The problem:
"Let f(x)= (ax+b)/(cx+d), f(f(f(1)))=1 and
f(f(f(2)))=3. Prove that f(1)=1."

I proved it in part, but I can't continue.

-Peter

I'm guessing you already know about Mobius maps from the question. Briefly a Mobius map is a function f: C* -> C* of the form z -> (pz + q)/(rz + s). (Note: C* is the extended complex plane; that is the complex numbers with an extra point at infinity.)

It is well known that the set of Mobius maps form a group so z -> f(f(f(z))) is also a Mobius map. It is also well known every Mobius map has either one or two fixed points unless f(z) = z for all z (in which case everything is fixed).

Now z -> f(f(f(z))) fixes 1,f(1),f(f(1)). It cannot have three seperate fixed points as it is not the identity function so at least two of 1,f(1),f(f(1)) are the same. If f(1) = 1 we're done. If f(f(1)) = 1 then f(f(f(1))) = f(1) so again f(1) = 1. If f(f(1)) = f(1) then f(f(f(f(1)))) = f(f(f(1))) so again f(1) = 1. So we're done.

Thanks for the solution. I'd like to learn more about Mobius maps. Can you suggest me any book or hyperlink?

Peter

Mathworld has quite a useful page here . They call it a Fractional Linear Transformation, but this seems to be the same as a Mobius map.

The only two real facts I've used about Mobius maps above are:

i) they form a group (in particlar the composition of two Mobius maps is also a Mobius map)

ii) a Mobius map with more than two fixed points is the identity (in fact a Mobius map which isn't the identity has exactly one or two fixed points in the extended complex plane).

i) is easy to prove, straight from the definitions. ii) I think is less obvious - it is clear approximately how the proof will go, but I'm worried there may be lots of special cases which may make the proof tedious to write out. I'll see if I can find a nice proof.