>>148
ほんとに？

f(x)=(x-a)/(ax+1)
f(f(x))=((x-a-aax-a)/(ax+1))/((ax-aa+ax+1)/(ax+1))=((1-aa)x-2a)/(2ax+(1-aa))
f(f(f(x)))=((x-aax-2a-2aax-a+aaa)/(2ax+1-aa))/((ax-aaax-2aa+2ax+1-aa)/(2ax+1-aa))=((1-3aa)x-(3a-aaa))/(3a-aaa)x+(1-3aa))

((1-3aa)x-(3a-aaa))/(3a-aaa)x+(1-3aa))=x
⇔(aaa-3a)xx+(aaa-3a)=0
⇔aaa-3a=0
⇔a=0,±√3

a≠0よりa=±√3