http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1590903&sid=0f1a417eed6f70762e1182442aad553c#p1590903

See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=373792&start=0

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1879662#p1879662

WLOG we may assume that $a\ge b$. If $a=b$ or $a=b+1$,it's quite simple.(So I skip it ) If $a\ge b+2$,then suppose $a=pb+q$,$q\in [1,b-1]$. $\frac{a+1}{b}+\frac{b+1}{a}=p+\frac{q+1}{b}+\frac{b+1}{pb+q}$. It's easy to find that both $\frac{q+1}{b}$ and $\frac{b+1}{pb+q}$ $\in (0,1]$. So $\frac{q+1}{b}+\frac{b+1}{pb+q}=1$ $\iff (p-1)(b-q-1)=0$. If $b=q-1$,the rest is very easy. Else $p=1$,it's clear that $\frac{q+1}{b}+\frac{b+1}{pb+q}\ge 1$. We R done! I hope that I'm right.

DelicateFlower wrote: WLOG we may assume that $a\ge b$. If $a=b$ or $a=b+1$,it's quite simple.(So I skip it ) If $a\ge b+2$,then suppose $a=pb+q$,$q\in [1,b-1]$. $\frac{a+1}{b}+\frac{b+1}{a}=p+\frac{q+1}{b}+\frac{b+1}{pb+q}$. It's easy to find that both $\frac{q+1}{b}$ and $\frac{b+1}{pb+q}$ $\in (0,1]$. So $\frac{q+1}{b}+\frac{b+1}{pb+q}=1$ $\iff (p-1)(b-q-1)=0$. If $b=q-1$,the rest is very easy. Else $p=1$,it's clear that $\frac{q+1}{b}+\frac{b+1}{pb+q}\ge 1$. We R done! I hope that I'm right. Wrong. Take $a=6$ and $b=3$. Moreover, transforming the equation $\frac{q+1}{b}+\frac{b+1}{pb+q}=1$ you make gross mistake.

But where is wrong(I agree with that I'm wrong.)?

DelicateFlower wrote: But where is wrong(I agree with that I'm wrong.)? For example, you write that $q \in [1,b-1]$. But may be $q=0$.

Answer: 3 and 4 Fix $k$. Among all pairs of positive integers $(a,b)$ satisfying $\frac{b+1}a+\frac{b+1}a=k$, take the pair with minimum sum. We will prove that such pair satisfies $a=b$, otherwise assume WLOG that $a>b$. Consider the equation $\frac{x+1}b+\frac{b+1}x=k$, which is equivalent to $x^2-x(kb-1)+b^2+b=0$. One of the equation's roots is $x_1=a$, the other one is $x_2=kb-1-a=\frac{b^2+b}a$ which is a positive integer. So $x_2+b\ge a+b$, $\frac{b^2+b}a\ge a$, $b^2+b\ge a^2\ge(b+1)^2=b^2+2b+1$, which is impossible. Therefore we must have $a=b$, and $k=2+\frac2a$ which equals 3 and 4 for $a=2$ and $a=1$ respectively.

nnosipov wrote: DelicateFlower wrote: But where is wrong(I agree with that I'm wrong.)? For example, you write that $q \in [1,b-1]$. But may be $q=0$. OIC.

What it is OIC?

nnosipov wrote: What it is OIC? OIC=Oh,I see.

Thank you .

Haha~ I like speaking llike that.

This problem is the further research of All-Russian 1994 Grade 11 P1