Có \(y^4-y^2=y^2\left(y^2-1\right)=y^2\left(y-1\right)\left(y+1\right)\)
\(xy^3-xy=xy\left(y^2-1\right)=xy\left(y+1\right)\left(y-1\right)\)
\(x^3y-xy=xy\left(x^2-1\right)=xy\left(x-1\right)\left(x+1\right)\)
\(x^4-x^2=x^2\left(x^2-1\right)=x^2\left(x-1\right)\left(x+1\right)\)
Do đó
\(y^4-y^2=xy^3-xy=x^3y-xy=x^4-x^2=0\)
\(\Leftrightarrow y^2\left(y-1\right)\left(y+1\right)=xy\left(y+1\right)\left(y-1\right)=xy\left(x-1\right)\left(x+1\right)=x^2\left(x-1\right)\left(x+1\right)=0\)Khi đó ta có các cặp số (x;y) tmđb là \((0;0);(1;1);(-1;1);(-1;-1);(1;-1)\)
P/s : Linh tinh ạ!!
