Bai toan 1. Cho hai so nguyen duong a, b nguyen to cung nhau.
£>dt A — a?'^ + 6^" vdi n la so nguyen ducfng. Chiing minh rang moi i^dc le cua A c6 dang 2"+^^; + 1 vdi keW.
Chiirng minh. Ta chi can chiJng minh moi i/dc nguyen to le cua A 1^ du. Goi p la mot ufdc nguyen to le bat ky cua A thi ta c6
^ a ^ " ^ - ^ ^ " ( m o d p ) . (1)
68 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic
Do (a, h) = 1 nen (a, p) = (6, p) = 1 suy ra ton tai b' e { 1 , 2,... , p - 1} de bb' = 1 (mod p). Vay txi (1) ta c6 a^"^' = ^j^^d p) suy ra (a6')'"^' = (66')"" (mod p), d i n tdi: >
^ ^, (a6')'"^' = 1 ( m o d p ) . Dat /i = ordp(ab')- Tiif (2) dinh ly Fermat ta c6
(2)
,n+l
/i I 2
h\p-l
/i = 2°, a e [0, n + 1
Neu 0 ^ a ^ n , ta c6 (a6')'" = (a6')'"""
(66')^" = 1 (mod p) nen suy ra:
{abT^ibbT ( m o d p ) ,
2° = 1 (mod p) ma
hay:
a^" = b^ (mod p). (3) Tiif (1) va (3), ta suy ra 2a^" = 0 (mod p), ma (a, p) = 1 nen p | 2, v6 ly. Vay a - n + 1 suy ra 2"+^ | p - 1, hay la p = 2"+^/!; + 1
(A; e N*). Ta c6 dieu phai chufng minh. • NhSn xet. Bai toan 1 c6 the coi la mot bai toan tdng qudt cho
mot so' bai toan c6 dang tren. Chang han bang each suf dung Bai toan 1 ta c6 the giai difdc bai toan sau: Chiing minh rang ton tgi vd han cap so nguyen to {p, q) thoa man q \^ -lvdp\ - 1.
B a i toan 2. Cho n nguyen duang le khdc 1. Chiing minh rang:
(a) Neu n I (6" + 7") thi 13 | n .
(b) Ton tgi vd so so nguyen duang n sao cho n | (6" + 7").
Chufng minh. (a) Goi p la ufdc nguyen to' nho nhat cua n thi p le.
Tir gia thiet,
p I (6" + 7"), (1) ta c6 (6, p) = (7, p) = 1 nen ton tai duy nha't x G { 1 , 2, . . . , p - 1}
sao cho 7x = 1 (mod p). (2)
Cap so nguyen, cdn nguyen thuy vd ling dung 69
Ket hdp (1) va (2), ta diTcJc p | a;"(6" + 7") suy ra p | [(6a;)' (72;)"], hay:
p I [{QxT + 1]. \
V i n le nen ta thu dufdc: i ; ( - 6 x ) " = 1 ( m o d p ) .
Bay gicằ, dat:
da
(1;
ordp(—6a;) = h.
(3)
(4)
Ta c6
' (-6a;)" = 1 (mod p)
< (-6a;)P~^ = 1 (mod p) ^
^ (-6a;)'' = 1 (mod p)
h n
h\p-l h ^ p - 1 <p,
t. J
suy ra /i = 1 do each chpn p. Dieu nay dan den - 6 x = 1 (mod p) ma 7x = 1 (mod p) nen p | (6a; + 7x) hay p | 13a;, suy ra p | 13 hay p = 13. Vay 13 I n .
(b) Ta CO the chufng minh bang quy nap theo n rang ton tai vo
so so nguyen difdng n c6 dang 3*^ thoa man n | (6" + 7"). • Nhan xet. Ta c6 bai toan tdng qudt nhiT sau: Cho n, a, b Id
cdc so nguyen duang thoa man {a, b) = I, a + b Id so nguyen to vd n\ + b. Chiing minh rang:
(a) n\{a + b).
(b) Ton tgi vo so so nguyen duang n sao cho n \" + 6") va n {a + b).
Bai toan 3 (THTT 6/2006). Cho a, b Id hai so nguyen duang thoa man cdc so 2a - 1 2b - I, a + b la cdc so nguyen to. Chiing minh rang cdc so 0^ + 6°, a" + b'' deu khong chia het cho a + b. ^ Chufng minh. V i cac so' 2a - 1, 2 b - 1 deu la cac so' nguyen to' nen a, > 1, v i the a + 6 la so' nguyen to' le. Khong mat tinh tdng quat gia suf a chan, b le, suy ra {a + b)\" + b'').
70 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic Bay gi5, gia suf {a + b) \ + b"). Ta xet a > b, neu 6 > a ta
lap luan Wdng W. Do: ' ab + ^ a'' + b'' + b^b"-'' - I ) , nen ta c6
,,,,. „ a + b\ - 1).
Lai c6 {a + b) \ do {a, b, a + b) = I nen:
' ( a + 6) I (b'^-'-l). (1) Do a + 6 la so nguydn t6' n6n theo dinh ly Fermat ta c6
(a + 6 ) | ( 6 ° + ' ' - ^ - l ) . (2)
Dat h = OTda+b{b) thi = 1 (mod a + b). TO (1) va (2), ta c6
h\{a-b)
h I ( 2 a - 1).
h\{a + b-l)
Ma 2a - 1 la so nguyen to nen / i = 1 hoac h = 2a-l.
• Neu /i = 1 thi a = 1 (mod a + 6) suy ra 0 < a + 6 < a - 1, v6
• N e u = 2 a - 1 thi ( 2 a - l ) | (a + 6 - l ) h a y ( 2 a - l ) | {a-b), v6 ly v i 2a - 1 > a - b.
Vay dieu gia suf 1^ sai ta c6 {a + b) f (a*" + b"). TOdng ttf,
(a + 6 ) t ( a " + 6''). •
Bai toan 4. C/io p^b Id mot so nguyen to.
(a) Chiing minh rang ton tai so nguyen to q khdc p thoa q \ ly + i].
(b) Gid su[ {p-iy + 1 = nr=i PT Pi ^ô nguyen to vd aj G N*. Chiing minh rdng
n o
Z^Pi^i > y -
Cap nguyen, cdn nguyen thuy vd ling dung 71
ChiJng minh. (a) Ta c6 p | [(p - 1)'' + l] nen (p - 1 ) ^ + 1 la hdp
s6', dong thai: . pp-' <{p-iy + i<jf,
nen ton tai ufdc nguyen to q khac p cua (p - I ) ' ' + 1.
(b) Dat A = E r= i va 5 = E H i ^ i -
• Neu B thi theo bat dang thiJc A M - G M ta c6 '
2
A ^ 5 ^ ( p - 1)P + 1 > B > ^ p ' > ^ -
• Neu 2±i < 5 < p thi cung theo A M - G M ta c6 •
A ^ 5 ^ ( p -l)p + l > B = Bp^ > ^ i i p > ^ .
Neu 5 ^ p thi do g I [(p - 1)" + l ] nen q\{p- if^ - 1. Dat ord,(p - 1) = / i thi ta CO
h^p
h\{q-l) ^ h = 2Vh = 2p.
h I 2p
Neu / i = 2 thi 9 I [(p - 1)^ - l ] suy ra g | p - 2 v i g p. K h i do:
0 = ( p - l f + l = 2 ( m o d g ) , nen q = 2, wo ly v i g le.
Neu h = 2p thi 2p | (9 - 1), suy ra <? - 1 ^ 2p > p. Tiif day ta
CO Pi > p,yi = 1, n , suy ra:
A ^ p(ai + a2 + ---+an) = Bp^p^> —.
Bai toan dufcfc chiJng minh hoan toan. •
72 , Cdc phucfng phdp gidi todn qua cdc ky thi Olympic
NhSn xet. Bai todn vdi danh gia tot hdn nhu" sau: Cho p ^ 5 Id mot so nguyen to. Gid su: {p - 1)P + I = lYi=iPT Pi Id cdc so nguyen to vd Oj G N * . Chiing minh rang '
Bai toan 5. Chiing minh rang vdi moi m , n nguyen duang thi ton tai X nguyen duang thoa man:
• \ '^f-r'' {2^ = 1999 (mod T) , i
, ^ ^ I 2^ = 2009 (mod 5")
ChuTng minh. Ta c6 the chi^ng minh diTdc 2 la can nguyen thuy modulo 3"" va modulo 5". TO do suy ra ton tai x i , 2:2 de:
2^1 = 1999 (mod 3"^) J 2"' = 1 (mod 3) 2^= = 2009 (mod 5") | 2^= = 4 (mod 5)
TO day, ta thay ngay xu X2 chan. NhiT vay, he phu'dng trinh dong du:
i = ^ ( m o d 3 " ' - i )
t = y (mod 2 • 5"-^)
CO nghiem theo dinh ly du" Trung Hoa. Chpn x = 2t thi ta c6 x = xi (mod 2 • 3"^-^ = ^(3'")) (2^ = 1999 (mod 3^") x = X2 (mod 4 • 5" = (/P(5")) ^ | 2^ = 2009 (mod 5")
Bai todn dUdc chtfng minh xong. • B a i toan 6. Cho p la so nguyen to, p = 3 (mod 8) hogc p = 5
(mod 8), p = 2q + 1, trong do q cung Id so nguyen to. Chiing minh rang + u'^ + u>^ -\ 0;^""' = - 1 vdi u Id mot nghiem khdc 1 ciia phaang trinh — I.
Chtfng minh. Ta c6 p = ± 3 (mod 8) nen 2 khong chinh phiTdng modp, hay la:
2V = - 1 (mod p),
dan tdi 229 = 2P-1 = 1 (mod p). Dat h = ordp(2) thi h\{p-l) hay h \ V i q nguyen to nen h = l hoac h = 2 hoac h = 2q.
Cup so nguyen, can nguyen thuy vd ling dung 73
• Neu / i = 1 thi 2^ = 1 (mod p) suy ra p | 1, v6 l y . • *
• Neu h=^2 thi 2^ = 1 (mod p) suy ra p = 3, 9 = 1, v6 l y .
• Neu h = 2qiKih = p-l = (p{p), suy ra 2 la can nguyen thuy modulo p. Nhir the, tap {2\2^,2^"^} la he thang dif thu gpn modp, hay la:
{ 2 \ 2 2 , . . . , 2 ^ - ^ } = { l, 2, . . . , p - l } .
Tijf day, ta c6 ' v
Bdi tpan diTdc chuTng minh xpng. •