■タクシー数(その15)
x+y=±(2m^2−4m−1)
x−y=±(2m^2+2m−1)
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[1](+/+)では
2x=4m^2−2m−2,x=2m^2−m−1=(m−1)(2m+1)
2y=−6m,y=−3m
2x=A±(2m^2+1)
2(2m^2−m−1)=2m^2−4m−1+(2m^2+1)=4m^2−4m
−2m−2=−4m,m=1→(x,y)=(0,−3)
2(2m^2−m−1)=2m^2−4m−1−(2m^2+1)=−4m−2
4m^2−2m−2=−4m−2,4m^2=−2m (NG)
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[2](+/−)では
x+y=2m^2−4m−1
x−y=−2m^2−2m+1
2y=4m^2−2m−2,y=2m^2−m−1=(m−1)(2m+1)
2x=−6m,x=−3m
2x=A±(2m^2+1)
−6m=2m^2−4m−1+(2m^2+1)=4m^2−4m
4m^2+2m=0,m=0→(x,y)=(0,−2)
−6m=2m^2−4m−1−(2m^2+1)=−4m−2
2m=2,m=1→(x,y)=(−3,0)
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[3](−/+)では
x+y=−2m^2+4m+1
x−y=−2m^2−2m+1
2x=−4m^2+2m+2,x=−2m^2+m+1
2y=6m,y=3m
2x=A±(2m^2+1)
−4m^2+2m+2=2m^2−4m−1+(2m^2+1)=4m^2−4m
8m^2−6m−2=0,m=(6±5)/8 (NG)
−4m^2+2m+2=2m^2−4m−1−(2m^2+1)=−4m−2
4m^2−6m−4=0,m=(6±5)/4 (NG)
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[4](−/−)では
x+y=−2m^2+4m+1
x−y=2m^2+2m−1
2x=6m,x=3m
2y=−4m^2+2m+2,y=−2m^2+m+1
2x=A±(2m^2+1)
6m=2m^2−4m−1+(2m^2+1)=4m^2−4m
4m^2−10m=0,m=0→(x,y)=(0,1)
6m=2m^2−4m−1−(2m^2+1)=−4m−2
10m=−2 (NG)
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[まとめ]結局,何を求めているのかわからない.
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