XÐT SÈ D− HAI VÕ.VÝ DÔ 5

1. XÐt sè d− hai vÕ.

VÝ dô 5: T×m nghiÖm nguyªn cña ph−¬ng tr×nh:

9x+ =2 y

2

+y

(*)

Gi¶i:

Ta cã:

VT =9x+ ≡2 2 mod 3

( )

VP= y

2

+ ≡y 2 mod 3

( )

y y

(

+ ≡1

) (

2 mod 3

)

( )

1 mod 3⇒ y

( v× nÕu y=3k hoÆc y = 3k+2 th×

VP0 mod 3

( ) ).

⇒ = +

(trong ®ã k

Z

) thay vµo pt(*) ta cã :

3 1y k

( ) (

2

)

2

2

9x+ =2 3k+1 + 3k+ ⇔1 9x=9k +9k⇔ =x k +k = +

2

x k k = +

VËy

 ∈k Z

VÝ dô 6: Gi¶i ph−¬ng tr×nh nghiÖm nguyªn kh«ng ©m sau:

(

2

x

+1 2

)(

x

+2 2

)(

x

+3 2

)(

x

+ −4

)

5

y

=11879

Ta cã

2 ; 2

x

x

+1; 2

x

+2; 2

x

+3; 2

x

+4

lµ 5 sè tù nhiªn liªn tiÕp

2

x

(

2

x

+1 2

)(

x

+2 2

)(

x

+3 2

)(

x

+4 5

)

MÆt kh¸c ¦CLN(

2

x

;5) = 1 nªn (

2

x

+1 2

)(

x

+2 2

)(

x

+3 2

)(

x

+4 5

)

Víi

y≥1

th×

VT =

(

2

x

+1 2

)(

x

+2 2

)(

x

+3 2

)(

x

+ −4

)

5 5

y

cßn

VP=118794 mod 5

( ) suy ra ph−¬ng

tr×nh kh«ng cã nghiÖm.

Víi y =0 ta cã :

(

2

x

+1 2

)(

x

+2 2

)(

x

+3 2

)(

x

+ − =4

)

5

0

11879

(

2

x

+1 2

)(

x

+2 2

)(

x

+3 2

)(

x

+ =4

)

11880

(

2

x

1 2

)(

x

2 2

)(

x

3 2

)(

x

4

)

9.10.11.12 2

x

1 9 2

x

8 2

x

2

3

x 3⇔ + + + + = ⇒ + = ⇔ = ⇔ = ⇔ =

VËy ph−¬ng tr×nh ®; cho cã nghiÖm duy nhÊt ( ) ( )

x y; = 3; 0

VÝ dô 7: T×m x, y nguyªn d−¬ng tho¶ m;n :

3

x

+ =1

(

y+1

)

2

( )

2

( )

3

x

+ =1 y+1 ⇔3

x

= y y+2

(**)

Ta cã VT = 3

x

1 mod 2 ( ) VP = y y ( + ≡ 2 ) ( 1 mod 2 )

Suy ra y lµ sè lÎ mµ y vµ y+2 lµ hai sè lÎ liªn tiÕp

 =

m

3y⇒ + =

Tõ pt(**)

n

2 3

Ta cã y +2 > y

n > m

≥1

NÕu m > 1 th× y vµ y+ 2 ®Òu chia hÕt cho 3 ( v« lÝ v× ( y; y+2) =2 )

VËy m =1

n = 0

x=1

y =1