■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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