$f: \Bbb N→ \Bbb N$ , $f\big(f(m) +f(n)\big)=m+n$ $\space$ for all $m,n∈\Bbb N$ How to find all functions $f: \Bbb N→ \Bbb N$ which satisfy $f\big(f(m) +f(n)\big)=m+n$ $\space$ for all $m,n∈\Bbb N$
($\Bbb N$ is the set of all natural numbers, i.e. positive integers)  ?
 A: Claim: $f$ must be the identity map.
Proof: Evidently, $f$ is injective, because $\forall m\ne m'$, 
$$f(f(m)+f(1))\ne f(f(m')+f(1)).$$
To show $f$ is surjective, let us denote $E=f(\mathbb{N})$ and 
$$F=E+E=\{f(m)+f(n):m,n\in \mathbb{N}\}.$$
By definition and the assumption of $f$,
$$E=f(\mathbb{N})\supset f(F)=\{n\in\mathbb{N}:n\ge 2\}.\tag{1}$$ 
If $1\notin E$, then 
$$F=E+E\subset\{n\in\mathbb{N}:n\ge 4\}.$$
Combining the equation above with the injectivity of $f$, we can conclude that 
$$\#f(\mathbb{N}\setminus F)\ge 3,$$
which contradicts $(1)$. Therefore, $1\in E$, i.e. $f$ is surjective. It follows that $$F=\{n\in\mathbb{N}:n\ge 2\}=f(F).\tag{2}$$
From $(2)$ we know that $f(1)=1$, so by the assumption of $f$,
$$f(f(n)+1)=n+1,\quad\forall n\in\mathbb{N}.\tag{3}$$
The the conclusion follows from $(3)$ and induction.
A: We first show that $f$ is an injective function , in fact,
$\begin{align} f(m)=f(n) &\implies f(m)+f(n)=f(n)+f(n)\\
 &\implies f(f(m)+f(n))=f(f(n)+f(n))\\ 
&\implies m+n=n+n. \end{align}$
Hence $f(m)=f(n)$ $\implies m=n$, i.e. $f$ is injective.
Let $k$ be a positive integer less than $n$ , then $n-k$ is a positive integer and from given functional equation we will have , $f(f(m+k)+f(n-k))=(m+k)+(n-k)=m+n=f(f(m)+f(n))$ , using the injectivity of $f$ this gives ,
$$f(m+k)+f(n-k)=f(m)+f(n), \forall m,n,k\in\Bbb N, k<n. \tag i$$
Suppose $f(1)=b>1$. $b≥2$. We get $f(2b)=f(b+b)=f(f(1)+f(1))=1+1=2$ and $f(b+2)=f(f(1)+f(2b))=1+2b$. If $b=2$,then the previous two relations lead to $f(4)=2$ and $f(4)=1+2×2=5$, absurd!
Hence $b>2$, so $b-2$ is a positive integer, also $b-2<2b$; hence using (i), we obtain $f(b-1)+f(b+2)=f(1+b-2)+f(2b-(b-2))=f(1)+f(2b)$. This, along with $f(1)=b,f(2b)=2$ and $f(b+2)=1+2b$, yield $f(b-1)+1+2b=b+2$ , giving $f(b-1)=1-b$ , impossible since $1-b<0$ but $f(b-1)≥1$.We conclude that $b=f(1)=1$. Now we use induction, if $f(m)=m$ for some $m ∈\Bbb N$ , then from the given equation and $f(1)=1$ , we obtain $m+1=f(f(m)+f(1))=f(m+1)$ , hence by principle of mathematical induction $f(n)=n$ , for all $n∈\Bbb N$ .
A: A bit different from the previous solutions
Note that the set ${(2,3,4,5\dots)}$ is a subset of the range of the function, since in the original equation, the RHS can take these values
Put $m=n=1$ in the equation to get $$f(2f(1))=2$$
Now, put $m=2f(1)$. This yields $$f(f(n)+2)=n+2f(1)$$
If $f(1)\geq 2$, then, select $n$ such that $f(n)=k$ where $k\geq 2$, $f(k+2)=n+2f(1)\geq 5$
Since $2,3,4$ are also in the range, $f(1),f(2),f(3)$ must be a permutation of $2,3,4$
If $f(1)=2$, then $f(4)=5$, but since $f(4)=f(2f(1))=2$, we get a contradiction.
If $f(1)=3$ or $f(1)=4$, select $n$ such that $f(n)=k$ where $k\geq 2$, $f(k+2)=n+2f(1)\geq 7$, which implies that the range cannot contain all of $2,3,4,5,6$, a contradiction.
Thus, $f(1)=1$, and we can show by induction that $f(n)=n$ for all natural numbers $n$, which is indeed a solution to the functional equation.
