YLG_123 wrote: Let $\mathbb{Z}_{>0}$ be the set of positive integers. Find all functions $f : \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for all $m, n \in \mathbb{Z}_{>0 }$: $$f(mf(n)) + f(n) | mn + f(f(n)).$$ $P(1,n): f(f(n))+f(n)|n+f(f(n))\Rightarrow f(n)\le n$ Since $f(1)\le 1\Rightarrow f(1)=1$. $P(m,1): f(m)+1|m+1$. We will prove with induction that $f(n)=n$. Assume that $f(n)=n \forall n < k$. We will prove that $f(k)=k$. Assume otherwise: $f(k)= k-q$ $P(1,k): 2(k-q)|2k-q\Rightarrow 2(k-q)|q$ $P(k,1):k-q+1|k+1\Rightarrow k-q+1|q$ Let $(k,q)=d, k=da, q=db(a>b,(a,b)=1)\Rightarrow 2(a-b)|b, da-db+1|b$. $\Rightarrow (a-b)(da-db+1)|b$ $\Rightarrow (a-b)(da-db+1)=hb$ $\Rightarrow da^2+a=db^2+b+hb$ $\Rightarrow a(da+1)=b(db+1+h)$. Since $(a,b)=1\Rightarrow da+1=mb, db+h+1=ma$. $\Rightarrow (d-m)(a-b)=h, (d-m)(a+b)=-h-2$. This implies both that $d>m$ and $d<m$ which is a contradiction. So $f(x)=x\forall x \in \mathbb{N}$.