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Romanian Mathematical Olympiad

Algebra Difficulty 4.9 AIME Prove it Romania

Find all functions f:N∗→N∗f : \mathbb{N}^* \to \mathbb{N}^* such that
f(n)+f(n+1)+f(f(n))=3n+1, for all n∈N∗. f(n) + f(n+1) + f(f(n)) = 3n + 1, \text{ for all } n \in \mathbb{N}^*.

Solution

From f(1)+f(2)+f(f(1))=4f(1) + f(2) + f(f(1)) = 4 follows f(1)∈{1,2}f(1) \in \{1, 2\}.

If f(1)=1f(1) = 1, then f(2)=2f(2) = 2 and an easy induction shows that f(n)=nf(n) = n.

If f(1)=2f(1) = 2, then f(2)=1f(2) = 1 and, inductively,
f(n)={n+1,if n is oddn−1,if n is even. f(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases}.
Both the above found functions fulfill the initial condition.

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