■連続数のピタゴラス三角形(その28)

 さらに,もう一度,pをp+2q,qをp+qで置き換えれば,

  p/q→(p+2q)/(p+q)→(p+2q+2p+2q)/(p+2q+p+q)=(3p+4q)/(2p+3q)→

(3p+6q+4p+4q)/(2p+4q+3p+3q)

=(7p+10q)/(5p+7q)

  P^2−2Q^2=(7p+10q)^2−2(5p+7q)^2

=−p^2+2q^2=±1

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 さらに,もう一度,pをp+2q,qをp+qで置き換えれば,

(7p+10q)/(5p+7q)

=(7p+14q+10p+10q)/(5p+10q+7p+7q)

=(17p+24q)/(12p+17q)

  P^2−2Q^2=(17p+24q)^2−2(12p+17q)^2

=p^2−2q^2=±1

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