sororak wrote: a.) Let $m>1$ be a positive integer. Prove there exist finite number of positive integers $n$ such that $m+n|mn+1$. $m+n|m(m+n)$ so $m+n|m^2+mn$ So $m+n|mn+1\iff m+n|(m^2+mn)-(mn+1)=m^2-1$ There are finitely many factors of $m^2-1$ where $m>1$ is fixed, hence there are finitely many natural numbers $n$ s.t. $m+n|mn+1$

sororak wrote: b.) For positive integers $m,n>2$, prove that there exists a sequence $a_0,a_1,\cdots,a_k$ from positive integers greater than $2$ that $a_0=m$, $a_k=n$ and $a_i+a_{i+1}|a_ia_{i+1}+1$ for $i=0,1,\cdots,k-1$. We call two integers "connected" if they satisfy this property. Note that $(2p-1)+(2p+1)|(2p-1)(2p+1)+1$, so any consecutive odd integers are connected and hence any pair of odd integers are connected. Also note that $2p+(4p^2-2p-1)|2p(4p^2-2p-1)+1$, hence any even integer $2p$ is connected to an odd integer $4p^2-2p-1$. Therefore any two integers are connected.

sororak wrote: a.) Let $m>1$ be a positive integer. Prove there exist finite number of positive integers $n$ such that $m+n|mn+1$. b.) For positive integers $m,n>2$, prove that there exists a sequence $a_0,a_1,\cdots,a_k$ from positive integers greater than $2$ that $a_0=m$, $a_k=n$ and $a_i+a_{i+1}|a_ia_{i+1}+1$ for $i=0,1,\cdots,k-1$. Solution of problem 01: From the question we get that, $m+n\mid mn+1$ and it is obvious that,$m+n\mid m(m+n)=m^2+mn$ So,we can say that,$m+n\mid mn+m^2-(mn+1) =m^2-1$ Now we can say that,$m^2-1$ has finitly some factors (m+n).... Where m>1. Hence,we can say there exists finitely some natural numbers n which satisfy this divisibility

jgnr wrote: We call two integers "connected" if they satisfy this property. Note that $(2p-1)+(2p+1)|(2p-1)(2p+1)+1$, so any consecutive odd integers are connected and hence any pair of odd integers are connected. Also note that $2p+(4p^2-2p-1)|2p(4p^2-2p-1)+1$, hence any even integer $2p$ is connected to an odd integer $4p^2-2p-1$. Therefore any two integers are connected. I'm new to NT, but what was the motivation behind considering $2p$ and $4p^2 - 2p - 1$?

The motivation is to conduct a integeral polynomial and by comparing the coefficient and the root ，you can get the exact form