■整数三角形の話(その9)

 したがって,(a,b,c)の場合

→(a+b,b,c),(a,a+b,c)

になる.

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 0<n<mなる互いに素な整数m,nにより,

  {a,b}={m^2−n^2,2mn+n^2}

  c=m^2+mn+n^2,s=m^2+2mn

  s^2−sa+a^2=s^2−sb+b^2=c^2

s=a+b=m^2+2mn

より,

[1]c=m^2+mn+n^2,a=m^2+2mn,b=m^2−n^2

  a^2−ab+b^2=m^4+4m^3n+4m^2n^2−m^4+m^2n^2−2m^3n+2mn^3+m^4−2m^2n^2+n^4

=m^4+2m^3n+3m^2n^2+2mn^3+n^4

=(m^2+mn+n^2)^2=c^2

[2]c=m^2+mn+n^2,b=m^2+2mn,a=2mn+n^2

  a^2−ab+b^2=m^4+4m^3n+4m^2n^2−2m^3n−m^2n^2−4m^2n^2−2mn^3+4m^2n^2+4mn^3+n^4

=m^4+2m^3n+3m^2n^2+2mn^3+n^4

=(m^2+mn+n^2)^2=c^2

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[まとめ]a^2−ab+b^2=c^2一般解は,

c=m^2+mn+n^2,a=m^2+2mn,b=m^2−n^2

または

c=m^2+mn+n^2,b=m^2+2mn,a=2mn+n^2

の形になる.

 一方,a^2+ab+b^2=c^2の一般解は

a=(m^2−n^2),b=(2mn+n^2),c=(m^2+mn+n^2)

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