■奇数の2乗−1の整除性(その4)

[Q](b+c)/a+(c+a)/b+(a+b)/c=nの自然数解(a,b,c)を求めよ.

[A]a=F2k-1,b=F2k+1,c=F2k-1L2kF2k+1が知られている.

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 確認してみたい.

F1=1,F2=1,F3=2,F4=3,F5=5,F6=8,F7=13,F8=21,F9=34

L1=1,L2=3,L3=4,L4=7,L5=11,L6=18,L7=29,L8=47

[1]k=1

 F1=1,F3=2,L2=3→ a=1,b=2,c=6

 (a+b+c)(1/a+1/b+1/c)

=9(1/1+1/2+1/6)

=9(6/6+3/6+1/6)=9・10/6=15  (OK)

n=12=0  (mod4)

[2]k=2

 F3=2,F5=5,L4=7→ a=2,b=5,c=70

 (a+b+c)(1/a+1/b+1/c)

=77(1/2+1/5+1/70)

=77(35/70+14/70+1/70)=77・50/70=5  (OK)

n=2=2  (mod4)

[3]k=3

 F5=5,F7=13,L6=18→ a=5,b=13,c=1170

 (a+b+c)(1/a+1/b+1/c)

=1188(1/5+1/13+1/1170)

=1188(234/1170+90/1170+1/1170)=1188・325/1170=330  (OK)

n=327=3  (mod4)

[4]k=4

 F7=13,F9=34,L8=47→ a=13,b=34,c=20774

 (a+b+c)(1/a+1/b+1/c)

=20821(1/13+1/34+1/20774)

=20821(1598/20774+611/20774+1/20774)=20821・2210/20774=2215  (OK)

n=2212=0  (mod4)

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