Is this graph connected Define the following graph on the vertex set ${\mathbb N}_{\geq1}\&gt;$:

Two numbers $a$, $b\in {\mathbb N}_{\geq1}$ are connected by an edge (written  $a \ \mathcal{R} \ b)$ if and only if $a+b \ | \ ab-1$.

Clearly $1$ is isolated. Can we connect all integers greater than $2$ to $2$?
For example: $$2014 \ \mathcal{R} \ 147 \ \mathcal{R} \ 4175 \ \mathcal{R} \ 3891 \ \mathcal{R} \ 142 \ \mathcal{R} \ 43 \ \mathcal{R} \ 7 \ \mathcal{R} \ 3 \ \mathcal{R} \ 2.$$ Therefore $2014$ can be connected to $2$ (written $2014\sim2$).


  Question: Is this graph connected?


Motivation :

I worked on the Machin formula and I wondered if we had $$\arctan \frac{1}{a} + \arctan \frac{1}{b} = \arctan \frac{1}{c}$$ where $a,b,c$ are integers and this happens if  $c=\frac{ab-1}{a+b}$ is an integer.

EDIT : My apologie I forgot to mention that the graph is restricted to positive integer.
A partial answer :

We have $a \sim b$ if and only if there exist a sequence of integers $a_1, \ldots, a_n$ such that  $a \ \mathcal{R} \ a_1 \ \mathcal{R} \ \cdots \ \mathcal{R} \ a_n \ \mathcal{R} \ b$. The relation $ ab-1 = c (a + b) $ can be written as $(a-c)(b-c)=c^2+1$ and can be solves $a=c+d$ and $b=c+ \dfrac{c ^ 2 + 1} d$ where d is a divisor of $c^ 2 +1$.

If $c$ is even: All divisors of $c^2+1$ are congruent to $1$ modulo $4$ then $a$ is congruent to $b$ modulo $4$.

If $c$ is odd: $d$ is congruent to $1$ modulo $4$ and $\dfrac{c^2}d+1$ is congruent to $2$ modulo 4 or vice versa.
If $c=4k+1$ then $a$ is congruent to $2$ modulo $4$ and $b$ is congruent to $3$ modulo $4$ or vice versa.
If $c = 4k +3$ then $a$ is congruent to $0$ modulo $4$ and $b$ to $1$ modulo $4$ or vice versa.

Therefore there is probably three components in the graph (but I did not prove this):

The first formed only by $1$, the second one formed by integer congruent to $0$ modulo $4$ or $1$, and the last by integer congruent to $2$ or $3$ modulo $4$.
If you told me to find for a given set. $c$ - integer of any sign.

Then for the equation.  $$\frac{ab-1}{a+b}=c$$

using factorization. In the following manner.

$$(k-n)(k+n)=4(c^2+1)$$

Then the solutions are.

$$a=c+\frac{k+n}{2}$$

$$b=c+\frac{k-n}{2}$$

$$.........$$

$$a=c-\frac{k+n}{2}$$

$$b=c-\frac{k-n}{2}$$