MathDB

Iran NMO 2008 (Second Round) - Problem3

Source:

September 22, 2010

functionalgebra proposedalgebra

Problem Statement

Let

a,b,c,

and

d

be real numbers such that at least one of

c

and

d

is non-zero. Let

f:\mathbb{R}\to\mathbb{R}

be a function defined as

f(x)=\frac{ax+b}{cx+d}

. Suppose that for all

x\in\mathbb{R}

, we have

f(x) \neq x

. Prove that if there exists some real number

a

for which

f^{1387}(a)=a

, then for all

x

in the domain of

f^{1387}

, we have

f^{1387}(x)=x

. Notice that in this problem,

f^{1387}(x)=\underbrace{f(f(\cdots(f(x)))\cdots)}_{\text{1387 times}}.

Hint. Prove that for every function

g(x)=\frac{sx+t}{ux+v}

, if the equation

g(x)=x

has more than

2

roots, then

g(x)=x

for all

x\in\mathbb{R}-\left\{\frac{-v}{u}\right\}

.

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