菲波那契数列满足毕达哥拉斯定理的推广(英文)

Pythagoras Meets Fibonacci & Generalization

（１）若 ｎ＝ ２ｋ（偶数）（Ｆ_（ｎ＋３）~２—Ｆ_（ｎ＋１）~２）~２＋（２Ｆ_（ｎ＋４）· Ｆ_ｎ）~２＋（４Ｆ_（ｎ＋２））~２＝（Ｆ_（ｎ＋３）~２＋Ｆ_（ｎ＋１）~２）~２． （２）若ｎ＝２ｋ＋１（奇数） （Ｆ_（ｎ＋４）~２—Ｆ_ｎ~２）~２＋（２Ｆ_（ｎ＋３）·Ｆ_（ｎ＋１））~２＋（４Ｆ_（ｎ＋２））~２=（Ｆ_（ｎ＋４）~２＋Ｆ_ｎ~２）~２． 在此：Ｆ_ｎ，Ｆ_(ｎ＋１)，Ｆ_(ｎ＋２)，Ｆ_(ｎ＋３)，Ｆ(ｎ＋４)连续五项．

In [1] Fibonacci: F_0=0, F_1=F_2 = 1, F_n = F_(n-1)+F_(n-2) (n≥2) If n is an even,then (3) (F_(n+4)·F_(n+1)~2 + (2F_(n+2)~2 = (F_(n+1)· F_(n+3))~2; If n is an odd,then (4) (F_n+3)· F_(n+1)~2 + (2F_(n+2)~2=(F_(n+4)F_n)~ 2. F. i. n=3, (8. 3)~2+ (2·5)~2=(13·2)~2,i. e. 24~2+10~2=26~2. We get Pythagoras Meets Fibonacci & Generalization. If n is an even,then (1) (F_(n+3)~2-F_(n+1)~2)~2 + (2F_(n+4)·F_n)~2 + (4F_(n+2))~2= (F_(n+3)~2+F_(n+1)~2)~2. If n is an odd,then (2) (F_(n+4)~2-F_n^2)~2 + (2F_(n+3)·F_(n+1))~2 + (4F_(n+2))~2=(F_(n+4)~2+F_n^2)~2.