Tìm x,y nguyên biết: 5x^2-2xy+y^2=17

`5x^2 -2xy+y^2 =17`

`<=>4x^2 +(x^2 -2xy+y^2)=17``

`<=>4x^2 +(x-y)^2 =17`

`<=>4x^2 =17-(x-y)^2`

Do `(x-y)^2 >=0=>17-(x-y)^2 <=17=>4x^2 <=17`

`<=>x^2 <=4=> -2<=x<=2`

- Nếu `x= -2=> 5.(-2)^2 +2.2y+y^2 =17<=>20+4y+y^2 =17<=>(y+2)^2 =1`

`<=>` \(\left[ \begin{array}{l}y+2=1\\y+2=-1\end{array} \right.\)  `<=>` \(\left[ \begin{array}{l}y=-1\\y=-3\end{array} \right.\) 

- Nếu `x=-1=>5.(-1)^2 +1.2y+y^2 =17<=>5+2y+y^2 =17<=>(y+1)^2 =13` ( Vô lý `AAy\in ZZ` )

- Nếu `x=0=>5.0^2 -2.0y+y^2 =17<=>y^2 =17` ( Vô lý `AAy\in ZZ` )

- Nếu `x=1=>5.1^2 -2.1y+y^2 =17<=>5-2y+y^2 =17<=>(y-1)^2 =13` ( Vô lý `AAy\in ZZ` )

- Nếu `x=2=>5.2^2 -2.2y+y^2 =17<=>20-4y+y^2 =17<=>(y-2)^2 =1`

`<=>` \(\left[ \begin{array}{l}y-2=1\\y-2=-1\end{array} \right.\)  `<=>` \(\left[ \begin{array}{l}y=3\\y=1\end{array} \right.\) 

Vậy `(x;y)\in {(-2;-1),(-2;-3),(2;3),(2;1)}`