if the function such $f(m+nf(m))=mf(n+1)$ find the f if $f:N^{+}\to N^{+}$,and such for any postive integer $m,n$ have
$$f(m+nf(m))=mf(n+1)$$
find the $f$
I guess this function is $f(n)=n$,But How to solve it?Thanks.
 A: Following is rephrased solution by user pslove at corresponding AoPS topic (also swapping $m,n$ to match OP). 
Let $P(m,n)$ be the claim that $f(m+nf(m))=mf(n+1)$. Then for $n,a \in \mathbb{N}$ we have:
\begin{eqnarray*}
(n+af(n))f(f(n)+1)&=&f(n+af(n)+f(n+af(n))f(n))\hspace{10mm}&P(n+af(n), f(n))\\
&=&f(n+(a+nf(a+1))f(n))&P(n,a)\\
&=&nf(a+nf(a+1)+1)&P(n, a+nf(a+1))\\
&=&n(a+1)f(n+1)&P(a+1,n)\\
\end{eqnarray*}
This with little algebra gives
$$
\frac{nf(n+1)}{f(f(n)+1)}-f(n)=\frac{n-f(n)}{a+1}.
$$
Now right side has to be independent of $a$ for all $n$, which is only possible if $n-f(n)=0$. More specifically, if $n-f(n)\neq 0$ for any $n$, then for that particular $n$ being fixed, we can vary $a$, and both sides should vary, but we know left side is constant (with respect to fixed $n$).
So $f(n)=n$ for all $n$.
A: Function $f(x)$ is defined on the countable set of points and can be considered as the countable sequence of values
$$\{f(1),f(2),f(3),\dots\},$$
with the system of the equations
$$f(nf(m)+m) = mf(n+1),\quad (m,n)\in \mathbb N^2. \tag1$$
Let us research the structure of this sequence.
The value $f(1)=p$ should satisfy the system of
$$f(np+1) = f(n+1),\quad n\in \mathbb N.\tag2$$
If $\mathbf{p=1}$ then the equation $(1)$ is satisfied for each value of $n$.
If $\mathbf{p>1}$ then $f(x)$ is a periodic function with the period $p$,
$$f(m+p) = f(m).\tag3$$
At the same time, the value $f(2)=q$ should satisfy the system of
$$f(f(m)+m) = mq,\quad m\in \mathbb N.\tag4$$
In particular, for the values $m\in\{1,2,1+q,1+q+q^2,1+q+q^2+q^3\dots\},$
$$\begin{cases}
f(p+1)=q\\
f(2+q) = 2q\\
f(2+q+q^2) = 4q\\
f(2+q+q^2+q^3) = 8q\\
\dots.
\end{cases} \tag5$$
If $f(1) = p>1,$ then the system $(4)$ contradicts with the periodiity condiion $(2).$
If $f(1) = p = 1,$ then for $m=1$ from $(1)$ should the identity
$$f(n+1) = f(n+1).\quad n\in\mathbb N,$$
then $f(1) = 1.$
Since $f(x)$ is a countable set of points, it can be presented in the polynomial form of
$$f(x) = 1+a_1(q-1)(x-1)+a_2(q-1)^2(x-1)^2+...,\quad a_i\in \mathbb R.\tag6$$
wherein from $(5)$ should
$$\begin{cases}
1+a_1(q-1)+a_2(q-1)^2+a_3(q-1)^3+\dots = q\\
1+a_1(q^2-1)+a_2(q^2-1)^2+a_3(q^2-1)^3+\dots = 2q\\
1+a_1(q^3-1)+a_2(q^3-1)^2+a_3(q^3-1)^3+\dots = 4q\\
1+a_1(q^4-1)+a_2(q^4-1)^2+a_3(q^4-1)^3+\dots = 8q\\
\dots\\
1+a_1(q^{k+1}-1)+a_2(q^{k+1}-1)^2+a_3(q^{k+1}-1)^3+\dots = 2^kq\\
\dots.
\end{cases}\tag7$$
Transitional limit $k\to \infty$ leads to the solution
$$q=2,\quad a_1=1,\quad a_2=a_3=\dots =0,$$
$$f(x)= x.\tag8$$
Substitution of $(8)$ to $(1)$ leads to the identity
$$nm+m=m(n+1).$$
Thus, $(8)$ is the single solution.
A: $f(m+nf(m))=mf(n+1)
$
Putting $n=0$,
this becomes
$f(m) = mf(1)
$.
Putting $k=f(1)$
so $f(n) = kn$,
this becomes
$k(m+kmn) = mk(n+1)
$
or
$1+kn = n+1
$
so
$k=1$.

If $f(n) = an+b$ then
$a(m+n(am+b))+b=m(a(n+1)+b)
$
or
$am+a^2mn+abn = amn+am+bm
$
or
$a^2mn+abn 
= amn+bm
= m(an+b)
$
or
$abn
=m(an+b-a^2n)
$
Putting $n=1$,
this is
$ab = m(a+b-a^2)
$.
The left side is
independent of $m$,
and the right side is not
if $a+b-a^2 \ne 0$.
Therefore
$ab = a+b-a^2 = 0$
so
$b = a^2-a$
and
$0 = a(a^2-a)
=a^2(a-1)
$.
If $a=0$
then $b=0$.
Otherwise
$a=1, b=0$.
Therefore the only linear solutions are
$0$ or $n$.
A: If we set $n=0$, you get that for every $m \in N^+$, $$f(m)=mf(1).$$
So we just have to find $f(1)$. Fixing $m$ and $n$ and using the last equality, 
\begin{align*}
f(m+nf(m)) &= f(m+nmf(1))\\
&=(m+nmf(1))f(1),
\end{align*}
now because $f(m+nf(m))=mf(n+1)=m(n+1)f(1)$, we get that $(m+nmf(1))f(1)=m(n+1)f(1).$ Simplifying the last equality, we see that $f(1)^2 = 1$; because $f$ is restricted to be positive then $f(1)=f(1)$ and $f(n)=n$ for very $n\in N^+$.
