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Let $f:\mathbb N\rightarrow \mathbb N$ be an injective function such that $$f^{f(a)}(b) f^{f(b)} (a) = [f(a+b)]^2$$ for all $a,b \in \mathbb N$. Let $S$ be the sum of all possible values of $f(2017)$. Find $S$ (mod $1000$).

Write

$$f^{k}(n) =f(f(f(\cdots k \text{ times }\cdots f(n)))\cdots).$$

So I put $a=b$ and found out that $f^{f(a)}(a) = f(2a)$.

Since $f$ is injective $f^{f(a)-1}(a)=2a$. I found out that $f(1)=2$.

Now $f(2)=3$ by putting value of $a=1$ in initial equation and solving for $f(b)=3$. It seems by guessing that $f(n)=n+1$ but can't the function be periodic after certain interval and is unbounded and repeat its value? It can still be injective.

Any help here to clarify this doubt?

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An injective function 'cannot repeat its value' - if $f$ is injective, then if $f(a)=f(b)$, it must be the case that $a=b$. Because the problem condition tells you that $f$ is injective, it follows that it is not periodic.

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