Plugging $n=m$ leads us to $(1)$ \[f(2f(m))=2m\] We have then \[f(2f(m))=2m\] \[f(2f(n))=2n\] By adding: \[f(2f(m))+f(2f(n))=2m+2n\] Taking both sides as arguments for $f$, we obtain $(2)$ \[2f(m)+2f(n)=f(2m+2n)\] Now assume we have $f(1)$ and $f(2)$. Let $k \in N$ and $k>1$. Then from the condition $(2)$ we obtain: \[2f(k)+2f(k)=f(2k+2k)=f(2(k+1)+2(k-1))=2f(k+1)+2f(k-1)\] which means that \[f(k+1)=2f(k)-f(k-1)\] so our function is an arithmetic sequence of form \[f(m)=am+b.\] Plugging this into our equation we get \[a^{2}(m+n)+2ab+b=m+n\] for every $m,n \in N$. Thus $a=1$ and $b=0$ and finally \[f(m)=m\]

Sorry to revive, I have a very short and nice solution. Clearly $ f$ is injective. Take the smallest $ a$ for which $ f(a)\ne a$. If $ a>1$, $ f(a)=f(f(1)+f(a-1))=a$, contradiction. So $ a=1$. Let $ f(1)=n>1$. $ f(n+f(n-1)) = f(f(1)+f(n-1))=n=f(1)$. So $ n+f(n-1)=1$, which is impossible. Thus $ f(x)=x,\forall x\in\mathbb{N}$.

Johan Gunardi wrote: Sorry to revive, I have a very short and nice solution. Clearly $ f$ is injective. Take the smallest $ a$ for which $ f(a)\ne a$. If $ a > 1$, $ f(a) = f(f(1) + f(a - 1)) = a$, contradiction. So $ a = 1$. Let $ f(1) = n > 1$. $ f(n + f(n - 1)) = f(f(1) + f(n - 1)) = n = f(1)$. So $ n + f(n - 1) = 1$, which is impossible. Thus $ f(x) = x,\forall x\in\mathbb{N}$. Very nice your solution,I like it

jastrzab wrote: which means that \[f(k+1)=2f(k)-f(k-1)\]so our function is an arithmetic sequence of form \[f(m)=am+b.\] I don't really get this part...

soluiton; it's clear that $f$ is injective. we have $f(f(n+1)+f(n-1))=2n$ and $f(2f(n))=2n$ thus from injectivity $f(n+1)+f(n-1)=2f(n)$ ie $f(n+1)-f(n)=f(n)-f(n-1)$ ie $\frac{f(n+1)-f(n)}{(n+1)-n}=\frac{f(n)-f(n-1)}{n-(n+1)}=cte$ thus $f(n)=n+c$ replacing we get $c=0$

Oh yeah so next is $f(n+1)=f(n)+c$ thus $f(n)=f(1)+(n-1)c\implies f(n)=nc+f(1)-c=an+b$ Where $a=c$ and $b=f(1)-c$

let $f(1)=c$ and we can easily see that $f$ is injective also $\forall i>1 \exists y : f(y)=i$ so we shall define for $i>1, a_i$ such as that $f(a_i)=i$ and then for $m>2, n>1 \implies$ $f(f(a_m)+f(a_n))=f(f(a_{m-1})+f(a_{n+1}))=f(m+n) \implies a_{n+1}-a_n=a_m-a_{m-1}$ now put $m$ a constant which will imply that $a_{n+1}-a_n = c_0, n \ge 2$ for some constant $c_0$. now combining $(1,1)$ gets us $f(f(1)+f(1))=2 \implies f(2c)=2$ again combining $(2c,2c)$ get us $f(4)=4c$, combing $(4,4)$ would imply $f(8c)=8$, now we see that $a_8=8c,a_2=2c, a_8-a_2=6c_0 \implies c_0=c$ and now because $a_2=2c$ then $a_3=3c$ and then $\forall i>1 : a_i=ic$ assume that $c>1$ and $f(c-1)=k \implies c-1=a_k=kc$ which is a contradiction so $c=1$ and $\forall i \in N: f(i)=i$

Also K27