Show that infinitely many positive integer pairs $(m,n)$ exist s.t $\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{N}$ Show that infinitely many positive integer pairs $(m,n)$ exist such that $\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{N}$.
I couldn't solve it but I did make an observation which might or might not be helpful.
WLOG assume that $m < n$ (since the term is symmetric wrt $m,n$ along with the fact that ignoring the $m=n$ case wouldn't create any trouble). Write $n=mq+r$ for some $r\in \{[0,m-1]\cap \mathbb{N}\}$. Now, the question boils down to showing $\frac{m+1}{n}+\frac{r+1}{m} \in \mathbb{N}$. Note that each of the summand is $\leq 1$. Since the equality case wouldn't be helpful in generating infinitely many pairs of $(m,n)$, we can safely say that the sum is equal to $1$.
Now I don't know how to proceed from here.
 A: I shall find all pairs $(m,n)$ of positive integers such that $$\frac{m+1}{n}+\frac{n+1}{m}\in\mathbb{N}.$$
Let $k$ be such a positive integer for which $$\dfrac{m+1}{n}+\dfrac{n+1}{m}=k\tag{1}$$ for some $m,n\in\mathbb{N}$.  Note that $t=m$ is a solution to
$$t^2-(kn-1)t+(n^2+n)=0.$$
However, there is another root $t=kn-1-m=\dfrac{n^2+n}{m}$, which is an integer as $kn-1-m\in\mathbb{Z}$, and which is positive since $\dfrac{n^2+n}{m}>0$.  Thus, if $(m,n)$ is a positive integer solution to (1), then  $(n,kn-1-m)=\left(n,\dfrac{n^2+n}{m}\right)$ is also a positive integer  solution.
Now, suppose that $(m_0,n_0)$ is a solution to (1) such that $m_0\geq n_0$ and $m_0+n_0$ is smallest possible.  If $m_0>n_0$, we see that $\left(n_0,\dfrac{n_0^2+n_0}{m_0}\right)$ is also a solution, but
$$n_0+\frac{n_0^2+n_0}{m_0}=n_0+n_0\left(\frac{n_0+1}{m_0}\right)\leq n_0+n_0<m_0+n_0\,.$$
This contradicts the minimality of $m_0+n_0$, and so $m_0=n_0$ must hold.  Thus,
$$k=\frac{m_0+1}{m_0}+\frac{m_0+1}{m_0}=2+\frac{2}{m_0}\,.$$
That is, $(m_0,n_0)=(1,1)$ (which gives $k=4$), or $(m_0,n_0)=(2,2)$ (which gives $k=3$).
In the first case, $m_0=n_0=1$ and $k=4$.  Define a sequence $(a_0,a_1,a_2,\ldots)$ by taking $a_0=1$, $a_1=1$, and
$$a_{r}=4a_{r-1}-a_{r-2}-1$$
for $r=2,3,4,\ldots$.  It follows that all solutions $(m,n)$ with $k=4$ such that $m\leq n$ are of the form $(a_{r},a_{r+1})$ for some $r=0,1,2,\ldots$.  For example, $a_2=2$, $a_3=6$, $a_4=21$, and $a_5=77$.
In the second case, $m_0=n_0=2$ and $k=3$.  Define a sequence $(b_0,b_1,b_2,\ldots)$ by taking $b_0=2$, $b_1=2$, and
$$b_r=3b_{r-1}-b_{r-2}-1$$
for $r=2,3,4,\ldots$.  It follows that all solutions $(m,n)$ with $k=3$ such that $m\leq n$ are of the form $(b_{r},b_{r+1})$ for some $r=0,1,2,\ldots$. For example, $b_2=3$, $b_3=6$, $b_4=14$, and $b_5=35$.
A: Equation:  $$\frac{m+1}{n}+\frac{n+1}{m}=a$$
You can solve using the Pell equation: $p^2-(a^2-4)s^2=1$
Then the solutions are:
$$n=2(p-(a+2)s)s$$
$$m=-2(p+(a+2)s)s$$ 
And more solutions:
$$n=\frac{2p(p+(a-2)s)}{a-2}$$
$$m=\frac{2p(p-(a-2)s)}{a-2}$$
You can also write the solution formula if the coefficient is such that the equation $p^2-(a^2-4)s^2=4$ and taking advantage of his decisions.
Then the formula has the form:
$$n=\frac{p-(a-2)s+2}{2(a-2)}$$
$$m=\frac{p+(a-2)s+2}{2(a-2)}$$
