Is there exist a function $f:\mathbb {N}\to \mathbb {N}$ with for $\forall m,n \in \mathbb {N}$ $$f\left(mf\left(n\right)\right)=f\left(m\right)f\left(m+n\right)+n ?$$
Problem
Source: Kazakhstan MO 2018 final round.Grade 11;Problem 3
Tags: function, algebra, Kazakhstan
04.05.2018 14:32
25.10.2021 15:31
$P(1,n): f(f(n))=f(1)f(n+1)+n$. Hence $f(1)|f(f(n))-n$. $P(f(m),f(n)): f(f(m)f(n))=f(1)f(m+1)f(f(m)+n)+mf(f(m)+n)+n$. Hence $mf(f(m)+n)+n \equiv f(f(m)f(n)) \equiv nf(m+f(n))+m \pmod{f(1)}$. Take $n=1,m\rightarrow mf(1): 1\equiv f(mf(1)+f(1))$. Thus $f(1)|f((m+1)f(1))-1$. $P(f(1),n): 1\equiv 1\cdot f(n+f(1))+n \pmod{f(1)} \implies f(1)|f(n+f(1))+n-1$ However, $P(f(1)+2,1): f(1)|f(f(1)+2)f(f(1)+3) \implies f(1)|2$. If $f(1)=1$ or $2$, $P(1,1)$ gives a contradiction.
31.12.2021 16:36
I have a rather short solution. Can someone check? It is obvious that $f$ is not constant. Take $n$ such that $f(n) \neq 1$. $P(\frac{n}{f(n)-1}, n)$ gives contradiction.
31.12.2021 16:39
electrovector wrote: I have a rather short solution. Can someone check? It is obvious that $f$ is not constant. Take $n$ such that $f(n) \neq 1$. $P(\frac{n}{f(n)-1}, n)$ gives contradiction. You should prove, that $\frac{n}{f(n)-1}$ is natural
31.12.2021 16:41
Yes you are right sorry