 | ■No28047に返信(takuさんの記事)
> (√2+i)x^2+√3y+y^2i=4√2-√3+5i(i^2=1)を満たす有理数x,yを求めよ。
In[1]:=
Expand[(Sqrt[2] + I)*x^2 + Sqrt[3]*y +
y^2*I - (4*Sqrt[2] - Sqrt[3] + 5*I)]
Out[1]=
-5*I - 4*Sqrt[2] + Sqrt[3] + I*x^2 +
Sqrt[2]*x^2 + Sqrt[3]*y + I*y^2
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(-5 + x^2 + y^2)*I + (-4 + x^2)*Sqrt[2] + (1 + y)*Sqrt[3]]=0
で Q上 I,Sqrt[2] ,Sqrt[3] が線型独立故
In[3]:=Solve[{-5 + x^2 + y^2 == 0, -4 + x^2 == 0, 1 + y == 0}, {x, y}]
Out[3]={{x -> -2, y -> -1}, {x -> 2, y -> -1}}
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