回復 1# kuen 的帖子
先考慮\({{x}^{2}}+xy+{{y}^{2}}=1\)
由於過\(\left( 1,0 \right)\),令\(y=k\left( x-1 \right)\)
\(\begin{align}
& {{x}^{2}}+kx\left( x-1 \right)+{{k}^{2}}{{\left( x-1 \right)}^{2}}=1 \\
& \left( {{k}^{2}}+k+1 \right){{x}^{2}}-\left( 2{{k}^{2}}+k \right)x+\left( {{k}^{2}}-1 \right)=0 \\
& x=\frac{{{k}^{2}}-1}{{{k}^{2}}+k+1} \\
& y=\frac{-{{k}^{2}}-2k}{{{k}^{2}}+k+1} \\
\end{align}\)
再令\(k=\frac{m}{n}\)
\(\begin{align}
& x=\frac{{{m}^{2}}-{{n}^{2}}}{{{m}^{2}}+mn+{{n}^{2}}} \\
& y=\frac{-{{m}^{2}}-2mn}{{{m}^{2}}+mn+{{n}^{2}}} \\
\end{align}\)
最後取\(\left( x,y,z \right)=\left( {{m}^{2}}-{{n}^{2}},-{{m}^{2}}-2mn,{{m}^{2}}+mn+{{n}^{2}} \right)\)
[ 本帖最後由 thepiano 於 2014-9-26 04:40 PM 編輯 ]