回復 1# anson721 的帖子
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