Find all $f : \mathbb{N} \to \mathbb{N}$ such that $f(a) + f(b) \mid a+b, \ \forall \ a, b \in \mathbb{N}$

Question

Find all $f : \mathbb{N} \to \mathbb{N}$ such that $$f(a) + f(b) \mid a+b, \ \forall \ a, b \in \mathbb{N}$$

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All I can find is the following:

If we put $a=b=n$ we get $f(n)\mid n$, so $f(n)\leq n$ for all $n \in \mathbb{N}$.

So $f(1)=1$ and we have $f(n)+1\mid n+1$.

Any suggestion how to proceed?

Answer 1

As mentioned in the comments we have that $f(p-1) = p-1$ for a prime $p$. Now let $n \in \mathbb{N}$. There exists prime $p$ s.t. $p-1 > n$. Then we have:

$$f(n) + p - 1 \mid n + p-1$$

We have that the greatest proper divisor of $n+p-1$ is bounded from above by $\frac{n+p-1}{2} < p-1$

But $f(n) + p - 1 > p -1$, so it must be $n+p-1$. Hence $f(n) = n$ and the proof.

Answer 2

I'll use $f(a)+f(b)\mid b-f(b)+a-f(a)$ to prove by induction $f$ is the identity function. Assume instead $b$ is the least counterexample. Since $a+f(b)\mid b-f(b)$ for $a<b$, all integers from $f(b)+1$ to $f(b)+b-1$ inclusive divide $b-f(b)$, which by hypothesis is positive and exceeds its factors. These include $f(b)+b-1\ge b$, a contradiction.