if the function such $f(m+nf(m))=mf(n+1)$ find the f

Question

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.

Answer 1

$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 $k=1$.

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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 $n$.

Answer 2

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 $n$.

Answer 3

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.

Answer 4

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 $n\in N^+$.