Question to find all strictly increasing $f:\mathbb{N}\to\mathbb{N}$ mapping only primes to primes and satisfying $f(2n)=f(n)+n$ for all $n$

Question

Let $f:\mathbb{N}\to\mathbb{N}$ be a function such that

1. $f(m)<f(n)$ whenever $m<n$;
2. $f(2n)=f(n)+n$ for all natural numbers $n$;
3. $n$ is a prime number whenever $f(n)$ is a prime number.

Find all functions $f$ satisfying above conditions with proof.

I have noticed that the identity function is a solution. I am however, unable to prove this result. To start off I am unable to prove $f(1)=1$, so I'm unable to proceed by induction. I am also not able to effectively use the third condition of the question. I request somebody to please provide a solution for this. Any help is much appreciated. Thanks a lot :)

Answer 1

We will prove $f(1)=1$, hence by induction you can take it further to show that $f(n)=n ~\forall n \in \mathbb{N}$

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Observe that $$f(n)=f(1)+n-1$$ satisfies the conditions of the functional equation. Now, suppose $f(1)=k>1$.

Next observe that all of the numbers $(k+1)!+2,(k+1)!+3,\cdots,(k+1)!+(k+1)$ are composite numbers (why?). Now, if $p$ is the smallest prime greater than $(k+1)!+(k+1)$, then plugging $n=p-k+1$ yields $f(n)=p$. Therefore, $n$ is a prime number. Now because $k>1$, we have $n=p-k+1<p$ and we conclude $(k+1)!+2<n<p$.

Now note that $p$ was defined to be the smallest prime exceeding $(k+1)! + (k+1)$, and since $(k+1)!+2,(k+1)!+3,\cdots,(k+1)!+(k+1)$ are all composite, means it is also the smallest prime exceeding $(k+1)!+1$. But $n$ was supposed to also be a prime less than $p$ but more than $(k+1)! + 1$. This contradicts that $p$ is the smallest prime exceeding $(k+1)!+(k+1)$ which implies it is impossible to have $f(1)>1$.

Answer 2

We know that $$f(2)=f(1)+1, \quad f(4)=f(2)+2=f(1)+3$$ and, being $f$ strictly increasing $f(3)=f(1)+2$. By induction you can then prove that $$f(n)=f(1)+n-1.$$

But then, if $f(1)>1$, $n$ should be prime whenever $n+f(1)-1$ is prime, which is impossible.