Câu hỏi của Park Soyeon

Question

$x^2-xy+y^2=3$ ..Giải phương trình nghiệm nguyên trên ???

Mong m.n giúp đỡ !!

alibaba nguyễn · Accepted Answer

$x^2-xy+y^2=3$

$\Leftrightarrow2x^2-2xy+2y^2=6$

$\Leftrightarrow\left(x-y\right)^2+x^2+y^2=1+1+4$

$\Rightarrow\left(\left(x-y\right)^2,x^2,y^2\right)=\left(1,1,4;1,4,1;4,1,1\right)$

$\Rightarrow\left(x,y\right)=\left(-1,-2;1,2;2,1;-2,-1;-1,1;1,-1\right)$

Câu trả lời:

$x^2-xy+y^2=3$

$\Leftrightarrow2x^2-2xy+2y^2=6$

$\Leftrightarrow\left(x-y\right)^2+x^2+y^2=1+1+4$

$\Rightarrow\left(\left(x-y\right)^2,x^2,y^2\right)=\left(1,1,4;1,4,1;4,1,1\right)$

$\Rightarrow\left(x,y\right)=\left(-1,-2;1,2;2,1;-2,-1;-1,1;1,-1\right)$