IMO shortlist). Cho a, b nguyn dng. Chng minh rng nu
4ab 1 | (4a2 1)2, th a = b.
Gi . Dng phn chng, gi s a > b vi mi a, b tha mn. T gi thit, hy chng minh:
1. 4ab 1 | (a b)2
2. a0 b0 > (a0 + b0)(4a0b0 1) vi a0, b0 l nghim nh nht theo ngha a0 + b0 min, t suy ra mu thun
r
Bi ton 21 (**, Kiran Kedlaya). Chng minh rng (xy + 1)(yz + 1)(zx + 1) l s chnh
phng khi v ch khi xy + 1, yz + 1, zx+ 1 l s chnh phng.
Gi . 1. Nu xy + 1, yz + 1, zx + 1 l s chnh phng th hin nhin ta c tch ba s
chnh phng.
2. Nu (xy + 1)(yz + 1)(zx+ 1) l s chnh phng. Hy chng minh tn ti t tha mn h
sau:
(x+ y z t)2 = 4(xy + 1)(zt+ 1)(x+ z y t)2 = 4(xz + 1)(yt+ 1)(x+ t y z)2 = 4(xt+ 1)(yz + 1)
(t chnh l nghim ca phng trnh t2 + x2 + y2 + z2 2(xy + yz + zt+ tx+ zx+ ty)4xyzt 4 = 0). Xt nghim t nh nht.S dng phn chng: gi s xy + 1 khng chnh phng. T y, hy chng minh:
(a) t > 1max{x,y,z} > 1 nn t > 0.
(b) Xt hai trng hp t = 0 v t > 0, dn n mu thun.
r
Ti liu tham kho
1. ng Hng Thng, Nguyn Vn Ngc, V Kim Thy, Bi ging s hc. NXB Gio dc,
1996.
2. Kiran S. Kedlaya, When Is (xy + 1)(yz + 1)(zx + 1) a Square. Mathematics Magazine,
Vol. 17, No.1, Feb., 1998.
3. Authur Engel, Problem Solving Strategies. Spinger Verlag, 1998.
4. Trn Nam Dng, Li gii v bnh lun VMO 2012. Din n Mathscope, 2012.
5. Site: http://www.artofproblemsolving.com/Forum
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Chuyn 2:Vn dng phng php LTEvo gii cc bi ton s hc
Phm Quang Ton 1
B v s m ng (Lifting The Exponent Lemma) l mt b rt hu dng
trong vic gii cc bi ton s hc v rt c bit n trong lch s Olympiad.
Thc cht l n c m rng ra t b Hensel. Ta thng vit tt tn ca b
l LTE, tn Ting Vit th c th gi l b v s m ng. Bi vit ny xin
c gii thiu vi bn c v b v nhng ng dng c sc ca n vo cc bi
ton l thuyt s.
Bi vit ch yu da vo ti liu ca thnh vin Amir Hossein bn trang mathlinks.ro
(v mt l thuyt th mnh gi nguyn bn bi vit ca Amir Hossein sang bi vit
ny) v c km theo mt s v d c ly t cc k thi Olympic ton trn th gii.
Mt s khi nim
y, thay v k hiu a...b ngha l a chia ht cho b, ta s k hiu b|a. V a 6 ...b s c thay bng
b - a.nh ngha 1. Cho p l s nguyn t, a l s nguyn v l s t nhin. Ta c p l ly tha
ng (exact power) ca a v l s m ng (exact exponent) ca p trong khai trin ca
nu p|a v p+1 - a. Khi ta vit p a hay vp(a) = .V d. Ta c v5(5400) = 3 hay 5
3 5400 v 5400 = 53 32 22.Sau y l mt s tnh cht. Chng minh tnh cht ny khng kh, xin dnh cho bn c.
Tnh cht 1. Cho a, b, c l cc s nguyn. Ta c
1. vp(ab) = vp(a) + vp(b)
2. vp(an) = n vp(a)
3. min{vp(a), vp(b)} 6 vp(a+ b)Du ng thc xy ra khi vp(a) 6= vp(b).
4. vp (gcd(|a|, |b|, |c|)) = min{vp(a), vp(b), vp(c)}5. vp (lcm(|a|, |b|, |c|)) = max{vp(a), vp(b), vp(c)}
Ch . vp(0) = vi mi s nguyn t p.1Lp 9C THCS ng Thai Mai, Tp Vinh
17
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Hai b
u tin, xin gii thiu vi bn c hai b . V hai b ny s gip ta tm cch chng
minh c cc nh l khc ca LTE.
B 1. Cho x, y l hai s nguyn v cho n l s nguyn dng. Cho s nguyn t p bt k
sao cho p|x y v p - x, p - y. Ta c
vp(xn yn) = vp(x y).
Chng minh. Ta c p|x y nn
xn1 + xn2y + + xyn2 + yn1 nxn1 6 0 (mod p)
M xn yn = (x y) (xn1 + xn2y + + xyn2 + yn1) nn ta suy ra iu phi chngminh. r
B 2. Cho x, y l hai s nguyn v n l s nguyn dng l. Cho s nguyn t p bt k tha
mn p|x+ y v p - x, p - y. Khi
vp(xn + yn) = xp(x+ y).
Chng minh. p dng b 1 ta c vp(xn(y)n) = vp(x(y)) nn vp(xn+yn) = vp(x+y).
(v n l). B c chng minh. r
Lifting The Exponent Lemma (LTE)
nh l 1. Cho x v y l cc s nguyn (khng nht thit phi nguyn dng), n l mt s
nguyn dng v p l mt s nguyn t l tha mn p|x y v p - x, p - y. Ta c
vp (an bn) = vp(a b) + vp(n)
Chng minh. Ta s i chng minh quy np theo vp(n). Trc ht, ta s i chng minh khng
nh sau:
vp(xp yp) = vp(x y) + 1
chng minh iu th ta cn ch ra rng
p|xp1 + xp2y + + xyp2 + yp1 (1)
v
p2 - xp1 + xp2y + + xyp2 + yp1 (2)Vi (1), nh p dng x y (mod p) ta suy ra
xp1 + + yp1 pxp1 0 (mod p)
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19
Vi (2), ta t y = x+ kp vi k N. Khi vi 1 6 i 6 p 1 (i N) th
yixp1i (x+ kp)ixp1i
xp1i(xi + i(kp)xi1 +
i(i 1)2
(kp)2xi2 + )
xp1i (xi + i(kp)xi1) xp1 + ikpxp2 (mod p2).
Do ,
xp1 + xp2y + + yp1 xp1 + (xp1 + kpxp2) + (xp1 + 2kpxp2) + + (xp1 + (p 1)kpxp2) pxp1 + p 1
2 kp2xp2
pxp1 6 0 (mod p2)
Nh vy vp(xp yp) = vP (x y) + 1.
Quay li bi ton, t n = pk h vi b, k N, b > 1 gcd(b, p) = 1. Khi th
vp(an bn) = vp((apk)h (bpk)h)
= vp
(ap
k bpk)
= vp((apk1)p (bpk1)p)
= vp(apk1 bpk1) + 1 = vp((apk2)p (bpk2)p))
...
= vp(x y) + k = vp(x y) + vp(n)
nh l c chng minh. r
nh l 2. Cho hai s nguyn x, y, n l s nguyn dng l, v p l c nguyn t l sao cho
p|x+ y v p - x, p - y. Khi
vp(xn + yn) = vp(x+ y) + vp(n).
Chng minh. p dng nh l 1 ta c
vp(xn (y)n) = vp(x (y)) + vp(n)
hay
vp(xn + yn) = vp(x+ y) + vp(n)
r
nh l 3. (cho trng hp p = 2) Cho x, y l hai s nguyn l tha mn 4|x y. Khi
v2(xn yn) = v2(x y) + v2(n).
Chng minh. Theo b 1 th nu p nguyn t, gcd(p, n) = 1, p|x y v p - x, p - y th
vp(xn yn) = vp(x y)
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20
Do ta ch cn xt ti trng hp n l ly tha ca 2, tc cn chng minh
v2(x2n y2n) = v2(x y) + n
Tht vy, ta c
x2n y2n = (x2n1 + y2n1)(x2n2 + y2n2) (x2 + y2)(x+ y)(x y)
V x y 1 (mod 4) nn x2k y2k 1 (mod 4). Do
v2(x2n1 + y2
n1) = v2(x
2n2 + y2n2
) = = v2(x+ y) = 1
Nh vy v2(x2n + y2
n) = n+ vp(x y), ta c iu phi chng minh. r
nh l 4. (cho trng hp p = 2) Cho hai s nguyn l x, y, n l s nguyn dng chn v
2|x y. Khi v2(x
n yn) = v2(x y) + v2(x+ y) + v2(n) 1.
Chng minh. Ta c 4|x2 y2 nn t n = 2k h vi k, h N, gcd(h, 2) = 1. Khi ta c
v2(xn yn) = v2(xh2k yh2k)
= v2((x2)2
k1 (y2)2k1)...
= v2(x2 y2) + k 1
= v2(x y) + v2(x+ y) + v2(n) 1r
Ta c h qu sau:
H qu. Cho a, n l hai s nguyn dng:
i) p l hai s nguyn t l sao cho vp(a 1) = N, khi vi mi s t nhin ta cv(a
n 1) = + vp(n) = .
ii) n chn sao cho v2(a2 1) = N, khi vi mi s nguyn dng th v2(an 1) =
+ v2(n) = + 1.
Ch .
a) Nu trong cc bi ton i hi vn dng phng php LTE, ta nn ti cc iu kin
t ra ca n, x, y, la chn nh l ph hp a vo li gii bi ton.
b) Nu d liu bi ton cho a|b vi a, b N th vi mi p l c nguyn t ca b, ta lun cvp(b) > vp(a). Ngc li, nu vp(b) > vp(a) th a|b. Nh vy
a|b vp(b) > vp(a)
y l mt tnh cht rt thng c dng trong cc bi ton s dng phng php
LTE.
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21
Mt s v d
Sau y mnh xin a ra mt s v d v cc ng dng ca phng php ny.
V d . Tm s nguyn dng n nh nht tha mn 22013|1999n 1.Li gii. p dng nh l 4 ta c
v2(1999n 1) = v2(n) + v2(2000) + v2(1998) = v2(n) + 5
tha mn 22013|1999n 1 th v2(n) + 5 > 2013 hay v2(n) > 2008.Vy s nguyn dng n nh nht tha mn ra l 22008.
V d . (IMO Shortlist 1991) Tm s nguyn dng k ln nht tha mn 1991k l c ca
199019911992
+ 199219911990
.
Li gii. t a = 1991 th a l s nguyn t l. Do theo nh l 2 th
va
((a 1)aa+1 + (a+ 1)aa1
)= va
((a 1)a2)aa1 + (a+ 1)aa1
)= va
((a 1)a2 + a+ 1
)+ va(a
a1)
= a 1 + va(
(a 1)a2 + a+ 1)
Cng theo nh l 2 th va
((a 1)a2 + 1
)= va(a)+va(a
2) = 3 nn va
((a 1)a2 + a+ 1
)= 1.
Vy, va
((a 1)aa+1 + (a+ 1)aa1
)= a. Ta thu c max k = a = 1991 .
V d 1. (Italy TST 2003) Tm b s nguyn nguyn (a, b, p) sao cho a, b l s nguyn dng,
p l s nguyn t tha mn 2a + pb = 19a.
Li gii. V a nguyn dng nn 17|19a 2a. Vy p = 17. p dng nh l 1 ta c
v17(19a 2a) = v17(17) + v17(a)
b = 1 + v17(a) 6 1 + a
1. Nu b < 1 + a hay 1 6 b 6 a. D dng chng minh quy np rng 19a 2a > 17a via > 1. M 17a > 17b. Vy a = b = 1 trng hp ny.
2. Nu b = 1 + a th d dng chng minh quy np 19a 2a < 17a+1 = 17b, mu thun.
Vy (a, b, p) = (1, 1, 17) l p n duy nht bi ton.
V d . (IMO 1990) Tm s nguyn dng n sao cho n2|2n + 1.Li gii. Vi n = 1 tha mn. Vi n > 2, nhn thy n l.Gi p l c nguyn t l nh nht ca n. Khi ta suy ra 22n 1 (mod p). Gi k l s nguyndng nh nht tha mn 2k 1 (mod p). Khi k|2n. Theo nh l Fermat nh th 2p1 1(mod p) nn k|p 1. Nh vy ta suy ra gcd(n, k) = 1 nn k|2. Vi k = 1 th p|1, mu thun.Vy k = 2. Do p = 3 hay 3|n.t v3(n) = k (k N). p dng nh l 2 th ta c
v3(2n + 1) = v3(3) + v3(n) = 1 + k
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22
Li c v n2|2n + 1 nn v3(2n + 1) > v3(n2) k + 1 > 2k. Vy k = 1. t n = 3m vi m Nv gcd(m, 3) = 1.
Gi p1 l c nguyn t nh nht ca m. Khi ta c 26m 1 (mod p1). Gi k1 l s nguyn
dng nh nht tha mn 2k 1 (mod p1). Tng t th ta d dng suy ra k|6. V p1 > 5 nnk = 3 hoc k = 6.
Vi k = 3 th p1|7 nn p1 = 7. Vi k = 6 th p1|63 m p1 > 5 nn p1 = 7. Tuy nhin2n + 1 = 23m + 1 = 8m + 1 2 (mod 7) m 7|n2, mu thun.Vy c nguyn t duy nht ca n l 3 m 3 n nn n = 3.S nguyn dng n tha mn bi l n {1; 3}.V d 2. (European Mathematical Cup 2012, Senior Division) Tm s nguyn dng a, b, n v
s nguyn t p tha mn
a2013 + b2013 = pn
Li gii. t a = px y, b = pz t vi x, y, z, t N; t, y > 1 v gcd(y, p) = 1, gcd(t, p) = 1.Khng lm mt tnh tng qut, gi s rng x > z. D nhn thy rng n > 2013x > 2013z. Khi phng trnh ban u tng ng vi
t2013 + p2013(xz) y2013 = pn2013z
Nu x > z th p - V T . Do p - pn2013z suy ra n = 2013z. Vy ta c phng trnh
t2013 + p2013(xz) y2013 = 1,
mu thun v V T > 2 (do , ty > 1). Vy x = z. Phng trnh tr thnh
t2013 + y2013 = pn2013z = pk (k = n 2013z N) (3)
Nu p|2013 th theo nh l Fermat nh ta suy ra t2013 + y2013 2 (mod p), mu thun v p|pk.Vy gcd(p, 2013) = 1.
D thy theo (3) th p|t+ y. Do bng vic p dng nh l 2 ta c
vp(t2013 + y2013
)= vp(t+ y)
Ta li c t+ y|t2013 + y2013 v (3) nn ta suy rapk = t+ y = t2013 + y2013
t2012(t 1) + y2012(y 1) = 0V t, y > 1 nn t phng trnh ta suy ra t = y = 1. Do p = 2, t suy ra a = b = 2h, n =2013h+ 1 vi h N.Nhn xt. Ta c th tng qut bi ton ln thnh: Gii phng trnh nghim nguyn dng
an + bn = pk
vi p nguyn t.
V d 3. (Romanian IMO TST 2005) Gii phng trnh nghim nguyn dng
3x = 2x y + 1
Li gii. Ta xt hai trng hp:
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23
1. Nu x l th p dng nh l 1 ta c v2(3x 1) = v2(3 1) = 1 hay v2(2x y) = 1. Do
x = 1. T phng trnh ta suy ra y = 1.
2. Nu x chn th p dng nh l 1 ta c
v2(3x 1) = v2(3 1) + v2(3 + 1) + v2(x) 1 = 2 + v2(x)
v2(2x y) = 2 + v2(x) x+ v2(y) = v2(x) + 2 (1)
t x = 2m k vi m,n N. Ta d dng chng minh bng quy np rng 2m k > m+ 2vi m N, m > 3. Do x > v2(x) + 2 vi v2(x) > 3 hay vi x > 2v2(x) = 8. Nh vyx > 8 th (1) khng xy ra. Vy x 6 8, x chn nn x {2; 4; 6}. T y ta tm c(x, y) = (2; 2), (4; 5).
Vy phng trnh c nghim nguyn dng (x, y) = (1; 1), (2; 2), (4; 5).
Nhn xt. Qua bi ton trn, ta lu mt s tng c dng trong phng php ny: Vi
p l mt c nguyn t ca a = pm k vi m, k N th:
i) a > pvp(a).
ii) pm k > m+ vi m > . T y suy ra a > vp(a) + vi vp(a) > hay a > p.
Cc bi trn ch yu l cc bi khng kh vn dng b LTE v ta xc nh c cc
yu t p, a, b mt cch d dng. Tuy nhin, vn c mt s bi ton i hi ta phi i tm ra
cc yu t p, a, b ...
V d 4. (IMO 1999) Tm tt c cc cp (n, p) nguyn dng sao cho p l s nguyn t v
(p 1)n + 1 chia ht cho np1.Li gii. D thy vi n = 1 th p l s nguyn t bt k u tha mn ra. Vi n > 2, ta ccc trng hp:
Trng hp 1. Nu p = 2 th n|2. Do n = 2.Trng hp 2. Nu p l. Ly q l c nguyn t nh nht ca n, khi (p 1)n 1(mod q) hay (p 1)2n 1 (mod q) v gcd(p 1, q) = 1. Ta ly o l s nguyn dng nh nhttha mn (p 1)o 1 (mod q). Khi th ta suy ra o|2n. p dng nh l Fermat nh ta c(p 1)q1 1 (mod q). Do o|q 1.Nh vy, o|2n v o|q 1. Nu gcd(o, n) > 1 hay o, n chia ht cho s nguyn t r, khi ta suyra r|n v r 6 o. M o|q 1 nn o < q, do r < q. M r v q u l c nguyn t ca n, muthun vi iu kin nh nht ca q. Vy gcd(n, o) = 1. Do 2|o. Vy (p 1)2 1 (mod q)hay q|p(p 2).
1. Nu q|p 2 th ta c (p 1)n + 1 1n + 1 2 (mod q). Vy q = 2. Ta c (p 1)n + 1chia ht cho 2 nn p = 2, mu thun v p l.
2. Nu q|p. D nhn thy n phi l (v nu n chn th (p 1)n + 1 0 (mod 4), mu thunv p l). Ta p dng nh l 2 ta c
vq ((p 1)n + 1) = vq(n) + vq(p) > vq(n) (p 1) (4)
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24
t p = qa b vi a, b N. D dng chng minh bng quy np qa b > a + 2 (ch vq|p nn q > 3), du bng xy ra khi a = b = 1, q = 3. Do p > vq(p) + 2. Kt hp vi(3) ta suy ra
p 2 > vq(p) > vq(n)(p 2)Vy q = p = 3 v v3(n) = 1. t n = 3k vi k N, gcd(k, 3) = 1, gcd(k, 2) = 1. Nhvy t bi ta s c 9k2|8k + 1.Hin nhin 9|8k + 1. Ta ch cn i tm k sao cho k2|8k + 1. Vi k = 1 th n = 3, tha mn.Vi k > 2, hon ton tng t, ly r l c nguyn t nh nht ca k v s l s nguyndng nh nht sao cho 8s 1 (mod r). Ta suy ra s|2 nn s = 2. Khi r|82 1 hayr|7, iu ny mu thun v 8k + 1 2 (mod 7).
Vy, cp s (n, p) tha mn bi l (1, p), (2, 2), (3, 3).
V d 5. (Brazil XII Olympic Revenge 2013) Tm cc b ba s (p, n, k) nguyn dng tha
mn p l s nguyn t Fermat v
pn + n = (n+ 1)k (5)
S nguyn t Fermat l s nguyn t c dng 22x
+ 1 vi x t nhin.
Li gii. t = 2x. Nu n = 1 th (5) p = 2k 1 = 2 + 1. Do k = 2, = 1 nn p = 3.Nu n > 2. Ta gi r l mt c nguyn t ca n. T phng trnh ta suy ra pn 1 (mod n)hay pn 1 (mod r). Do gcd(p, r) = 1. t k l s nguyn dng nh nht tha mn pk 1(mod r). Ta cng c theo nh l Fermat nh th pr1 1 (mod r). Vy ta suy ra k|r 1 vk|n. V gcd(r 1, n) = 1 nn k = 1. Ta c r|p 1 hay r|2. Vy r = 2 hay 2|n. Ta c
(5) pn 1 = (n+ 1) [(n+ 1)k1 1]T phng trnh dn n v2(p
n 1) = v2((n+ 1)k1 1).
Nu k 1 l th
v2((n+ 1)k1 1) = v2(n) < v2 (p2 1)+ v2(n) 1 = v2(pn 1),
mu thun. Vy k 1 chn. p dng nh l 4 ta cv2(p
n 1) = v2((n+ 1)k1 1)
v2(p2 1) + v2(n) 1 = v2(n) + v2(n+ 2) + v2(k 1) 1v2(p 1) + v2(p+ 1) = v2(n+ 2) + v2(k 1)
Nu v2(k 1) > v2(p 1) th p 1|k. Do (n + 1)k n + 1 (mod p) theo nh l Fermatnh. Tuy nhin theo (5) th n (n + 1)k (mod p) nn n n + 1 (mod p), mu thun. Vyv2(k 1) < v2(p 1). Khi theo phng trnh ta c
1 6 v2(p+ 1) = v2(2 + 2) < v2(n+ 2)
Do v2(n+ 2) > 2. Ta suy ra n 2 (mod 4).1. Nu p > 5 th 22
x+ 1 > 5 nn x > 2. Do p 2 (mod 5). p dng n 2 (mod 4) th
ta suy ra pn 4 (mod 5). Do 4 + n (n + 1)k (mod 5). V n + 4 6 n + 1 (mod 5)nn k 6 1 (mod 4). V k l nn k 3 (mod 4). Vy 4 + n (n+ 1)3 (mod 5).
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25
Nu n 0 (mod 5) th 4 + n (n+ 1)3 3 (mod 5), mu thun. Nu n 1 (mod 5) th 4 + n (n+ 1)3 2 (mod 5), mu thun. Nu n 2 (mod 5) th 4 + n (n+ 1)3 4 (mod 5), mu thun. Nu n 3 (mod 5) th 4 + n (n+ 1)3 3 (mod 5), mu thun. Nu n 4 (mod 5) th 4 + n (n+ 1)3 3 (mod 5), mu thun.
Vy vi mi n N th n+ 4 6 (n+ 1)3 (mod 5). Ta loi trng hp p > 5.
2. Nu p = 5 th = 2. Khi th 3 = v2(n + 2) + v2(k 1). V v2(n + 2) > 2 nn ta suyra v2(n+ 2) = 2, v2(k 1) = 1. Ta cng c 5n + n = (n+ 1)k.
Vi n = 2 th k = 3. Vi n > 3. Gi q l c nguyn t l ca n th q|5(n,q1) 1 = 52 1 = 24. Vy q|3
nn q = 3. Do n 0 (mod 6). Kt hp vi n 2 (mod 4) ta suy ra 5n 1(mod 13) nn n 1 (n+ 1)k (mod 13). p dng nh l 1 ta c
v3(5n 1) = v3
((n+ 1)k1 1) 1 + v3 (n
2
)= v3(k 1) + v3(n)
Vy 3|k 1. Ta cng c k 3 (mod 4) nn k 7 (mod 12). Theo nh l Fermatnh ta suy ra (n + 1)k (n + 1)7 (n + 1) (mod 13). Nh vy n 1 n 1(mod 13) dn n n 0 (mod 13), v l. (v vi 13|n th 5n 1 (mod 13), muthun do 5n 5 (mod 13)).
Vy (p, n, k) = (3, 1, 2), (5, 2, 3).
V d . Tm b ba s nguyn dng (a, b, c) sao cho ab + 1 = (a+ 1)c.
Li gii. Gi p l mt c nguyn t l ca a. Khi th theo nh l 1 ta c
vp((a+ 1)c 1) = vp(a) + vp(c) > vp(a) b
vp(c) > vp(a)(b 1) (6)
1. Nu c l th ta c v2 ((a+ 1)c 1) = v2(a). Do b = 1. Nh vy th ta c a+1 = (a+1)c
suy ra c = 1.
2. Nu c chn th v2(c) > 1 v b > 2. Theo nh l 4 th
v2((a+ 1)c 1) = v2(a) + v2(a+ 2) + v2(c) 1 = v2(a) b (7)
Nu v2(a) = 1 th ta lun c v2(c) > v2(a). Kt hp vi (6) ta suy ra c > a(b1) > b,mu thun v lc th (a+ 1)c > ab + 1.
Nu v2(a) > 2 th (7) v2(c) = v2(a) (b 1). Kt hp vi (6) ta dn nc > a(b 1) > b, mu thun
Vy phng trnh c nghim (a, b, c) = (k, 1, 1) vi k l s nguyn dng ty .
Nhn xt. T bi ton trn, ta c thm mt s m rng sau:
M rng 1. Tm cc s nguyn dng m, l, n, k tha mn (1 +mn)l = 1 +mk.
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26
M rng 2. (IMO Shortlist 2000) Tm b ba s nguyn dng (a,m, n) tha mn am+1|(a+1)n.
Ngoi vic phng php LTE c ng dng trc tip vo li gii th phng php ny cn
c dng tm dng v hn ca bi ton chia ht.
V d 6. Chng minh tn ti v hn s t nhin n tha mn n|3n + 1.Phn tch v nh hng li gii. iu by gi ta cn lm v i tm mt trong cc dng ca n
tha mn n|3n + 1.Trc ht, nhn thy 5|32 + 12. By gi ta n cc iu kin a, b, p trong nh l 2, pdng v ta s c 5k+1|325k + 125k . Do 2 5k|325k + 1. Vy ta ch cn chng minh n = 2 5kvi k N th n|3n + 1.Li gii. Trc ht, ta s i chng minh 345
k1 1 (mod 5k). p dng nh l 1 ta c
v5
(345
k1 1)
= v5(34 1) + v5(5k1) = k
Vy 345k1 1 (mod 5k) hay 5k|
(325
k 1)(
325k
+ 1). Do 5k|325k + 1. Li c 2|325k + 1
nn 2 5k|325k + 1.V k N nn tn ti v hn s t nhin n = 2 5k sao cho n|3n + 1.V d 7. (Romanian Master of Mathematics Competition 2012) Chng minh tn ti v hn
s nguyn dng n tha mn 22n+1 + 1 chia ht cho n.
Phn tch v nh hng li gii. Ta s tm mt s n tha mn iu kin trn. D thy n = 3
tha mn. Ta mnh dn th vi n = 9, 27 cng u tha mn. T y ta d dng tm cmt dng ca nn l n = 3k. y mnh xin gii thiu hai li gii:
Li gii 1. Ta s i chng minh s nguyn dng an = 3n tha mn yu cu bi ton. Tht
vy, theo nh l 2 ta c
v3(2an + 1) = v3(3) + v3(an) = k + 1
V
v3(22an+1 + 1) = v3(3) + v2(2
an + 1) = k + 2
Vy an|22an+1 + 1.Li gii 2. Ta s i chng minh s nguyn dng an =
23n+19
tha mn yu cu ra.
p dng nh l 2 ta c
v3(an) = v3(3) + v3(3n) 2 = n 1
t an = 3n1m vi m N, gcd(3,m) = 1. Ta c
v3(22an+1 + 1) > v3(2
an + 1) > v3(an) = n 1.
Vy 3n1|22an+1 + 1. Mt khc, tip tc p dng nh l 2 th
v3(2an + 1) = v3(3) + v3(an) = n
Do 3n|2an + 1. Vy ta suy ra 23n + 1|22an+1 + 1. M m|2an + 1 nn m|22an+1 + 1.V gcd(m, 3) = 1 nn an|22an+1 + 1.
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Bi tp vn dng
1. Chng minh phng trnh x7 + y7 = 1998z khng c nghim nguyn dng.
2. Tm tt c s nguyn dng n tha mn 72013|5n + 1.
3. Tm s nguyn dng n ln nht sao cho 2n|2011201320161 1.
4. Chng minh tn ti v hn s nguyn dng n N tha mn n2|2n + 3n + 6n + 1.
5. (Japan MO Finals 2012) Cho p l s nguyn t. Tm mi s nguyn n tha mn vi mi
s nguyn x, nu p|xn 1 th p2|xn 1.
6. Cho a > b > 1, b l mt s l, n l mt s nguyn dng. Nu bn|an 1. Chng minhab > 3
n
n.
7. Tm s nguyn dng n tha mn 9n 1 chia ht cho 7n.
8. (IMO Shortlist 2007) Tm mi hm s ton nh f : N N sao cho vi mi m,n N vvi mi p nguyn th, f(m+ n) chia ht cho p khi v ch khi f(m) + f(n) chia ht cho p.
9. (IMO 2000) Tn ti hay khng s nguyn n tha mn n c ng 2000 c nguyn t v
2n + 1 chia ht cho n ?
10. Vi mt s t nhin n, cho a l s t nhin ln nht tha mn 5n 3n chia ht cho 2a.Ly b l s t nhin ln nht tha mn 2b 6 n. Chng minh rng a 6 b+ 3.
11. Chng minh rng nu n > 2 sao cho n|7n 3n th n chn.
12. Tm s nguyn dng n tha mn
i) n|5n + 1.ii) n2|5n + 1.iii) n3|5n + 1.
13. Tm mi s nguyn dng k sao cho k s nguyn t l u tin p1, p2 , pk u tn tihai s nguyn dng a, n tha mn
p1 p2 pk 1 = an
14. (MOSP 2001) Tm cc s nguyn dng (x, r, p, n) tha mn xr 1 = pn.
15. Tm tt c cc b s (m, p, q) vi p, q nguyn t v m nguyn dng sao cho 2mp2+1 = q5.
16. (Iran TST 2009) Cho n l mt s nguyn dng. Chng minh rng
352n1
2n+2 (5) 32n12n+2 (mod 2n+4)
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28
17. (IMO Shortlist 2010) Tm cc cp s nguyn khng m (m,n) tha mn
m2 + 2 3n = m (2n+1 1) .18. (Iran Third Round 2011) Cho s t nhin k > 7. C bao nhiu cp nguyn dng (x, y)
tha mn
7373x 99y (mod 2k)?
19. Gii phng trnh nghim nguyn dng trong p l s nguyn t:
pa 1 = 2n(p 1)
Ti liu tham kho
[1] Amir Hossein Parvardi, Lifting The Exponent Lemma: (ti liu pdf)
[2] Cc din n ton:
diendantoanhoc.net/forum
forum.mathscope.org
mathlinks.ro
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Chuyn 3:Cc bi ton s hc hon v vng quanh
Nguyn Anh Huy, Nguyn Vit Tm 1
S bnh ng v hon v gia cc n s l mt np p ca Ton hc, thng c
gp trong cc bi ton h phng trnh v bt ng thc. Tng t, S hc cng
c nhng bi ton hon v vng quanh m cc n l s nguyn, s nguyn dng,
tuy nhin li gii a dng v phc tp hn rt nhiu. Bi vit ny s cp n
hai hng i c th gii dng bi trn, l i xng ha v bt ng thc.
Trong bi vit c s dng mt s kin thc v bc nhy Viete v hm vp(n), bn
c c th tham kho hai chuyn trc.
Phng php i xng ha
Phng php ny thng c dng trong cc bi ton chia ht hon v vng: a | f(b); b | f(a).Nu c (a; b) = 1 ta xy dng hm g tha g(x)
... x x Z v h thah(a; b) = f(a) + g(b) = f(b) + g(a) i xng theo a, b. Khi
a | h(a; b); b | h(a; b) ab | h(a; b) h(a; b) = kaby l phng trnh nghim nguyn vi a, b i xng. Ta c th dng bt ng thc nu
deg h = 1 hoc bc nhy Viete nu deg h = 2.
Nu khng c (a; b) = 1 th t gi thit ta suy ra ab | f(a)f(b), sau khai trin v phi v bcc hng t chia ht cho ab c phng trnh nghim nguyn c dng tng t.
Bi tp 3.1. Tm cc s nguyn t p > q thaq 1 | 3p 1p 1 | 3q 1Li gii
t a = p 1, b = q 1 (a > b > 1), ta ca | 3(b+ 1) 1b | 3(a+ 1) 1 ab | (3a+ 2)(3b+ 2) () 6a+ 6b+ 4 ... ab1Lp 12CT THPT chuyn L Hng Phong
29
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30
T ta c
6a+ 6b+ 4 > ab 6a
+6
b+
4
ab> 1
Li c12
b+
4
b2> 6a
+6
b+
4
ab
Suy ra12
b+
4
b2> 1 b2 12b 4 6 0 1 6 b 6 12
Do b + 1 nguyn t nn b {1; 2; 4; 6; 10; 12}. Xt cc trng hp sau, vi lu a + 1 nguynt:
* Nu b = 1 : () 6a+ 6 + 4 ... a 10 ... a a {1; 2; 10}.* Nu b = 2 : () 6a+ 12 + 4 ... 2a 8 ... a a {2; 4}.* Nu b = 4 : () 6a+ 24 + 4 ... 4a a+ 14 ... 2a a {6; 10}.* Nu b = 6 : () 6a+ 36 + 4 ... 6a 40 ... 6a (loi).* Nu b = 10 : () 6a+ 60 + 4 ... 10a 3a+ 32 ... 5a a = 16.* Nu b = 12 : () 6a+ 72 + 4 ... 12a (loi).T ta tm c cp (p, q) nguyn t tha bi ton l
(2, 2), (3, 3), (5, 3), (7, 5), (17, 11). 2
Bi tp 3.2. Tm s b s nguyn dng (a; b; c) i mt nguyn t cng nhau tho a < b < c
v
a | bc 31; b | ca 31; c | ab 31
Li gii
Do a | bc 31 v a | a(b+ c) nn ta c
a | ab+ bc+ ca 31.
Tng t vi b v c. Li do (a; b) = (b; c) = (c; a) = 1 nn
abc | ab+ bc+ ca 31 ()
Xt cc trng hp sau:
~ Trng hp 1: a > 3Suy ra b > 4; c > 5, do ab+ bc+ ca > 31. Ta cng c
abc > 3bc > ab+ bc+ ca > ab+ bc+ ca 31
iu ny mu thun vi (*).
~ Trng hp 2: a = 2Suy ra b | 2c 31 v c | 2b 31, do
bc | 2b+ 2c 31
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31
Chng minh tng t ta c b = 3; c = 5.
~ Trng hp 3: a = 1Suy ra b | c 31 v c | b 31. Nu b+ c = 31 th hai iu kin trn hin nhin tho. Khi tac a = 1 < b < c v b+ c = 31. Do a < b < c nn 2 6 b 6 15. D thy c 14 b s (a; b; c) tho.Nu b+ c 6= 31 th ta chng minh 1 < b < c < 31. Nu ngc li:* Nu b > 31 c > b > b 31 > 0. Do c khng l c ca b 31.* Nu b = 31 th 31 | c 31 31 | c, loi do (b; c) = (31; c) = 1.* Nu b < 31 < c th |c| > |31 b| > 0 do c khng l c ca b 31.Nh vy 1 < b < c < 31. Ngoi ra, t b | c 31 v c | b 31 vi (b; c) = 1 ta cng c
bc | 31 b c.
* Nu b > 5 c > 6 bc > 30 > 11 > 31 (b+ c), v l.* Nu b = 4 th 4 | c 31 v c | 27 c = 27 (loi do b+ c 6= 31).* Nu b = 3 th 3 | c 31 v c | 28, suy ra c {4; 7}.Vy trng hp 3 cho 16 b (a; b; c) tho bi ton.
Kt lun: C 17 b s (a; b; c) tho bi ton. 2
Bi tp 3.3. Tm cc s nguyn t p, q tho p < q < 1000 v
q | p3 1; p | q3 1
Li gii
Ta c q | p3 1 = (p 1)(q2 + q + 1) m q > p 1 nn
q | p2 + p+ 1.
V p | q3 1 = (q 1)(q2 + q + 1). Do ta xt 2 trng hp:
~ Trng hp 1: p | q 1t p2 + p+ 1 = nq (n N). Do p2 + p+ 1 q 1 (mod p) nn n 1 (mod p).Li c
n =p2 + p+ 1
q6[p2 + p+ 1
p+ 1
]=
[p+
1
p+ 1
]= p (do q > p+ 1)
Suy ra n = 1, vy q = p2 + p+ 1. Do p 6 31, v 372 + 37 + 1 > 1000.Th vi tt c s nguyn t p < 31 ta tm c cc b s tho bi ton l
(p; q) = (2; 7), (3; 13), (5; 31), (17; 307).
~ Trng hp 2: p | q2 + q + 1Do p | q2 + q + 1 v q | p2 + p+ 1 nn ta c
p | q2 + q + 1 + p2 + p; q | p2 + p+ 1 + q2 + q
hay
pq | p2 + q2 + p+ q + 1
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Nh vy ta xt phng trnh nghim nguyn dng
a2 + b2 + a+ b+ 1 = mab () (m N)
Vit li (*) di dng
a2 + (1mb)a+ b2 + b+ 1 = 0 ()Gi S l tp cc b s nguyn dng (a; b) tho (**). Theo nguyn l cc hn trong S tn ti
cp s (a0; b0) tho a0 + b0 nh nht. Khng gim tng qut gi s a0 > b0.Theo nh l Viete, (**) cng c nghim (a; b0) trong a thoa0 + a = mb0 1a0a = b20 + b0 + 1T phng trnh u suy ra a Z. T phng trnh sau suy ra a > 0. Vy (a; b0) S. Do
a0 + b0 6 a + b0 a0 6 a = b20 + b0 + 1
a0 b20 + b0 + 1 > a20
Nu a0 > b0 th a0 > b0 + 1 a20 > (b0 + 1)2 > b20 + b0 + 1, mu thun. Vy a0 = b0, suy ra
a20 + (1 a0).a0 + a20 + a0 + 1 = 0
Suy ra 1... a0 a0 = b0 = 1. Vy ta c m = 5.
Khi (*) tr thnh
a2 + b2 + a+ b+ 1 = 5ab ()Theo chng minh trn, nu (a0; b0) l nghim ca (*) th (5a0 b0 1; a0) cng l nghim ca(*). Do t nghim (1; 1) ta c nghim (3; 1), sau l (13; 3), ...
Tm li cc nghim ca (*) c cho bi cng thc
(ai; bi) = (xi+1;xi) vi (xn) :
x0 = x1 = 1xn+1 = 5xn xn1 1 n > 1Vic chng minh iu trn hon ton tng t bi s hc VMO 2012, v xin c dnh cho
bn c.
Tip theo ta tm cc nghim l s nguyn t nh hn 1000. D thy ch cn xt i 6 4, dox6 > 1000. T ta tm c (p; q) = (3; 13), (13; 61).
Kt lun: Cc b s (p; q) tho bi ton l (2; 7), (3; 13), (5; 31), (13; 61), (17; 307). 2
Phng php dng bt ng thc
Ta c th sp xp cc n s i xng theo th t hoc ch ra mt n c mt i lung no
ln nht hoc b nht. Lu vic sp xp y khng ch gii hn a > b > c, m cng cth l f(a) > f(b) > f(c), trong f l mt hm s hc.
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33
Bi tp 3.4. Cho cc s nguyn dng a, b, c. Chng minh
(a; b).(b; c).(c; a)
(a; b; c)2=
[a; b].[b; c].[c; a]
[a; b; c]2
Trong (a; b) v [a; b] ln lt l l c chung ln nht v bi chung nh nht ca a, b.
Li gii
Gi p l s nguyn t bt k. Nu p khng l c ca [a; b; c] th d thy vp(V T ) = vp(V P ) = 0.
Nu p | [a; b; c] th p l c ca a, b hoc c. t a = px.a1; b = py.b1; c = pz.c1 v khng gimtng qut gi s x > y > z. Khi ta c
vp(V T ) = y + z + z 2z = y
v
vp(V P ) = x+ y + x 2x = ySuy ra vp(V T ) = vp(V P ) p P. Do V T = V P. 2Nhn xt l trong li gii trn ta gi s vp(a) > vp(b) > vp(c), v s dng nh l c bnca s hc: Mi s nguyn ln hn 1 u c mt v ch mt cch phn tch ra tha s nguyn t.
Bi tp 3.5. Tm cc s nguyn t p, q tha
(5p 2p)(5q 2q) ... pq ()
Li gii
Gi s p 6 q. Nhn xt p = q = 3 tha.> Nu p = 3, q > 3 th ta c
117(5q 2q) ... 3q 39(5q 2q) ... q
Do 5q 2q 5 2 3 [q] theo Fermat nn q | 39, do q = 13.> Nu 5 < p < q ( V T () khng chia ht cho 5 nn p, q 6= 5) th tng t ta c5p 2p 5 2 3 [p], do t gi thit suy ra
5q 2q ...p
Li c 5p1 2p1 1 [p] nn 5p1 2p1 ... p.Do q > p 1 nn (q, p 1) = 1. Theo nh l Bezout, tn ti m,n N tha
mq (p 1)n = 1 hoc m(p 1) nq = 1 ()
Xt ng d mod p ta c 5p1 2p15q 2q 5n(p1) 2n(p1)5mq 2mq
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34
Suy ra
5n(p1).2mq 2n(p1).5mq
Kt hp vi (**), sau khi rt gn hai v ta c 5 2 [p] hay p = 3 (loi v ang xt p > 5).Kt lun: (p; q) = (3; 3), (3; 13), (13; 3). 2
Bi tp 3.6. Tm cc cp s nguyn t (p, q) tha
2p + 2q... pq
Li gii
Nhn xt l (x 1, x+ 1) = (2, x+ 1) 6 2 x Z.* Nu p = q th 2p+1
... p2 p = q = 2.* Nu p > q :
Xt q = 2 ta c 22 + 2p... 2p 2 + 2p1 ... p p {2; 3}.
Xt q > 2 th do 2p + 2q = 2q(2p1 + 1)... q nn
2p1 + 1... q (1) 22(pq) 1 [q]
Gi a l s nh nht tha 2a 1 [q]. Theo tnh cht ca ord suy ra q 1 ... a v 2(p q) ... a.t p q = 2k.m; q 1 = 2l.n; a = 2r.s (k + 1 > r; l > r;m ... s;n ... s;m,n, s l ).
> Trng hp 1: r = k + 1. Ta suy ra l > k + 1 v2(p q) = 2k+1.mq 1 = 2l.n 2(p q)s = am(q 1)m = (p q).n.2lk1
n.2lk1
2s=
(q 1).m2s(p q) =
q 1a
=2l.n
2r.s
2lk2 = 2lr r = k + 2 (mu thun)
> Trng hp 2: r 6 kSuy ra p q ... a 2pq 1 ... 2a 1 ... q (2)T (1) v (2) suy ra q | (2pq 1, 2pq + 1).M (2pq 1, 2pq + 1) 6 2 theo chng minh trn, do q = 2 (mu thun do ang xt q > 2).Kt lun: (p, q) = (2; 2), (2; 3), (3; 2). 2
Bi tp 3.7. Cho cc s nguyn dng x, y, z tho (xy+1)(yz+1)(zx+1) l s chnh phng.
Chng minh xy + 1, yz + 1, zx+ 1 u l s chnh phng.
Li gii
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35
Trong cc b s (x; y; z) tho bi ton, xt b (x; y; z) c x+ y + z nh nht. (1)
Khng gim tng qut gi s z = max{x; y; z}.Gi t l s tha phng trnh bc hai
t2 + x2 + y2 + z2 2(xy + yz + zt+ tx+ zx+ ty) 4xyzt 4 = 0 ()t2 2t(x+ y + z + 2xyz) + x2 + y2 + z2 2(xy + yz + zx) 4 = 0
Nhn xt rng (*) tng ng vi 3 phng trnh sau:
(x+ y z t)2 = 4(xy + 1)(zt+ 1)(x+ z y t)2 = 4(xz + 1)(yt+ 1)(x+ t y z)2 = 4(xt+ 1)(yz + 1)
V t nguyn do (*) c 2 nghim nguyn
t1,2 = x+ y + z + 2xyz 2
(xy + 1)(yz + 1)(zx+ 1)
Nn nhn c 3 phng trnh trn v theo v, ta suy ra (xt + 1)(yt + 1)(zt + 1) l s chnh
phng.
Ngoi ra ta cng c
xt+ 1 > 0; yt+ 1 > 0; zt+ 1 > 0
Suy ra
t > 1max{x; y; z} > 1 (do x = y = z = 1 khng tho)
* Nu t = 0 th t (*) ta suy ra
(x+ y + z)2 = 4(xy + yz + zx+ 1) (x+ y z)2 = 4(xy + 1)
Suy ra xy + 1 l s chnh phng. Chng minh tng t ta cng c yz + 1, zx+ 1 l s chnh
phng. * Nu t > 0 th t (1) suy ra t > z vi mi t tho (*).Nhng t ch c th bng t1 hoc t2, v ta li c
t1t2 = x2 + y2 + z2 2(xy + yz + zx) 4 6 z2 x(2z x) y(2z y) < z2
Suy ra iu mu thun. Vy ta c pcm. 2
Bi tp 3.8. Tm cc b s nguyn (a; b; c) thaa2 bc = 91b2 ca = 91c2 ab = 91
Li gii
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36
Khng gim tng qut gi s a 6 b 6 c. Do 91 khng l s chnh phng nn a, b, c 6= 0.Nu a > 0 th b, c > 0 do a2 bc 6 0 < 91, vy a < 0.Nu (a; b; c) tha h th (a;b;c) cng tha h. Do ta xt cc b s c c > 0.Nu b = c th ta c
a2 b2 = 91 = b2 ab[a b = 0a = 2b
D thy a = b = c v l, v a = 2b cng dn n 5b2 = 91, v l.Do b 6= c. Tng t ta c a 6= b.
~ Trng hp 1: b > 0T h ta c
91b = b(b2 ac); 91c = c(c2 ab) 91b 91c = b3 c3 91 = b2 + bc+ c2 > 3b2 b {1; 2; 3; 4; 5}
Thay vo h ta tm c cc nghim (a; b; c) = (10; 1; 9), (11; 5; 6).
~ Trng hp 2: b < 0T h ta c
91b = b(b2 ac); 91a = a(a2 bc) 91(b a) = b3 a3 91 = a2 + ab+ b2 > 3b2 b {1;2;3;4;5}
Thay vo h ta tm c cc nghim (a; b; c) = (9;1; 10), (6;5; 11).Tm li cc b (a; b; c) tha gi thit l
(10; 1; 9), (10;1;9), (11; 5; 6), (11;5;6), (9;1; 10), (9; 1;10), (6;5; 11), (6; 5;11)v cc hon v. 2
Bi tp t luyn
Bi 1: (APMO 2002) Tm cc s nguyn dng a, b tho
b2 a | a2 + b; a2 b | b2 + aBi 2: Chng minh c v hn b s nguyn dng (a; b; c) tho ab + 1, bc + 1, ca + 1 u l
s chnh phng.
Bi 3: Tm cc s nguyn dng x, y, z i mt nguyn t cng nhau tho
x
y+y
z+z
x N
Bi 4: Tm cc b s nguyn (x; y; z) tho 2 6 x 6 y 6 z v
z | xy 1; y | zx 1;x | yz 1
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37
Bi 5: Tm cc s nguyn dng a, b, c > 1 i mt khc nhau tha
(a 1)(b 1)(c 1) | abc 1
Bi 6: Chng minh vi mi s nguyn dng a, b, n th
(36a+ b)(36b+ a) 6= 2n
Bi 7*: (VMO 2013) Tm s cc b (a; b; c; a; b; c) vi a, b, c, a, b, c {0; 1; 2; ...; 14} vtho
ab+ ab bc+ bc ca+ ca 1( mod 15)Bi 8: (VMO 2012) Tm cc s nguyn dng l a, b tho
a | b2 + 2; b | a2 + 2
Bi 9: Cho a, b, c Z thaa
b+b
c+c
a= 3
Chng minh abc l lp phng mt s nguyn.
Ti liu tham kho
1. Trang web brilliant.org.
2. Cc chuyn s hc - Phan Huy Khi.
3. Cc phng php gii ton qua cc k thi Olympic - Trn Nam Dng (Ch bin), V
Quc B Cn, L Phc L.
4. Problems in Elementary Number Theory - Peter Vandendriessche, Hojoo Lee
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38
-
Chuyn 4:DY S S HC
Ninh Vn T1
Dy s l mt vn kh thit yu trong gii tch v c ng dng vo kh nhiu
cc lnh vc khc nh phng trnh hm, t hp, s hc. . . Nhng bi ton v gii
hn dy s dng nh tr thnh vn kh quen thuc v xut hin nhiu trong
cc k thi hc sinh gii cp trng, cp tnh, cp thnh ph cng nh cc k thi
Olympic, VMO. . . Nhng mt mng kh c bit ca dy s trong vic ng dng
s hc cng nh cc bi ton s hc gii cc bi ton dy s l mt vn kh
th v trong mng dy s. Hi vng chuyn ny s gip ch cho cc bn trong vic
tip thm kinh nghim v mng th v ny.
Dy s nguyn v tnh cht s hc
Ta s ln lt xt tng quan cc bi ton vi dng tng qut c th c ci nhn
bao qut ht v th gii dy nguyn phong ph, a mu. T ta s c c cc
tng khi gp mt bi ton chng minh dy nguyn hoc cc bi ton lin quan
n dng y.
Bi tp 4.1. Cho a, b, c Z tha mn a2 = b+ 1. Dy s (un) c xc nh{u0 = 0
un+1 = aun +bun2 + c2, n N
Chng minh rng mi s hng ca dy u l s nguyn.
Li gii
Cch gii bi ton ny cng tng t nh cch tm cng thc truy hi ng ca mt dng cng
thc truy hi cn thc tm cng thc tng qut ca mt dng dy ny nhng vi iu kin
ngt. Ta c:
un+1 aun =bun2 + c2
un+12 2aunun+1 + a2un2 = bun2 + c2un+12 = 2aunun+1 un2 + c2 (1)
1Hc sinh THPT chuyn Trn i Ngha.
39
-
40
Gim n xung 1 n v ta c:
un2 = 2aunun1 un12 + c2 (2)
Ly (1) (2), ta c:
un+12 un12 = 2aun (un+1 un1)
[un+1 = un1
un+1 = 2aun un1
*Vi un+1 = un1, n > 1 th dy (un) l dy tun hon theo chu k 2. Nh th , ta suy ra{u0 = u2 = u4 = ... = u2k = 0
u1 = u3 = u5 = ... = u2k+1 = |c|
Nn (un) c cc s hng u l s nguyn.
*Vi (un) tha {u0 = 1; u1 = |c|un+1 = 2aun un1, n > 1
,
Ta c cc h s trong h thc truy hi u l s nguyn v 2 s hng u tin ca dy cng l
s nguyn nn t ta suy ra (un) c cc s hng u l s nguyn.
Tm li, cc s hng ca dy trn u l s nguyn. 2
Bi tp 4.2. Cho dy (an) tha mn iu kin{a0 = 2; a1 = 5
an+1an1 an2 = 6n1, n > 1
Chng minh rng tn ti duy nht dy s nguyn dng tha mn iu kin trn.
Li gii
tng ca bi ton ny l tm mt cng thc truy hi ng cho dy trn. Nh vy, ta s
dng phng php quen thuc nh gi bi ton, l phng php sai phn.
Ta c {an+1an1 an2 = 6n1anan+2 an+12 = 6n
an (an+2 + 6an) = an+1 (an+1 + 6an1)
(an+2 + 6an)an+1
=(an+1 + 6an1)
an
t vn =(an+1+6an1)
an, ta c
vn+1 = vn = vn1 = ... = v1 = 5 an+1 = 5an 6an1, n = 1, 2, 3...
Do cc s hng u ca dy l s nguyn: a0 = 2, a1 = 5 v cng thc truy hi ng ca dy
u c cc h s l s nguyn nn t ta suy ra dy trn l dy s nguyn.
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41
c thm iu kin dy trn c mi s hng l s nguyn dng, ta cn phi c thm yu
t n iu tng. Vic chng minh n iu khng th s dng cng thc truy hi m ta va
tm c v ta cha chng minh c cng thc y l duy nht. Chnh v th m ta s chng
minh trc tip bng quy np thng qua cch din t ca bi.
Ta thy iu trn ng vi n = 1 : a1 > a0 (5 > 2).
Gi s iu ny ng n n = k : ak > ak1.
Xt n = k + 1, ta c:
ak+1ak1 ak2 = 6k1 ak+1 = 6k1 + ak2
ak1>
6k1
ak+ ak > ak (ak > ak1)
Vy iu trn cng ng vi n = k + 1 nn theo nguyn l quy np ta suy ra (an) l dy n
iu tng vi mi n N.By gi ta s chng minh dy trn l dy duy nht tha mn iu kin bi:
Cch thng thng m chng ta s ngh n l dng phn chng chng minh tn ti duy
nht.
Gi s tn ti dy an sao cho vi n > 2 tn ti an+2 : an+2 > an+2 tha:{
anan+2 an+12 = 6nanan+2
an+12 = 6n
Suy ra
an (an+2 an+2) = 0 (v l do an+2 > an+2)Nh vy ta suy ra dy trn l dy duy nht tha mn iu kin bi. 2
Nhng bi ton tng t dng trn c th tn ti nhiu dng khc nhau. i lc
nhng cch din t bng truy hi khin ta mt i phng hng sai phn. Nhng
ta cn phi hiu mc ch ca vic chng minh dy s nguyn l tm c cng
thc truy hi ng ca n.
Bi tp 4.3. (GER 2003, Ngy 2) Cho dy s (an) c xc nh bi a1 = 1; a2 = 1; a3 = 2an+3 = an+2an+1 + 7an
, n N
Chng minh mi s hng ca dy trn u l s nguyn dng.
Li gii
Ta c:
an+3an = an+1an+2 + 7
Do lch cc th t ca phn t trong dy khng u nhau nn s kh cho ta ngh n vic
dng sai phn v nu dng sai phn s xut hin thm phn t th 5. Tuy vy, nhng vic p
dng sai phn trong dy s nguyn l mt cng c kh mnh v ta khng nn b qua n chp
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42
nhong trong ngh.
H n xung 1 bc, ta c:
an+2an1 = anan+1 + 7
Tr 2 biu thc trn cho nhau, ta c
an+3an an+2an1 = an+1an+2 anan+1an (an+3 + an+1) = an+2 (an+1 + an1)an+3 + an+1
an+2=an+1 + an1
an
t vn =an+1+an1
an, ta c vn+2 = vn.
Vy dy (vn) tun hon theo chu k 2.
Nh vy ta cn xt thm tnh chn l ca dy, iu ny dn n y l dy gm 2 cng thc
truy hi song song nhau.
*Vi n chn ta suy ra:
v2k+2 = v2k = ... = v2 = 3 an+1 = 3an an1*Vi n l ta suy ra:
v2k+3 = v2k+1 = ... = v3 = 5 an+1 = 5an an1T ta xc nh c cng thc truy hi ca dy l:
a1 = a2 = 1, a3 = 2
an+1 = 3an an1, n = 2kan+1 = 5an an1, n = 2k + 1
(k N)
Do cc s hng u ca dy u l s nguyn v h s ca h thc truy hi ca dy cng l s
nguyn nn ta suy ra mi s hng ca (an) u l s nguyn.
dy trn l dy s nguyn dng, ta cn chng minh dy trn l dy tng ngt vi mi
n > 2.D thy: a3 > a2 nn iu trn ng vi n = 2.
Gi s iu trn ng vi n = k : ak > ak1Xt n = k + 1, ta c:
ak+1 =akak1 + 7
ak2>akak1 + 7
ak1> ak +
7
ak1> ak
Vy iu ny cng ng vi n = k + 1 nn theo nguyn l quy np ta suy ra (an) l dy tng
ngt.
T , ta chng minh c dy (an) c mi s hng u l s nguyn dng. 2
Bi tp 4.4. (Croatia TST 2011) Vi a, b l 2 s nguyn t phn bit, cho dy (xn) tha
mn x1 = a, x2 = b
xn+2 =xn+1
2 + xn2
xn+1 + xn, n N
Chng minh rng xn khng l s nguyn vi mi n > 3.
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43
Li gii
Ta d dng chng minh c xn > 0, x N theo quy np.Ta s chng minh mt b ca dy ny:
B : Nu k t mt phn t n0 sao cho xn0 khng l s nguyn ca dy m c
dng phn thc, th s khng tn ti n xn N, n > n0.Gi s x3 l mt s hu t.
Ta c: x3 =
a2+b2
a+b= c
d(c, d N; (c, d) = 1)
x4 =( cd)
2+b2
cd+b
= c2+b2d2
d(c+bd)
Do (c, d) = 1 nn x4 / N, t
x4 =e
f
(e, f N; (e, f) = 1; f ...d
)
Ta c: f... d f > d. Ta chia lm 2 trng hp:
> Trng hp 1: f > d 1 > df, ta c:
x5 =
(cd
)2+(ef
)2cd
+ ef
=(cf)2 + (ed)2
(cf + de) df
Gi s x5 N, suy ra
(cf)2 + (de)2... df (de)2 ... df de2 ... f (v l)
Mt khc t s cng khng th bng mu s do f < (cf)2 + (de)2 nn f khng th chia ht
cho (cf)2 + (de)2.
iu ny v l do (e, d) = 1 v d < f .
Vy x5 l s v t.
> Trng hp 2: f = d, ta c:
x5 =
(cd
)2+(ef
)2cd
+ ef
=(cf)2 + (ed)2
(cf + de) df=
c2 + e2
f (c+ e)
Gi s x5 N, suy rac2 + e2
... c+ e 2ce ... c+ eiu ny hin nhin v l.
Vy tm li, ta chng minh rng, nu xut pht vi x3 l mt s hu t th ta s thu c
mt dy s hu t v khng tn ti bt k s nguyn no trong dy vi n > 3.By gi ta ch vic chng minh x3 khng th l s nguyn. Do cng thc truy hi c dng phn
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44
thc nn x3 s c dng phn s ti gin.
Gi s x3 N, suy raa2 + b2
... a+ b 2ab ... a+ bXt mt trong hai s l s nguyn t chn. Gi s l a, suy ra
a = 2 4b ... b+ 2 4 ... b+ 2 (v l do b > 2)
Suy ra 4b = b+ 2 (v l do b l s nguyn t l nn tnh chn l ca 2 v khng ng nht).
K hiu {a, b} l c ca a hoc b.Vy a+ b | {a, b, 2} do (a, b, 2) = 1.Ta c nhn xt rng nu a+ b | {a, b, 2} do:* a+ b | {a, b} th {
a > a+ bb > a+ b
{
0 > b0 > a
(sai do a, b > 0)
* a+ b | {2}: Do a, b l s nguyn t nn a+ b > 2. Suy ra
a+ b = 2ab a+ b
... a
a+ b... b cb
... a
a... b a = b (sai do a 6= b)
Nh vy, ta kt lun rng iu gi s l sai nn x3 khng th l s nguyn.
Vy theo b trn ta suy ra xn khng th l s nguyn vi mi n > 3. 2
Bi tp 4.5. Cho mt dy s nguyn dng sao cho ta c th chn 1998 phn t bt k ca
dy to thnh mt h thng d khng y v ring bit nhau khi xt modulo 1999. Liu c
tn ti cch chia 1998 phn t ny thnh 2 tp con A,B sao cho tch ca cc phn t trong
mi tp con bng nhau khng?
Li gii
Mt suy lun n gin cho bi ny: 2 tch bng nhau th chng phi c cng s d khi
xt modulo vi mt s nguyn dng.
S dng b quen thuc: Nu x2 + y2... p (p nguyn t ) th p = 4k + 3 hoc p = 4k + 1. Khi
p = 4k + 3 th x, y... p.
Gi 1998 phn t bt k ca dy tha iu kin bi l x1;x2;x3; ...;x1998.
Ta c: A B = {x1;x2; ...;x1998}.Gi s xi 0 (mod 1999). Do 1998 phn t ny lp thnh mt h thng d khng y vring bit nhau nn
xj r 6= 0 (mod 1999) j 6= iV vy nu xi A th A chia ht cho 1999 v B khng chia ht cho 1999. Tng t nu xi B.Vy khng c phn t no trong dy chia ht cho 1999. Vy h thng d ca dy: {1; 2; 3...; 1998}.Ta c: (
xkAxk
2
)=xiA
xi.xjB
xj 1.2.3...1998 (mod 1999)
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45
Suy ra (xkA
xk
)2 1998! (mod 1999)
Mt khc theo nh l Wilson ta c: 1998! 1 (mod 1999), do (xkA
xk
)2+ 1 0 (mod 1999)
p dng b trn ta suy ra iu v l do 1999 l s nguyn t dng 4k + 3 nn 1 khng th
chia ht cho 1999.
Vy khng th chia 1998 phn t bt k ca dy nguyn dng ny thnh 2 tp con c tch cc
phn t trong tp bng nhau. 2
Bi tp 4.6. Cho dy s (un) thau1 = 1, u2 = 2
un+1 =unun1(n+ 1)
Pn1
a) Tm s nguyn t p S = up21pp1 l s nguyn.
b) Chng minh[un1n
] 0 (mod n 1).Li gii
tng ca bi ny n gin l tm pl s nguyn t up2 1 0 (mod pp1). Qua , tad on cng thc tng qut ca un c th a v mt bi ton s hc n thun.
> Cu a:Ta chng minh theo qui np un = n! n 1, 2, 3....Tht vy iu ny ng vi n = 1; 2 :
{u1 = 1 = 1!
u2 = 2.u1 = 2!
Gi s iu ny ng n n = ktc uk = k! Xt uk+1, ta c:
uk+1 =(k + 1)ukuk1
Pk1=
(k + 1)!(k 1)!(k 1)! = (k + 1)!
Vy iu ny ng n n = k + 1nn theo nguyn l quy np ta suy ra un = n!n = 1; 2; ....Vy S = (p2)!1
pp1 .Gi s S l s nguyn, ta suy ra S.pp1 + 1 = (p 2)!.
Xt p > 5. ta c:
(p 2)! 0 (mod p 1) S.pp1 + 1 0 (mod p 1) S + 2 0 (modp 1)
Mt khc: S + 2 < p 1 (p 2)! < 1 + pp2(p 3) (ng).T ta suy ra iu v l. Vy p 6 5. D thy p = 3 tha mn iu kin bi.
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46
> Cu b:+Trng hp 1: n l s nguyn t.
Theo nh l Wilson ta c:
(n 1)! 1 (mod n) (n 1)! + 1n
N
Li c [un1n
]=
[(n 1)!
n
]=
[(n 1)! + 1
n 1n
]=
(n 1)! + 1n
1 (do 0 < 1n< 1)
Suy ra [un1n
]=
(n 1)! (n 1)n
=(n 1) [(n 2)! 1]
n
Mt khc, do (n;n 1) = 1 (n2)!1n
N nn ta suy ra[un1n
] 0 (mod n 1)
+ Trng hp 2: n khng l bnh phng ca 1 s nguyn t n = rs vi 1 < r < s 6 n 1.Do r < s 6 n 1 nn ta suy ra r < s 6 n 2 nn r, sl c s ca mt trong cc s{1; 2; 3...;n 2}.Vy
(n 2)!n
=(n 2)!rs
N [un1n
]=
(n 1).(n 2)!n
= (n 1).k 0 (mod n 1)
+Trng hp 3: n l bnh phng ca 1 s nguyn t. t n = p2 vi 3 6 p < n 1 v nup = 2 th hin nhin ta c iu phi chng minh.
Lp lun tng t nh trng hp 2, p l c ca mt trong cc s {1; 2; 3...;n 2}. By gita xt xem liu p cn l c ca s no na khng.
Tht vy, ta c: n 1 = p2 1 = (p 1)(p+ 1) > 2p > p.Vy 2p cn l c ca mt s bt k trong cc s {1; 2; 3...;n 2}. Chnh v th m
(n 1)! 0 (mod 2p.p.(n 1)) (n 1)! 0 (mod 2n(n 1))
(n 1)!n
0 (mod n 1)[un1n
] 0 (mod n 1)
T cc iu trn ta c pcm. 2
Bi tp 4.7. Cho dy s xc nh bi:
an+1 = {an} [an] , n > 0
Chng minh rng an = an+2 khi n ln.
Li gii
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47
> Xt 0 < a0 < 1, ta c
[a0] = 0 an = [an] {an} = 0, n = 1, 2, 3...
> Xt a0 > 1, ta c:
an+1 = {an} [an] < [an] < an 0 < [an+1] < [an] < ... < [a1] [an0 ] = c1 khi n0 ln
> Xt a0 < 0 an < 0, n = 1, 2, 3... ta c:
an+1 = {an} [an] > an 0 > [an+1] > [an] > ... > [a1] [an0 ] = c2 khi n0 ln
Nh vy [an0 ] = k vi n0 ln.
Ta c:
an+1 = {an} [an] = an [an] ([an])2 = kan k2, n > n0S dng phng php sai phn vi tng hng t ca dy vi h s tng ng ta c:
kn1an+1 = knan kn+1knan+2 = k
n1an+1 kn...
kai+n1 = k2ai+n2 k3ai+n = kai+n1 k2
ai+n = knan k2(1 kn)1 k = k
n
(an +
k2
1 k) k
2
1 k
Nu |k| > 1 th
an+i = kn
(an +
k2
1 k) k
2
1 k (v l do dy trn b chn)
Suy ra 1 6 k 6 1 k {1; 0; 1}.* Nu k = 0 : an = 0, n > n0* Nu k = 1 : an+1 = an 1, n > n0 (v l do [an0 ] = k n > n0 m khong cch gia 2 hngt lin tip ca dy l 1).
* Nu k = 1 suy ra
an+1 = 1 an, n > n0 {an+1 = 1 anan+2 = 1 an+1
Tr nhau ta c an = an+2, n > n0 (pcm). 2
Dy s nguyn v tnh chnh phng
Bi tp 4.8.
a) Chng minh
(3 +
5
2
)n+
(35
2
)nl s nguyn.
b) Chng minh mi s hng l ca dy
(3 +
5
2
)n+
(35
2
)n 2 u l s chnh phng.
-
48
Li gii
> Cu a:Xt dy s sau: {
x1 = 3; x2 = 7
xn+2 = 3xn+1 xnn = 1; 2; ....Ta chng minh theo quy np
xn =
(3 +
5
2
)n+
(35
2
)nn = 1; 2; ...
Tht vy iu ny ng vi n = 1; 2 :
x1 =
(3 +
5
2
)+
(35
2
)= 3
x2 =
(3 +
5
2
)2+
(35
2
)2= 7
Gi s iu ny ng n n = ktc xk =
(3 +
5
2
)k+
(35
2
)k.
Xt xk+1, ta c:
xk+1 = 3xk xk1 = 3(3 +5
2
)k+
(35
2
)k(3 +5
2
)k1+
(35
2
)k1 xk+1 =
(3 +
5
2
)k1(7 + 3
5
2
)+
(35
2
)k1(7 35
2
)=
(3 +
5
2
)k+1+
(35
2
)k+1Vy iu ny ng n n = k + 1 nn theo nguyn l quy np ta suy ra
xn =
(3 +
5
2
)n+
(35
2
)nn = 1; 2; ...
Do xn =(3+5
2
)n+(352
)nn = 1; 2; ... c cng thc truy hi vi cc h s nguyn v x1;x2
nguyn nn t ta suy ra xn nguyn. Vy
(3 +
5
2
)n+
(35
2
)nnguyn n = 1; 2; ....
> Cu b:Tip tc vi tng bin lun dy nguyn theo cng thc truy hi v mt cht bin i kho
lo chng minh s chnh phng.
Ta c: (3 +
5
2
)n+
(35
2
)n 2
=
(5 + 12
)2n+
(5 12
)2n 2
=
[(5 + 1
2
)n(
5 12
)n]2
-
49
Vy bi ton tr thnh bi ton th nht v ta ch vic chng minh xn =
(5 + 1
2
)n(
5 12
)nnguyn vi mi n l.
Chng minh theo quy np xn =
(5 + 1
2
)n(
5 12
)nc cng thc truy hi l
{x1 = 1; x2 =
5
xn+2 =
5xn+1 + 2xn n = 1; 2; ...v x2n =
5;x2n+1 =
Tht vy iu ny ng vi n = 1; 2 :
x1 =
(5 + 1
2
)(
5 12
)= 1
x2 =
(5 + 1
2
)2(
5 12
)2=
4
5
4=
5
Gi s iu ny ng n n = 2k tc x2k+2 =
5x2k+1 x2kv x2k+1 nguyn do x1;x3; ...nguyn v x2k =
5.
Vy x2k+2 =
5 ( = x2k+1 )Xt x2k+3;x2k+4. Ta c:
x2k+3 =
5x2k+2 x2k+1
=
5
(5 + 12
)2k+2(
5 12
)2k+2(5 + 1
2
)2k+1(
5 12
)2k+1x2k+3 =
(5 + 1
2
)2k+1(5 +
5
2 1)(
5 12
)2k+1(55
2 1)
x2k+3 =(
5 + 1
2
)2k+1(3 +
5
2
)(
5 12
)2k+1(35
2
)
=
(5 + 1
2
)2k+3(
5 12
)2k+3(1)
V
x2k+4 =
5x2k+3 x2k+2
=
5
(5 + 12
)2k+3(
5 12
)2k+3(5 + 1
2
)2k+2(
5 12
)2k+2x2k+4 =
(5 + 1
2
)2k+2(3 +
5
2
)(
5 12
)2k+2(35
2
)
=
(5 + 1
2
)2k+4(
5 12
)2k+4(2)
Li c x2k+3 =
5x2k+2 x2k+1 = 5 x2k+1 l s nguyn (3).v x2k+4 =
5x2k+3 x2k+2 =
5( ) khng l s nguyn (4).
-
50
T (1);(2);(3);(4) ta suy ra nhng iu trn ng vi n = k+ 1 nn theo nguyn l quy np ta
chng minh c{x1 = 1; x2 =
5
xn+2 =
5xn+1 + 2xn n = 1; 2; ...v x2n =
5;x2n+1 =
Do x2n+1 = nn ta suy ra mi s hng l ca dy trn u l s chnh phng. 2
Da vo bi trn ta c th xt thm mt bi th v sau:
Bi tp 4.9. Cho dy s (xn) tha mn:{x0 = 1;x1 = 3
xn+2 = 6xn+1 xnChng minh rng vi mi n > 1 th (xn) khng l s chnh phng.
Li gii
Tng t tng ca bi 2, ta s bin i kho lo a v y nh dng ca bi
2. Nhng iu c bit y chnh l ta cn 2 dy chng minh song song cng
nhau bng quy np cng c cho li gii ca bi ton thm cht ch. (Lu : Li
gii ch b sung v cho ta thy mt ng dng kh mnh ca bi 2).
D dng chng minh theo quy np:
xn =
(3 + 2
2)n
+(3 22)n
2
Ta bin i xn theo 2 hnh thc sau:xn =
(2 + 1
)2n+(
2 1)2n2
=
[(2 + 1
)n+(
2 1)n]22
1
xn =
(2 + 1
)2n+(
2 1)2n2
=
[(2 + 1
)n (2 1)n]22
+ 1
V ta c th dung phng php truy hi tm cng thc truy hi ca(
2 + 1)n (2 1)n
ln lt c cng thc truy hi l:{v0 = 2; v1 = 2
2
vn+2 = 2
2vn+1 vnv
{z1 = 2; z2 = 4
2
zn+2 = 2
2zn+1 znTheo nh bi trn ta c th chng minh chng ln lt c tnh cht{
v2n = a
v2n+1 = b
2(a; b Z) v
{z2n = c
2
z2n+1 = d(c; d Z)
T ta rt ra c xnc dng A2 + 1 hoc B2 1 nn r rng xn khng th l s chnh
phng vi mi n > 1. 2
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51
Mt vn th v rng ti sao ta li phi cn n 2 dy song song chng minh.
Chnh 2 dy song song ny to nn s th v ca bi ton, khin n tr nn
cht ch hn trong li gii. V khi xt n chn v l ta c th din t xn theo 2
hnh thc v do 2 hnh thc trn u c dng 2x2
2 1 nn hin nhin n khng th
l s chnh phng vi mi n > 1.Mt li gii khc chung tng nhng li gii ngn gn hn:
Xt dy s nguyn {u0 = 0; u1 = 2
un+2 = 2un+1 + un
Ta d dng chng minh theo quy np
un =(
2 + 1)n (12)n
2
Mt khc, dy
{x0 = 1;x1 = 3
xn+2 = 6xn+1 xnc cng thc tng qut l
xn =
(3 + 2
2)n
+(3 22)n
2=
(2 + 1
)2n+(
2 1)2n2
Vy xn + (1)n+1 = un2. T ta kt lun rng xnkhng th l s chnh phng vi min > 1. 2
Bi tp 4.10. (CIS 1992)
Cho dy s an xc nh bi{a1 = 1
an+1 = a12 + a2
2 + a32 + ...+ an
2 + nn > 1
Chng minh rng an khng l s chnh phng vi mi n > 2.
Li gii
Vi nhng bi ton c cng thc truy hi phc tp nh trn, vic u tin ta cn ngh n l
s dng sai phn a v mt biu thc ngn gn v n gin hn bin lun. Qua , ta
vn dng cc tnh cht ca s chnh phng chng minh. C th s chnh phng ch c tn
cng l 0, 1, 4, 5, 6, 9.
Ta c: a2 = 2
Vi mi n > 2; ta c:an+1 =
nk=1
ak2 + n
an =n1k=1
ak2 + n 1
an+1 an = an2 + 1 > 0
-
52
Ta suy ra dy ny l mt dy tng ngt.
By gi ta c mt dy mi {a2 = 2
an+1 = an2 + an + 1 n > 2
Vn dng tnh cht trn, ta s chng minh an khng th l s chnh phng theo phng php
quy np n thun, v nu an l s chnh phng th n phi c tn cng l 1 trong cc s
0, 1, 4, 5, 6, 9
Ta thy r s hng trong dy c tn cng l 7 xut pht t n = 3: a3 = 7; a4 = 57
Gi s iu ny ng n n = k; tc akc tn cng l 7.
Xt n = k + 1, ta c: ak+1 = ak2 + ak + 1
Do ta ch xt ch s tn cng ca ak+1 nn ta s xt ch s tn cng ca tng s hng trong
biu thc trn.
Ta c ak c tn cng l 7nn ak2c tn cng l 9. Nh vy ch s tn cng ca
ak+1 = .b1b2...9 + c1c2...7 + 1 = d1d2...7(9 + 7 + 1 = 6 + 1 = 7
)iu ny cng ng vi n = k+ 1, theo nguyn l quy np ta suy ra an c tn cng l7n > 3.p dng tnh cht trn ta suy ra an khng th l s chnh phng vi mi n > 3.Mt khc a2 = 2 khng l s chnh phng.
Vy ta suy ra an khng l s chnh phng vi mi n > 2. 2
Bi tp 4.11. (Balkan 2002)
Cho dy s an tha mn {a1 = 20; a2 = 30
an+2 = 3an+1 an n > 1
Tm n A = 5an+1an + 1 l s chnh phng.
Li gii
Tng t, ta i tm cng thc tng qut ca dy s v bin lun. i vi nhng
bi tm gi tr n mt biu thc no l s chnh phng trong dy s. Ta cn
lu n nhng tnh cht ca dy chnh cng nh dy con (nu c) nh n iu,
ng d. . .
D dng chng minh theo quy np cng thc tng qut ca an l
an = 10
(3 +52
)n1+
(35
2
)n1
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53
Suy ra
A = 5an+1an + 1
A = 5.102.[(
3 +
5
2
)n+
(35
2
)n].
(3 +52
)n1+
(35
2
)n1+ 1A = 500
(3 +52
)2n1+
(35
2
)2n1+ 3
+ 1A = 500
(5 + 12
)4n2+
(5 12
)4n2+ 3
+ 1A = 500
(5 + 12
)2n1+
(5 12
)2n12 + 501Xt dy ph sau: {
x1 =
5; x2 = 6
xn+2 =
5xn+1 xn n > 1Dy trn c cng thc tng qut l
xn =
(5 + 1
2
)n+
(5 12
)nn N
Ta c:
xn xn1 =(
5 + 1
2
)n+
(5 12
)n(
5 + 1
2
)n1(
5 12
)n1
=
(5 + 1
2
)n2+
(5 12
)n2> 0
Suy ra xn > xn1 n N. Vy xn l dy tng ngt.Tng t bi 2, dy trn c mt tnh cht c th chng minh c bng quy np theo 2 dy
song song l: {x2n = c
x2n1 = d
5(c; d N)
Chnh v th m ta suy ra: A = 500(d
5)2
+ 501.
Gi s n0 N : A = 5an0+1an0 + 1 l s chnh phng, ta c:
(50d)2 + 501 = k2 (k N) 501 = (k 50d) (k + 50d)
Mt khc ta phn tch c 501 = 3.167, m{k + 50d+ k 50d = 2kk + 50d > k 50d
-
54
nn ta suy ra {k + 50d = 501
k 50d = 1 d = 5
Vy n0 N : xn0 = 5
5. Ta tnh ton c x3 = 5
5 v xn l dy tng ngt nn ta suy ra
n = 3.
Vy vi n = 3 th A = 5an+1an + 1 l s chnh phng. 2
Bi tp 4.12. Cho dy s tn tha mn:{t1 = 9; t2 = 25; t3 = 81
tn+3 = 7tn+2 14tn+1 + 8tn n > 1
a) Chng minh rng: vn = tn+1 2n+1tn+1 + 22n n > 1 l s chnh phng.b) Chng minh rng nu T = 2 + 2
12tn + 1 N th n l s chnh phng.
c) Gi s tn ti dy s xn sao cho xn.tn c l s c. Chng minh dy xn gm ton s chnh
phng.
Li gii
> Cu a:Ta thy biu thc cn chng minh c dng cn thc. iu ny gi cho ta vic chng minh tn
l mt dy s chnh phng.
Bng vic th ln lt cc s hng u tin ca dy, ta rt ra nhn xt rng dy tn gm ton
s chnh phng. By gi vic ta cn lm l chng minh dy tn gm ton s chnh phng.
Thng thng nhng bi ton chng minh dy l s chnh phng hoc khng l s chnh
phng ta thng phi tm ra cng thc tng qut ca dy a v bi ton s hc thng
thng. Nhng trong dy s ta c th ch ng bin i linh hot gia chng to nn li
gii hp l. in hnh vi bi ton ny ta thy vic chng minh tn = 22n + 2n+1 + 1 gm ton
s chnh phng r rng l mt cng vic khng kh thi.
Song ta c th da vo phng php ca dy s v chng minh thng d on v chng
minh tn l bnh phng ca mt ng thc no .
Nh phng php sai phn ta c nh gi nh sau:
3 5 9 17 33
21 22 23 24
Hay t1
t2
t3
t4
t5
21 22 23 24
Do , ta s chng minh theo quy np tn+1 = (2n +tn)
2.
Do t1 = 9 = 32 nn iu trn ng vi n = 2 : t2 = (2 + 3)
2 = 25.
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55
Gi s iu ny ng vi n = k; tc tk =(2k1 +
tk1
)2.
Ta xt n = k + 1; ta c:
tk+1 = 7tk 14tk1 + 8tk2 = 7tk 14tk1 + 8(
tk1 2k2)2
tk+1 = 7tk 6tk1 8.2k1tk1 + 8.22(k2) = 7tk 6
(tk 2k1
)2 8.2k1 (tk 2k1)+ 8.22(k2)tk+1 = tk + 2.2k
tk + 2
2k =(
tk + 2k)2
Vy theo nguyn l quy np ta suy ra
tn+1 =(2n +
tn)2 n N
Mt khc do t1 = 9 l s chnh phng nn t ta suy ra mi s hng ca dy u l s chnh
phng.
Vy vn = tn+1 2n+1tn+1 + 22n = (tn+1 2n)2 l s chnh phng.
> Cu b:Vic chng minh s chnh phng gip ta a bi ton v mt dng quen thuc hn:
Nu A = 2 + 2
12a2 + 1 N th n l s chnh phng".Tht vy, gi s A N 12tn + 1 = k (k N) 12tn = (k 1)(k + 1)Do k l nn ta suy ra k = 2l + 1 (l N). T ta suy ra: 3tn = l(l + 1).Do (l; l + 1) = 1 v tn l s chnh phng nn ta suy ra{
l = 3a2
l + 1 = b2hoc
{l = b2
l + 1 = 3a2(a; b N)
* Trng hp 1:
{l = b2
l + 1 = 3a2 3a2 = 1 + b2
Mt khc 3 = 4.0 + 3 nn t ta suy ra b2 + 1 c c dng 4k+ 3 nn ta suy ra 1... 3 (v l).
Vy trng hp ny khng th xy ra.
* Trng hp 2:
{l = 3a2
l + 1 = b2.
Ta c:
A = 2 + 2
12tn + 1 = 2 + 2k = 2 (k + 1) = 4(t+ 1) = 4b2
T ta suy ra A l s chnh phng.
Vy nu A = 2 + 2
12n2 + 1 N th n l s chnh phng.
> Cu c:Mt tnh cht c bn ca s chnh phng: Nu a l mt s chnh phng bt k th n lun
c l s c v ngc li. (tnh cht c chng minh nh vic s dng php m).
Chng minh: Gi s a l s chnh phng, ta vit dng tng qut ca a l
a = p121p2
22p323 ...pn
2n
-
56
S c ca a chnh l s tch s ca cc s nguyn t vi cc ln thay i khc nhau ca cc
s m i , i = 1;n. Chnh v th m theo quy tc nhn ta m c a c
(21 + 1) (22 + 1) (23 + 1) ... (2n + 1)
s c v hin nhin tng s c trn l mt s l.
Ngc li gi s s a c l s c, ta c dng tng qut ca a l
a = p1r1p2
r2p3r3 ...pn
rn
Theo quy tc nhn ta m c ac
(r1 + 1) (r2 + 1) (r3 + 1) ... (rn + 1)
Do a c l s c nn ri + 1, i = 1;n u phi l s l, suy ra
ri = 2k (k N) , i = 1;n
Vy a chnh l s chnh phng.
Tnh cht c chng minh hon tt. Quay li bi ton trn v s dng tnh cht ny
tm ra li gii thch hp cho bi ton.
Ta c: xn.tn c l s c. p dng tnh cht trn ta suy ra xn.tn l s chnh phng. Mt khc
tn l dy cc s chnh phng xn l dy cc s chnh phng.
Bi tp 4.13. Cho dy (un) tha mn{u0 = 2; u1 = 5
un+1 = 5un 6un1, n > 1
Xt hm s
f (x) =x3
3 5x
2un2
+(6un
2 + 6n)x+ C (C R)
c o hm ti x0 (a; b) vi a; b N.Chng minh rng f (x) c im cc i v cc tiu c honh lun l s nguyn.
Li gii
u tin, ta c nhn xt sau: 2 im cc tr ca hm s trn thuc phng trnh
f (x) = x2 5xun + 6un2 + 6n = 0
Nh th iu kin u tin nghim ca phng trnh trn l s nguyn th = un24.6n phi
l s chnh phng vi mi n > 1. Nh vy ta a bi ton v vic chng minh vn = un2 4.6nl s chnh phng vi mi n > 1.Trong bi ton ny, ta s chng minh bi ny bng phng php dng tam thc bc hai gii
quyt bi ton nhanh gn hn (ngha l ta s a bi ton v dng X2 + Xun + + 6n = 0
do s 6n xut hin mt cch khng t nhin v mt c s hng un bc nht trong bi ton).
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57
Nh vy dy s s c dng un+12 + unun+2 + 6
n + = 0 (do dng ny ta c th tch un+2
theo cng thc truy hi v a phng trnh trn v dng nu trn). Ngoi ra ta vn c th
chng minh biu thc trn thng thng theo quy np ca mt biu thc on c.
Nh din t trn, ta c:
un+12 5un+1un + 6un2 + 6n + = 0
Xem biu thc trn nh phng trnh bc 2 theo tham s un n un+1, ta c
= 252un2 24un2 4.6n 4 = 2un2 4.6n 4
ng nht h s vi biu thc cn chng minh ta c = 1; = 0.
Nh vy ta s chng minh theo quy np
unun+2 un+12 = 6n ,n N
Tht vy iu ny ng vi n = 0 do 2.13 52 = 60 = 1 v n = 1 do 5.35 132 = 6.Gi s iu ny ng n n = k, tc ukuk+2 uk+12 = 6k.Xt n = k + 1, ta c:
uk+1uk+3 uk+22 = uk+1 (5uk+2 6uk+1) uk+22 = 5uk+1uk+2 6uk+12 uk+22uk+1uk+3 uk+22 = 5uk+1uk+2 + 6k+1 6ukuk+2 uk+22=uk+2 (5uk+1 6uk) + 6k+1 uk+22 = 6k+1 (do uk+2 = 5uk+1 6uk)
Vy iu ny ng vi n = k + 1, theo nguyn l quy np ta chng minh c
unun+2 un+12 = 6n ,n N
Do , ta c:
u2n+1 unun+2 + 6n = 0 u2n+1 5unun+1 + 6u2n + 6n = 0 ()
Xem (*) nh phng trnh bc 2 theo tham s un n un+1, ta c:
X2 5unX + 6un2 + 6n = 0
Do dy s trn l dy cc s nguyn nn ta suy ra X cng phi l s nguyn. Nh vy
= 25un2 24u2n 4.6n = un2 4.6n = vn
phi l s chnh phng.
Xt phng trnh tng qut
x2 5unx+ 6un2 + 6n = 0 (x R)
Phng trnh c nghim khi v ch khi = vn = un2 4.6n > 0.
M vn l s chnh phng (chng minh trn) nn phng trnh ny lun c nghim v
N.Nh vy 2 nghim ca phng trnh trn ln lt l
x =5un
un2 4.6n2
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58
+Vi un l cng nh un chn, th 5un un2 4.6n lun l mt s chn nn hin nhin n
chia ht cho 2. Cho nn x lun l mt s nguyn.
Mt khc (x2 5xun + 6n + 6un2
)=x3
3 5x
2un2
+(6un
2 + 6n)x+ C (C R) = f (x)
Mt khc hm s ch c th t cc i hoc cc tiu ti a, b, x0 nn t ta suy ra iu phi
chng minh. 2
Bi tp 4.14. Xt hm s f(t) = t+[t]. Dy s (an) c xc nh{
a0 = m > 2, (m N)an+1 = f (an) , n = 0, 1, 2...
v dy (tb) : tb =[b
2] b N. Chng minh rng tn ti n0; b0 : an0 .tb0 Z.
Li gii
Vi iu kin bi, ta c th nhn ra rng tch s trn phi l s chnh phng. Nh vy hoc
an0 ; tb0 u l s chnh phng, hoc an0 = tb0 , hoc ch an0 .tb0 l s chnh phng. Vic ch ra
tn ti hoc an0 = tb0 , hoc ch an0 .tb0 l s chnh phng l cng on tnh ton vt v nht l
i vi hm phn nguyn. Nh vy ta s chng minh tn ti v s s hng l s chnh phng
khng nh s tn ti ca 2 bin c lp m khng cn ch ra c th iu kin ca 2 bin
bng vic tnh ton. Ta s tun t chng minh bi ton theo tnh cht ca hm phn nguyn
p dng khai trin Newton trong dy s cng nh lch ca cc hng t trong dy.
Xt dy (an), ta c: m < f (m) < f (f (m)) < ... nn dy trn l dy tng ngt. Vy ta chia
lm hai trng hp sau:
> Trng hp 1: Nu m khng phi l s chnh phng.Gi t2 l s chnh phng ln nht khng vt qu m = a0. Theo cch xc nh trn ta suy ra
c d = [m] .
t m = d2 + k. Ta c d2 < m 6 (d+ 1)2 0 < k 6 2d+ 1.
* a) Xt 0 < k < d+ 1.Theo cch xc nh dy ta c:
d2 < a1 = f (a0) = m+[m]
= d2 + k + d < (d+ 1)2
d Trng hp 2: Nu m l s chnh phng th hoc dy (an) c v hn s chnh phnghoc tn ti n0 N : an0khng l s chnh phng. Khi y ta quay li trng hp 1 v sau huhn bc ta s gp li phn t ca dy l s chnh phng.
Do qu trnh trn lp li v hn ln nn ta suy ra dy trn c v hn s chnh phng.
Xt dy (tb). Theo khai trin Newton, ta c:(
2 + 1)l
= zl
2 + yl(2 1
)l= zl
2 ylkhi l l
(2 1)l = 2zl2 yl2 1 + yl2 = 2zl2
yl4 + yl2 =(ylzl
2)2
yl4 + yl2 = ylzl
2
Mt khc:
yl2 4Mi s X1, X2, X3, X4 c 2 cch chn nn c tt c 2 2 2 2 = 16 cch chn 4 k t u.Mi s Xk, 4 < k 6 n c 26 cch chn nn c tt c 26n4 cch chn nhng tr trng hpton b l A i nn c tng cng 26n4 1 cch chn n 4 k t sau.Do , tng s mt khu hp l c th c l sn = 16(26
n4 1).
Cu b:
D dng thy rng tn chnh l s c ca sn. Ta cng c (16, 26n4 1) = 1 nn c th vit sn
di dng sn = 24 p vi p l mt s nguyn dng l.
Do n > 6 nn n 4 > 2 v 26n4 chia ht cho 4, tc l 26n4 1 c dng 4k + 3, khng th lmt s chnh phng, tc l n c s c chn.
Thm vo , 4 + 1 = 5 nn theo cng thc tnh s c ca mt s nguyn dng th tn va
chia ht cho 5 v va chia ht cho 2 nn n phi chia ht cho 10.
Ta c pcm.
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64
V d 4.
a) Cho n l mt s t nhin tha mn n + 1 chia ht cho 24. Chng minh rng tng cc c
dng ca n (k ca n) cng chia ht cho 24.
b) Xt s nguyn
A = 20102011 20112010 20521994
Hy chng minh rng A l mt hp s dng v tng cc c s dng ca A chia ht cho 24.
Li gii
a) Gi d(n) l tng cc c dng ca n. Trc ht, ta s chng minh rng d(n) chia ht cho
3.
Tht vy, gi a l mt c no ca n th nacng l c ca n v nu xt 0 < a 3 thyx > xy ()
Bt ng thc ny tng ng vi
x ln y > y lnx ln yy
>lnx
x
Hm s f(t) = ln tt, t > 3 c f (t) = 1ln t
t2< 0 nn y l hm nghch bin, suy ra f(x) < f(y)
hay (*) ng.
Cng bng cch dng hm s, ta c th chng minh rng vi n ln v 0 < a < n th c
nh ginn+a
(n+ a)n> 2
Do 20102011 > 2.20112010 > 20112010 + 20521994 nn A > 0. Xt trong modun 5 th
A (1 + 21994) (mod5)
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65
M
24 1(mod5) 21994 = 4(24)498 4(mod5)Suy ra A chia ht cho 5 hay A l mt hp s.
Tip theo, ta s chng minh rng A + 1 chia ht cho 24. Tht vy, do 2010, 2052 chia ht cho
3 nn
A+ 1 ((1)2010)+ 1 = 0(mod3)Hn na 2052 chia ht cho 8 v
A+ 1 32010 + 1 (32)1005 + 1 11005 + 1 = 0(mod8)Suy ra A+ 1 chia ht cho 3 v chia ht cho 8 nn A+ 1 chia ht cho 24.
Theo kt qu cu a, ta c tng cc c dng ca A chia ht cho 24, suy ra pcm.
Bi tp c hng dn, gi
Bi tp 5.1. Chng minh rng phng trnh (n) = k c v s nghim nguyn dng n vi
mi k, cn phng trnh (n) = k th c hu hn nghim nguyn dng n.
Gi .
Ta
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