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Tenth Philippine Mathematical Olympiad

Algebra Difficulty 5.5 AIME, harder Prove it Philippines

Problem:
Consider the function ff defined by
f(x)=1+2x f(x) = 1 + \frac{2}{x}
Find the roots of the equation
(f∘f∘⋯∘f⏟10 times)(x)=x (\underbrace{f \circ f \circ \cdots \circ f}_{10 \text{ times}})(x) = x
where "o" denotes composition of functions.

Solution

Solution:
−1-1 and 22

Let f(n)(x)=(f∘f∘⋯∘f⏟n times)(x)f^{(n)}(x) = (\underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}})(x). For allowed values of xx, note that f(n)(x)f^{(n)}(x) is of the form
f(n)(x)=anx+bncnx+dn f^{(n)}(x) = \frac{a_n x + b_n}{c_n x + d_n}
where an,bn,cn,dn∈Za_n, b_n, c_n, d_n \in \mathbb{Z} for all integers n≥1n \geq 1. The equation
anx+bncnx+dn=xorcnx2+(dn−an)x−bn=0 \frac{a_n x + b_n}{c_n x + d_n} = x \quad \text{or} \quad c_n x^2 + (d_n - a_n)x - b_n = 0
has at most two real roots. Since f(−1)=−1f(-1) = -1 and f(2)=2f(2) = 2, it follows that f(n)(−1)=−1f^{(n)}(-1) = -1 and f(n)(2)=2f^{(n)}(2) = 2 for all n≥1n \geq 1. Thus, the roots of f(10)(x)=xf^{(10)}(x) = x are −1-1 and 22.

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