Phương Pháp Gen trong giải phương trình nghiệm nguyên - Phan Đình Trung

8/10/2019 Phng Php Gen trong gii phng trnh nghim nguyn - Phan nh Trung

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TR NG THPT TN HNGChuyn :

PHNG PHP GEN TRONG GI I PHNG TRNHNGHI M NGUYN

Th c hi n : Phan nh TrungH c sinh l p 12A1, nin kha: 2014-2015

Trong Sinh h c, ta nh ngha gen l m t o n DNA mang c u trc xc nh thng tin di truy n. Tng t nh v y, phng php gen th c ch t l quy t c xy d ng c u trc nghi m t m t phng trnh khi bi t cc nghi m c s c a n. N u t m t nghi m c a phng trnh cho ta c qui t c xy d ng ra m t nghi m m i th qui t c g i l gen. Phng trnh Pell v phng trnh Markov chnh l hai v d i n hnh cho phng php ny.

I. Phng trnh Pell1. Phng trnh Pell lo i I:Phng trnh Pell lo i I l phng trnh nghi m nguyn c d ng:

x 2 dy2 = 1 , d Z (1)

Tnh ch t:

1. N u d l s chnh phng th (1) v nghi m.2. N u d l s nguyn m th (1) khng c nghi m nguyn dng.3. (i u ki n c nghi m c a phng trnh Pell lo i I) Phng trnh (1) cnghi m nguyn dng khi v ch khi d nguyn dng v khng chnh phng.

Cng th c nghi m c a phng trnh Pell lo i I:Ta xt trong tr ng h p nghim nguyn dng c a phng trnh Pell lo iI(t c l x, y Z + ). G i (a, b ) l hai nghi m nh nh t c a (1). Lc ny t tc nghi m c a n c vt s ch b i dy (xn ), (yn ):

xn = (a + b d)n + ( a b d)n

2yn =

(a + b d)n (a b d)n2 d

Ho c theo cng th c truy h i:x 0 = 1 , y0 = 0x 1 = a, y 1 = bx n +2 = 2ax n +1 x n , n = 0 , 1, 2,...yn +2 = 2ay n +1 yn , n = 0 , 1, 2, ...

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2. Phng trnh Pell lo i II:Phng trnh Pell lo i II l phng trnh nghi m nguyn c d ng:

x 2 dy2 = 1, d (2)

Tnh ch t:

1. N u d l s chnh phng th (2) v nghi m.2. N u d c c nguyn t d ng 4k + 3 th (2) v nghi m.3. G i (a, b ) l nghi m nh nh t c a (2). Khi , (2) c nghi m n u v chn u h

a = x 2 + dy2

b = 2xyc nghi m nguyn dng.

Cng th c nghi m c a phng trnh Pell lo i II:Xt phng trnh Pell lo i I:

x 2 dy2 = 1

. G i (a, b ) l hai nghi m nh nh t c a phng trnh ny. Xt h

a = x2

+ dy2

b = 2xy

Gi s h ny c nghi m duy nh t l (u, v ). Khi , t t c nghi m c a (2) c xc nh b i dy:

x 0 = u, y 0 = vx 1 = u 3 + 3 duv 2 , y1 = dv 3 + 3 u 2 vxn +2 = 2ax n +1 yn , n = 0 , 1, 2, ...yn +2 = 2ay n +1

yn , n = 0 , 1, 2, ...

3. Bi t p v d :

Bi ton 1 Tm t t c cc s nguyn dng x > 2 sao cho x 1, x ,x + 1l p thnh di ba c nh m t tam gic c di n tch l m t s nguyn.L i gi i:Gi s x l s nguyn dng th a mn bi ton. Khi , n a chu vi tam gic c tnh b i: p = x 1 + x + 1

2 =

32

x . G i y l di n tch tam gic . p

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d ng cng th c Heron, ta c:

y = 32x 32x x 32x x + 1 32x x 1 = 14x 3(x 2 4)Suy ra:

4y = x 3(x 2 4) 16y2 = 3 x 2 (x 2 4)Do 16y2 ... 2 3x 2 (x 2 4)

... 2 x ... 2. Nh v y, t n t i k N

sao chox = 2k. Theo :

16y2

= 3 .4k2

(4k2

4) y2 = 3 k 2 (k2 1)

y = k 3(k2 1)Ch r ng: : k, y Z + 3(k2 1) Z

+ . Do t n t i l Z + sao cho:3(k2 1) = l2

R rng l2 ... 3 l = 3m, m Z + . Khi phng trnh tr thnhk2

3m 2 = 1

y chnh l phng trnh Pell lo i I v cng th c nghi m c a n c xc nh b i:

k0 = 1 , m 0 = 0k1 = 2 , m 1 = 1kn +2 = 6kn +1 kn ,n = 0 , 1, 2, ...m n +2 = 6m n +1 m n ,n = 0 , 1, 2, ..

Suy ra:x 0 = 2 , y0 = 1

x 1 = 4 , y1 = 2x n +2 = 4x n +1 xn , n = 0 , 1, 2, ...yn +2 = 4yn +1 yn , n = 0 , 1, 2, ...

V y t t c gi tr c a x th a mn bi ton c xc nh b i dy:

x 0 = 2 , x 1 = 4xn +2 = 4xn +1 xn .n = 0 , 1, 2...

.

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Bi ton 2 Tm t t c cc s nguyn dng n sao cho trung bnh c ng c a n s chnh phng u tin l m t s chnh phng.L i gi i:B ng qui n p ta ch ng minh c r ng: V i m i S nguyn dng n , ta lunc:

12 + 2 2 + 3 2 + ... + n 2 = n(n + 1)(2 n + 1)

6Suy ra:

12 + 2 2 + 3 2 + ... + n 2

n =

(n + 1)(2 n + 1)6

Theo , ta c n tm n, y

Z + sao cho:(n + 1)(2 n + 1)

6 = y2

n 2 + 3 n + 1 = 6 y2

16n2 + 24 n + 4 = 48 y2

(4n + 3)2

48y2 = 1

t x = 4n + 3 , th th ta thu c:

x 2 48y2 = 1

y chnh l phng trnh Pell lo i I. B ng php th tu n t , ta tm c(7, 1) l nghi m nh nh t c a n. Theo , cng th c nghi m c a phngtrnh Pell ny l:

x 0 = 1 , y0 = 0x 1 = 7 , y1 = 1xk +2 = 14x k +1 xk , k = 0 , 1, 2, ...yk +2 = 14yk +1 yk , k = 0 , 1, 2, ...

B ng qui n p, ta ch ng minh c r ng:

x 2 k 1 (mod 4)x 2 k +1 3 (mod 4)

Ngoi ra:x 2 k +3 = 14x 2 k +2 x 2 k +1 = 14(14x 2 k +1 x 2 k ) x 2 k +1

= 196x 2 k +1 14x 2 k x 2 k +1= 195x 2 k +1 + x2 k +1 14x 2 k x 2 k +1= 194x 2 k +1 x 2 k 1

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ng th i, v x = 4n + 3 nn n = x 34

Z + , theo : nk = x2 k +1 3

4 n 0 = 1 , n 1 = 337, n k +1 = 194n k n k 1 + 144. V y, t t c gi tr n th a mnbi ton l:n 0 = 1 , n 1 = 337n k +1 = 194n k n k 1 + 144, k = 0 , 1, 2, ...

Bi ton 3 Tm c p s nguyn t p, q th a mn: p2 2q 2 = 1L i gi i:Xt phng trnh Pell lo i I:

x 2 2y2 = 1B ng php th tu n t , ta nh n th y (3, 2) l nghi m nh nh t c a phngtrnh ny ng th i y cng l c p s nh nh t th a mn bi ton. Theo, cng th c nghi m c a phng trnh Pell ny c xc nh b i:

x n = (3 + 2 2)n + (3 2 2)n

2yn =

(3 + 2 2)n (3 2 2)n2 2

Suy ra:

xn + yn = (3 + 2 2)n + (3 2 2)n

2 +

(3 + 2 2)n (3 2 2)n2 2

= ( 2 + 1) 2 n +1 + ( 2 1)2 n +1

2 2p d ng cng th c khai tri n Newton, ta c:

x n + yn = ( 2 + 1) 2 n +1 + ( 2

1)2 n +1

2 2=

2 n +1i =0 C

i2 n +1 ( 2)i + 2 n +1i =0 C i2 n +1 ( 2)i (1)2 n +1

i

2 2=

2 2 n j =0 C 2 j +12 n +1 2 j2 2 =

n

j =0

C 2 j +12 n +1 2 j

= C 12 n +1 +n

j =1

C 2 j +12 n +1 2 j = 1 +

n

j =1

C 2 j +12 n +1 2 j 1 (mod 2)

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i u ny ch x y ra khi p, q khc tnh ch n l hay ni cch khc phng trnh cho v nghi m n u min{ p, q } > 2. V y, (3, 2) l nghi m duy nh t c a biton.Bi ton 4 Tm t t c cc s t nhin n sao cho 2n + 1 v 3n + 1 u

l cc s chnh phng.L i gi i: t d = gcd(2 n + 1 , 3n + 1) , khi :

d | 2n + 1d | 3n + 1

d | 6n + 3d | 6n + 2

d | 1 d = 1

i u ny cho th y 2n + 1 , 3n + 1 ng th i l cc s chnh phng khi v chkhi t n t i y Z + sao cho:

(2n + 1)(3 n + 1) = y2

6n2 + 5 n + 1 = y2

144n2 + 120n + 24 = 24 y2

(12n + 5)2

24y2 = 1

t x = 12n + 5 , phng trnh tr thnh:

x2

24y2

= 1y chnh l phng trnh Pell lo i I, v 24 khng ph i l s chnh phngnn phng trnh ny ch c ch n c nghi m v cng th c nghi m c a n ccho b i:

x 0 = 1 , y0 = 0x 1 = 5 , y1 = 1xk +2 = 10x k +1 xk , k = 0 , 1, 2, ...yk +2 = 10yk +1 yk , k = 0 , 1, 2, ...

B ng qui n p ta ch ng minh c x2 k +1 5 (mod 12), do :n k =

x2 k +1 512Ngoi ra:

x 2 k +3 = 10x 2 k +2 x 2 k +1= 10(10x 2 k +1 x 2 k ) x 2 k +1= 99x 2 k +1 10x 2 k= 98x 2 k +1 + (10x 2 k x 2 k 1 ) 10x 2 k= 98x 2 k +1 x 2 k 1

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Suy ra: nk +1 = 98n k

n k 1 + 40. V theo t t c gi tr n th a mn bi

ton l:

n 0 = 0 , n 1 = 40n k + = 98n k n k 1 + 40, k = 0 , 1, 2, ...Nh n xt: V i cch lm tng t nh bi ton 4, ta c th gi i quy t cbi ton sau:Bi ton (Vi t Nam TST 2013):1. Ch ng minh r ng t n t i v h n s nguyn dng t sao cho 2012t + 1 v2013t + 1 u l cc s chnh phng.2. Xt m, n l cc s nguyn dng sao cho mn + 1 v (m + 1) n + 1 u l

cc s chnh phng. Ch ng minh r ng n chia h t cho 8(2m + 1)Bi ton 5 Ch ng minh r ng t n t i v h n s nguyn dng x, y, z sao

cho:x 2 + y3 = z 4

L i gi i:B : T n t i v h n s nguyn dng n, k sao cho:

n (n + 1)

2 = k2

Ch ng minh b :Bi n i phng trnh (1) v d ng tng ng:

(2n + 1) 2 8k2 = 1

t m = 2n + 1, phng trnh (1) tr thnh m 2 8k2 = 1 . y chnh lphng trnh Pell lo i I v cng th c nghi m c a n c cho b i:m 0 = 1 , k 0 = 0

m 1 = 3 , k 1 = 2m l+2 = 6m l+1 m lkl+2 = 6kl+1 kl

Ngoi ra, ta th y m l 1 (mod 2) nn gi tr n th a mn b c chob i: n 0 = 0 , n 1 = 1n l+2 = 6n l+1 n l + 2 , l = 0 , 1, 2, ...

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Dy s ny ch ng t kh ng nh c a b l ng. B c ch ng minh.Tr l i bi ton, B ng qui n p ta d dng ki m tra c r ng v i m i nnguyn dng, ta lun c:

13 + 2 3 + 3 3 + ... + n 3 =n (n + 1)

2

2

[13 + 2 3 + 3 3 + ... + ( n 1)

3 ] + n 3 =n (n + 1)

2

2

(n 1)n

2

2

+ n 3 =n (n + 1)

2

2

T y, p d ng b trn, ta ch n

x = n(n + 1)

2y = n

z 2 = n(n + 1)

2

R rng y chnh l ba b s th a mn bi ton. T suy ra pcm.

Bi ton 6 Ch ng minh r ng t n t i v h n s nguyn dng x, y sao cho:x + 1

y +

y + 3x

= 6

L i gi i:Phng trnh cho tng ng v i:

x 2 + (1 6y)x + y(y + 3) = 0Xem y nh m t phng trnh b c hai n x tham s y, ta c:

= (1 6y)2

4y(y + 3) = 32 y2

24y + 1By gi ta nh n th y r ng phng trnh cho c v h n nghi m nguyndng khi v ch khi t n t i v h n s nguyn dng y, k th a mn:

32y2 24y + 1 = k2

64y2

48y + 2 = 2 k2

(8y 3)2

2k2 = 7

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t m = 8y

3, phng trnh tr thnh:

m 2 2k2 = 7 ()

Xt phng trnh Pell lo i I: m 2 2k 2 = 1 , d th y phng trnh Pell nynh n nghi m nh nh t l (3, 2), b ng php th tu n t ta tm c (3, 1) lnghi m c s c a (). Theo , ta xt dy (x n ), (yn ):

x 0 = 3 , y0 = 2xn +1 = 3x n + 4yn , n = 1 , 2, 3, ...yn +1 = 2xn + 3 yn , n = 1 , 2, 3, ...

ng th i:

x 2n +1 2y2n +1 = (3 xn + 4 yn )

2

2(2x n + 3 yn )2 = x 2n 2y

2n = ... = x

21 2y

21 = 7

i u ny ch ng t phng trnh m 2 2k2 = 7 c v h n nghi m nguyndng. Ngoi ra ta cn nh n th y x2 k 3 (mod 8), th t v y ch r ng:xn +1 = 3x n + 4 yn

= 9 x n 1 + 12yn 1 + 4 yn= 9 x n 1 + 8 yn 1 + 4(yn + yn 1 )

= 9 x n 1 + 8 yn 1 + 8(xn 1 + 2 xn 1 ) xn 1 (mod 8)Nh v y,

x 2 k x 2 k 2 ... x 2 x 0 3 (mod 8)i u ny cho th y t n t i v h n y, k Z + sao cho:

(8y 3)2

2k2 = 7

T d dng suy ra pcm.

Bi ton 7 (Anh 2007) Ch ng minh r ng t n t i v s nguyn dng m, nth a mn :m + 1

n +

n + 1m

l m t s nguyn L i gi i: t m + 1

n +

n + 1m

= k (k Z + . Bi n i phng trnh v d ng:m 2 + m (1 kn ) + n

2 + n = 0

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Phng trnh ny c

= (1 kn )2 4(n 2 + n) = ( k2 4)n 2 2n (k + 2) + 1. Ch n k = 4. Ta c n ch ng minh t n t i v s s nguyn dng n sao cho = 12 n 2 12n + 1 l s chnh phng. Hay phng trnh:

12n 2 12n + 1 = t2

3(4n2

4n + 1) 2 = t2

t 2 3(2n 1)

2 = 2c v s nghi m nguyn dng.Ta xt hai dy (xn ), (yn ) c cho b i:

x 0 = 1 , y0 = 1xn +1 = 2xn + 3 ynyn +1 = xn + 2 yn

Khi y ta c :

x 2n +1 3y2n +1 = (2 x n + 3 yn )

2

3(xn + 2 yn )2 = x 2n 3y

2n = ... = x

20 3y

20 = 2

i u ny ch ng t phng trnh d ng:

X 2 3Y 2 = 2 (X = t, Y = 2n 1)

c v h n nghi m nguyn dng (xk , yk ) v i x k , yk l s h ng b t k c a haidy (xn )(yn )Ch r ng, ta c :

m = 4n 1 t

2Nn m nguyn th t ph i l . ng th i, b ng qui n p ta d dng ch ngminh c r ng c v s s h ng c a dy (xn ), (yn ) l s l . Nh v y phngtrnh:

m + 1n

+ n + 1

m = 4

c v h n nghi m nguyn dng. Bi ton c gi i quy t.Nh n xt: Khi tham kh o l i gi i trn, b n c s khng kh i th c m cr ng: "T i sao l c th l a ch n hai dy (xn ), (yn ) nh trn. Ph i chng ldo tnh c ?" Xin tr l i lun r ng l khng ph i tnh c , t t c u c s px p theo m t tr t t ring c a n c . Ta c phng php ch n ra cc dy(x n ), (yn ) nh sau :Xt phng trnh nghi m nguyn c d ng :

X 2 kY 2 = m (1)

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trong m l s nguyn v k l s nguyn dng khng chnh phng.N u (1) c nghi m nguyn dng l (A, B ). Xt phng trnh Pell d ngX 2 kY 2 = 1 c nghi m nguyn dng l (C, D ).Ta xt cc dy :

x 0 = A, y 0 = Bxn +1 = C x n + kDy nyn +1 = Dx n + Cyn

ng th i:

x 2n +1 ky 2n +1 = x 2n ky 2n = ... = x 20 ky 20 = m

cho nn phng trnh (1) ch c ch n c nghi m.II. Phng trnh Markov:

Phng trnh Markov c i n l phng trnh nghi m nguyn c d ng:

x 21 + x22 + x

23 + ... + x

2n = kx 1 x 2 x 3 ...x n (II)

Trong n, k l cc tham s nguyn dng.

Tnh ch t:1. N u phng trnh (II) c nghi m th n s c r t nhi u nghi m. Xem (II)

nh m t phng trnh b c hai n xn , cc s x1 , x 2 ,...,x n 1 l tham s . Khi:x 2 kx 1 x 2 ...x n 1 .x + x

21 + x

22 + ... + x

2n 1 = 0

G i x0 = y l nghi m th hai c a phng trnh ny. Theo nh l Vite tac:

x+ y = kx 1 x 2 ...x n 1 y = kx 1 x 2 ...x n 1xn = x21 + x22 + ... + x2n 1

x n(II.1)

Ch r ng u < x n khi v ch khi:

x21 + + x

2n 1 kx 1 x n

1 (II.2)Qu trnh ny c th th c hi n v i m i bi n s x j trong vai tr c a xn .Nhng ch i v i m t bi n - bi n l n nh t l c th x y ra (II.2) v ta thu c nghi m m i (x 1 , x 2 , . . . , x n ) nh hn nghi m c (th t theo t ng ccbi n). Nh v y, theo a s l cc nghi m tng ln v ta c cy nghi m.Ti p theo, tr nh ng tr ng h p c bi t, ta s gi s r ng x 1 x 2 xn . Ta s ni nghi m (x 1 , x 1 , , xn ) l nghi m g c (nghi m c s ), n u

x 21 + + x2n 1 x

2n 2xn kx 1 x n 1 (II.3)

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(t nghi m ny, t t c cc nhnh cy i n cc nghi m bn c nh, u tng)2.N u phng trnh (II) c nghi m nguyn dng th n c nghi m g c.3. N u n > 2, (x 1 , x 2 , , x n )) l nghi m g c, ngoi ra, x1 x 2 x n .Khi

x 1 xn 2 2(n 1)

k .

4. N u x1 x2 xn l cc s nguyn dng b t k tho mn i uki n 1 < x 2n x 21 + + x2n 1 , th t s R =

x21 + + x2nx 1 x 2 xn

khng v t qun + 3

2 ..

NH L: N u phng trnh (II) c nghi m v n = k, th n 2k 3 khin 5 v n > 4k 6 khi n = 3 , n = 4 .* Phng php Vite Jumping: nh l Vite: Xt phng trnh b c hai ax 2 + bx + c = 0 , a = 0 c hainghi m x1 , x 2 . Khi :

x 1 + x2 = ba

x 1 x 2 = ca

nh l o Vite: Xt tam th c b c haif (x) = ax 2 + bx + c, a

= 0 . G i

x 1 , x 2 l hai nghi m c a f (x). Khi , m t s th c th a mn x1 < < x 2khi v ch khi a.f ( ) < 0.

Chng ta b t u v i bi ton sau:

Bi ton 8 (IMO 1988) Cho a, b l cc s nguyn dng th a mn k =a 2 + b2

ab + 1 l m t s nguyn. Ch ng minh r ng k l m t s chnh phng.

L i gi i:Gi s k t lu n bi ton khng ng. Trong t p h p t t c cc s nguyndng (a, b ) th a mn bi ton, ta ch n ra hai ph n t a, b sao cho t ng a + bl nh nh t. Khng gi m tnh t ng qut, gi s a b > 0. Xt phng trnhb c hai n x:

x 2 + ( bkb)x + b2

k = 0R rng, phng trnh ny nh n m t nghi m l a. G i nghi m cn l i l x0 .Theo nh l Vite, ta c:

x 0 + a = kbbx 0 .a = b2 k

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T y, ta d dng suy ra c r ng x0

Z + .

N u x0 < 0 th x0 1, suy ra:x 2 (bk b)x + bk x

2 + ( bk b) + b2

k > 0, mu thu n N u x0 = 0 th k = b2 , mu thu n. N u x0 > 0 th (x 0 , b) l m t c p s th a mn bi ton. V lc ny:

x 0 + b = b2 k

a + b 0 k(1 b2 ) + 2 b2 > 0

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V b > 1

b2

1 > 0

b

2, lc ny ta c:

k < 2b2

b2 1 =

2

1 1b2

2

1 14

= 83

(1)

M t khc, ta l i c:1k

= ab 1a 2 + b2

ab 12ab

= 12

1ab

= 12 k > 2 (2)

T (1) v(2) suy ra i u mu thu n. Tm l i, ta c k = 5 l gi tr duy nh tth a mn bi ton (pcm).

Bi ton 10 (VMO 2012) Xt cc s t nhin l a, b th a mn a | b2 + 2 v b | a 2 + 2 . Ch ng minh r ng a, b l cc s h ng c a dy (x n ) c cho b i:

x 1 = x 2 = 1xn +2 = 4x n +1 x n

L i gi i:Ta c:

a

| b2 + 2

b | a 2 + 2 ab | (a 2 + 2)( b2 + 2) ab | (a 2 b2 + 2 a 2 + 2 b2 + 4)

Do a, b l nn ta c ngay ab | a 2 + b2 +2 . Tng t , ta cng c n u ab | a 2 + b2 +2th: a | b2 + 2b | a 2 + 2

. T c l:

a | b2 + 2b |

a 2 + 2 ab | a

2 + b2 + 2 ()

Trong cc ph n t (a, b) th a mn (), ta ch n ra m t c p (a, b ) sao cho a + bl nh nh t. Khng m t tnh t ng qut, gi s a b Xt phng trnh b chai n x sau:x 2 kbx + b

2 + 2 = 0R rng phng trnh ny nh n m t nghi m l a , g i nghi m kia l x 0 . Theo nh l Vite, ta c:

x 0 + a = bkx 0 .a = b2 + 2

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Ch r ng a l nh nh t, cho nn x0

a , suy ra:

x 0 + a 2a kb 2a ab

k2

N u trong hai s a, b c m t s b ng 1, gi s b = 1 th ka = a 2 +3 k = 4. N u min{a, b} > 1, th a b 2 nn:

k = ab

+ ba

+ 2ab

k2

+ 1 + 12 k 3

M t khc, theo AM-GM, ta c:

kab = a2

+ b2

+ 2 2(ab + 1) k 3Suy ra k = 3, v a2 + b2 + 2 = 3 ab. i u ny ch ng t r ng trong hai s a, bph i c m t s chia h t cho 3. Gi s 3 | b th th b 3. N u a = 1 th dth y ngay i u mu thu n, suy ra:

a 2 ab 6N u nh v y, th:

3 = ab +

ba

+ 2ab

32

+ 1 + 26

V l

V y ch c th l k = 4, t c l

a 2 + b2 + 2 = 4 ab ()

Gi s (y0 , y1 ) l m t c p s b t k th a (). Gi s y0 > y 1 , n u y0 = 1 thy1 = 1 t c l t n t in y0 = x1 , y1 = x 2 . Do , ta ch c n xt v i tr ngh p y0 , y1 > 1.Ch n c p (y1 , y2 ) = ( y1 , 4y1 y0 ), r rng y cng l m t c p s th a mn(). Lc ny ta ch t i 4y1 y0 < y 0 nn

y1 +

y2 =

y1 + (4

y1

y0 )

< y1 +

y0

Tng t , ta cng ch n c c p (y2 , y3 ) = ( y2 , 4y2 y1 ) cng th a () vy2 + y3 < y 1 + y2 < y 0 + y1 .Ti p t c qu trnh ny, ta c:

... < y i + yi +1 < ... < y 1 + y2 < y 1 + y0

M t khc, y1 + y0 > 2 nn t n t i k N sao choyk + yk +1 = 2 yk = yk +1 = 1

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T c l: yk = x 2 , yk +1 = x 1 . V nh v y, (yn ) c xc nh b i:

yo = y1 = 1yn +2 = 4yn +1 yn

Theo , ta c: xn +1 = yn . Bi ton c ch ng minh.

Bi ton 11 Xt phng trnh x2 + y2 + 1 = kxy .a) Tm t t c cc s k nguyn dng sao cho phng trnh trn c nghi m nguyn dng (x, y ).b) V i cc gi tr k tm c, hy tm t t c cc nghi m nguyn dng c a phng trnh.

L i gi i:Trong t p h p cc gi tr (x, y ) th a mn x

2 + y2 + 1xy Z ta ch n ra hai gi

tr (X, Y ) sao cho X + Y l nh nh t. Gi s lun X Y . Xt phng trnh:t 2 ktY + Y

2 + 1 = 0

D th y phng trnh ny c nghi m t1 = X , g i nghi m cn l i l t0 . Theo nh l Vite :

t 0 + X = kY t 0 X = Y 2 + 1T y d th y t0 cng nguyn dng, v tnh nh nh t c a X + Y nnt 0 X .Suy ra :

kY = t 0 + X 2X X Y

k2

V X, Y nguyn dng nn X, Y 1. Nh v y:

k = X Y +

Y X +

1XY

k2 + 1 + 1 k 4

D u b ng x y ra khi v ch khi: X = Y, 2X = kY, X = Y = 1, mu thu n.Nh v y k 3. Hn n a theo AM-GM ta d th y k 3.

V y k = 3. Th l i v i k = 3 th (x, y ) = (1 , 1) l m t nghi m c a phngtrnh.

b) Ta tm t t c cc nghi m c a phng trnh:

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x 2 + y2 + 1 = 3 xy (1)Xt dy s (un ) xc nh b i:

u0 = 1 , u 1 = 1un +1 = 3un u n 1 , n 1

Ta ch ng minh r ng (x, y ) l c p s nguyn dng b t k th a (1) khi v chkhi t n t i n th a x = xn , y = xn +1 .Th t v y, b ng qui n p ta d ki m tra c (xn , x n +1 ) th a (1) v i m i n.G i (w0 , w 1 ) l m t c p s nguyn dng b t k th a (1). N u w0 = 1 thw1 = 1, t c t n t i n = 0 w0 = u0 , w1 = w1 . Xt w0 , w1 > 1, gi s w0 > w 1 .Khi ta ch n w2 = 3w1 w0 . D th y w2 nguyn dng v c p (w1 , w2 ) =(w1 , 3w1 w0 ) lc ny cng th a (1). r ng ta c:

w2 = 3w1 w0 = w21 + 1

w0< w 0

Suy ra:w2 + w1 < w 1 + w0

Hon ton tng t ta ch n c c p (w2 , w3 ) = ( w2 , 3w2

w1 ) cng

th a w3 nguyn dng v w3 + w2 < w 2 + w1 .

C ti p t c qu trnh ny, ta c:

... < w i+1 + wi < ... < w 2 + w1 < w 1 + w0

Nhng w1 + w0 > 2 nn ph i t n t i k sao cho:

wk + wk +1 = 1 wk = u 1 = 1 , wk +1 = u 0 = 1T :

wk 1 = 3wk wk +1 = 3u 1 u 0 = u 2wk 2 = 3wk 1 wk = 3u 2 u 1 = u 3. . .w1 = 3w2 w3 = 3uk 1 uk 2 = u kw0 = 3w1 w2 = 3uk 3uk 1 = uk +1

Nh v y v i c p (w0 , w 1 ) b t k th t n t i n = k (w1 , w0 ) = ( u k , u k +1 ).Nh v y, t t c cc nghi m c a phng trnh l (x, y ) = ( u n , u n +1 ) v i dy(u n ) xc nh nh trn.

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Bi t p ngh :

1. Tm t t c cc gi tr nguyn dng p phng trnh x 2 + y2 + z 2 = pxyz c nghi m nguyn dng

2. (VMO 2002) Tm t t c cc gi tr nguyn dng n sao cho phngtrnh x + y + u + v = n xyuv c nghi m nguyn dng.3. (MOCP 03) Tm t t c cc gi tr nguyn dng n sao ch phngtrnh (x + y + z )2 = nxyz c nghi m nguyn dng.

4. (i Loan 1998) H i phng trnh x2

+ y2

+ z 2

+ u2

+ v2

= xyzuv 65c nghi m nguyn l n hn 1998 hay khng?5. (M 2002) Tm cc c p s (m, n ) nguyn dng sao cho m

2 + n 2

mn 1 l

m t s nguyn.

6. Tm s nguyn dng n nh nh t th a mn 19n + 1 v 95n + 1 ul cc s chnh phng.

Ti li u tham kh o:[1]: Phng trnh nghi m nguyn - Phan Huy Kh i.[2]: Phng trnh Diophant - Tr n Nam Dng[3]: nh l Vite v ng d ng - Tr n m nh Sang[4]: M t s ti li u t internet: diendantoanhoc.net, mathvn.com, mathscope.org,artofproblemsolving.com,...

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Phương Pháp Gen trong giải phương trình nghiệm nguyên - Phan Đình Trung