x^2-xy+y^2=3,且x,y為整數求x^2+y^2 (一題求解)

另解：

\(\displaystyle x^2-xy+y^2=3\)

\(\displaystyle \Rightarrow \left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}=3\)

\(\displaystyle \Rightarrow \left(2x-y\right)^2+3y^2=12\)

因為 \(x\) 與 \(y\) 皆為整數，可得

\(\displaystyle \left(2x-y, y\right)=\left(0, \pm2\right), (\left(3, \pm1\right)\) 或 \(\left(-3, \pm1\right)\)

\(\displaystyle \Rightarrow \left(x, y\right)=\left(2, 1\right), \left(-2, -1\right), \left(1, 2\right), \left(-1, -2\right), \left(-1, 1\right)\) 或 \(\displaystyle \left(1, -1\right)\)

\(\displaystyle \Rightarrow x^2+y^2 = 2\) 或 \(5\)

註：感謝 thepiano 老師私訊提醒小弟的條列疏漏（已修正），我真是粗心鬼。 XDDD