■連続数のピタゴラス三角形(その27)

  p/q→P/Q=(p+2q)/(p+q)

このとき,pをp+2q,qをp+qで置き換えています.

 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)

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【1】√2の近似値とヘロン数列

  p/q→(p+2q)/(p+q)

に次ぐ第3の近似分数は

  p/q→(p+2q)/(p+q)→(3p+4q)/(2p+3q)

となります.

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[1]ペル数列

p^2−2q^2=±1が成り立つとき

  p/q→P/Q=(p+2q)/(p+q)

  Q=p+q,P=q+Q=p+2q

  P^2−2Q^2=(p+2q)^2−2(p+q)^2

=−p^2+2q^2=±1

[2]ヘロン数列

p^2−2q^2=±1が成り立つとき

  p/q→P/Q=(3p+4q)/(2p+3q)

  Q=2p+3q,P=p+q+Q=3p+4q

  P^2−2Q^2=(3p+4q)^2−2(2p+3q)^2

=p^2−2q^2=±1

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