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Algebra Difficulty 4.7 AIME Prove it South Africa

Find all functions f:NNf : \mathbb{N} \to \mathbb{N} satisfying f(n+1)f(n+2)=f(n)2f(n+1)f(n+2) = f(n)^2 for all nNn \in \mathbb{N}.

Solution

Suppose that the function is non-constant. Then it will have a least element f(k)f(k). Then f(k+1)f(k)f(k+1) \neq f(k), or the function would be constant, and by definition f(k+1)>f(k)f(k+1) > f(k). But f(k+2)=f(k)2f(k+1)=f(k)f(k)f(k+1)<f(k)f(k+2) = \frac{f(k)^2}{f(k+1)} = f(k)\frac{f(k)}{f(k+1)} < f(k), which is a contradiction. So f(n)f(n) constant is the only allowed function.

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