<s>[INST] 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.
 [/INST] 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.
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