Prove that there are infinitely many $m,n$ for which $\frac{m+1}{n}+\frac{n+1}{m}$ is an integer Prove that there are infinitely many pairs of positive integers (m,n) such that 
$\frac{m+1}{n}+\frac{n+1}{m}$ 
is a positive integer.
I tried following:
clearly $(1,1)$ satisfies the condition. we assume that $(a,b)$ satisfies the condition. If we can find some $f(a)$ and $f(b)$ that satisfy the condition and they are increasing functions then we are done. But i didn't get such thing.

Apply the ideas of Vieta's root jumping to $ \frac{m+1}{n} + \frac{n+1}{m} = k $.
OP found $(1,1)$ as a solution, which give us the equation $ \frac{m+1}{n}+ \frac{n+1}{m} = 4$.
This simplifies to the quadratic $m^2 + (1-4n)m + n^2 + n = 0$.
Divide by $ m \neq 0$ to get: $ m - (4n-1) + \frac{n^2 + n}{m} = 0$.
So, if we have an integer solution $(m, n)$ , then

*

*$ \frac{n^2+n}{m}$ is an integer.

*$m+ \frac{n^2+n}{m} = 4n-1$.

Thus, for a fixed $n$, if $m$ is a solution to $X^2 + ( 1-4n)X + n^2+n = 0$, then so is $ \frac{n^2+n}{m}$.
This allows us to establish the Vieta root jumping: Define a recurrence via $ a_ 1 = 1, a _2 = 1, a_{n+2} = 4a_{n+1} - a_{n} - 1 = \frac{a_{n+1} ^2 + a_{n+1} } { a_n}$, and the above shows that $(a_{n}, a_{n+1})$ satisfies the conditions of the problem.
Observe that if $ a_{n+1} \geq a_{n}  > 0$, then $ a_{n+2} > \frac{a_{n+1}^2}{a_{n} } \geq a_{n+1}$.
Hence, we have a sequence of strictly increasing integers (apart from possibly the initial values), which gives us infinitely many solutions as desired.
Further Notes

*

*With $k=4$, this gives us the $1, 1, 2, 6, 21, \ldots$ solution found by paw.

*With $ k = 3$ (modify the above as needed), this gives us $2, 2, 3, 6, 14, 35, \ldots$ solution found by daniel.

*Furthermore, Vieta tells us that if we have any solution with $ m > n$, we can reverse engineer the sequence [IE Repeatedly apply $(m, n) \rightarrow (n, \frac{n^2+n}{m} )$] to end up with $ m=n$. Then, $ \frac{2(m+1)}{m}  = 2+\frac{2}{m}$ is an integer, which happens iff $ m = 1, 2$ giving us $k=4, 3$ respectively. Hence, we've found all solutions to the condition.

*We could show that $a_i $ was a strictly increasing sequence using just the linear recurrence via $ka_{n+1} - a_n - 1 > a_{n+1}$ for $ k = 4$. With $ k =3$, we have to be slightly careful.

