X2 + 5 XY + 6 Y2+ + X 2 Y − = ⇔ 2 0 ( X + 2 )( Y X + 3 ) Y + ( X +...

1) x

2

+ 5 xy + 6 y

2

+ + x 2 y − = ⇔ 2 0 ( x + 2 )( y x + 3 ) y + ( x + 2 ) y = ⇔ 2 ( x + 2 )( y x + 3 y + = 1) 2 (1)

Do ; x y ∈  suy ra x + 2 ; y x + 3 y + ∈ 1 

V ậ y t ừ (1) ta suy ra các trườ ng h ợ p sau

 + =  =

x y x

 + + = ⇔   = −

TH1: 2 2 6

 .

x y y

3 1 1 2

 + + = ⇔   =

TH2: 2 1 1

3 1 2 0

 + = −  = −

 + + = − ⇔   =

TH3: 2 2 2

3 1 1 0

 + = − ⇔  =

 + + = −   = −

TH4: 2 1 3

3 1 2 2

V ậ y các c ặ p s ố nguyên ( ; ) x y th ỏa mãn là (6; 2);(1;0);( 2;0);(3; 2) − − − .