Question

Tìm x, y nguyên thỏa mãn: y² - (y+2)x² = 1.

Answer 1

Nhận thấy \(y=-2\) không phải nghiệm

Với \(y\ne-2\):

\(y^2-1=\left(y+2\right)x^2\Rightarrow x^2=\frac{y^2-1}{y+2}=y-2+\frac{3}{y+2}\)

Do \(x^2\) nguyên \(\Rightarrow\frac{3}{y+2}\) nguyên \(\Rightarrow y+2=Ư\left(3\right)=\left(-3;-1;1;3\right)\)

\(y+2=-3\Rightarrow y=-5\Rightarrow x^2=-8< 0\left(l\right)\)

\(y+2=-1\Rightarrow y=-3\Rightarrow x^2=-8< 0\left(l\right)\)

\(y+2=1\Rightarrow y=-1\Rightarrow x=0\)

\(y+2=3\Rightarrow y=1\Rightarrow x=0\)

Vậy \(\left(x;y\right)=\left(0;-1\right);\left(0;1\right)\)