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.
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)&amp;=&amp;f(n+af(n)+f(n+af(n))f(n))\hspace{10mm}&amp;P(n+af(n), f(n))\\
&amp;=&amp;f(n+(a+nf(a+1))f(n))&amp;P(n,a)\\
&amp;=&amp;nf(a+nf(a+1)+1)&amp;P(n, a+nf(a+1))\\
&amp;=&amp;n(a+1)f(n+1)&amp;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 $n$.
$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 $n$.