Does $k,l \in\mathbb N\setminus \{1\},\dfrac{k}{l} ,\dfrac{l}{k}\notin\mathbb N$ imply $\dfrac{k}{l}+\dfrac{l}{k}-\dfrac{1}{kl} \notin \mathbb N$? I tried to solve one problem and that led me to this simple problem whose solution I do not see, and it goes like this:

Suppose $k,l \in \mathbb N \setminus \{1\}$ and $\dfrac {k}{l} \notin \mathbb N$ and $\dfrac {l}{k} \notin \mathbb N$. Do we have $\dfrac{k}{l} + \dfrac{l}{k} - \dfrac{1}{kl} \notin \mathbb N$?

 A: Simplify the expression to see  that $k^2+l^2-1$ must be a multiple of  $kl$ for the expression to be an integer.It is clear that this happens for $k=2,l=3$, for example. 
A more non-trivial observation gives the interesting fact that:
$$
\frac{(k+1)}{k} + \frac{k}{(k+1)} -\frac{1}{k(k+1)} = \frac{2k(k+1)}{k(k+1)} =2
$$
regardless of which $k$ you take. So there are infinitely many counterexamples to this proposition (of course, $\frac{k+1}{k}$ is not integer for all values of $k$ except $1$). 
EDIT: If you are asking for non-trivial values of the right hand side, here are some:
$$
k=55,l=21 \implies \frac{k}{l} + \frac{l}{k} -\frac{1}{lk} = 3
$$
$$
k=99,l=10 \implies \frac{k}{l} + \frac{l}{k} -\frac{1}{lk} = 10
$$
Here is something that may catch you off guard:
$$
\frac{k^2-1}{k} + \frac{k}{k^2-1} -\frac{1}{k(k^2-1)} = k
$$
So it follows that even non-trivial values can be attained.
One last to finish it:
$$
k=987,l=377 \implies \frac kl + \frac lk - \frac 1{lk} = 3
$$
EDIT OF EDIT: I must point this out, because I find it simply quite atrocious.
$$
\frac{F_n}{F_{n+2}} + \frac{F_{n+2}}{F_n} - \frac{1}{F_{n+2}F_n}
$$
where the $F_n$ are Fibonacci numbers,$n \geq 5$ (that is, $F_n \geq 3$), is always a natural number! (prove it yourself) For example, $\frac{21}{8} + \frac{8}{21}-\frac{1}{168} = 3$.
A: Let $k=2$ and $l=3$.  Then your expression is an integer.
A: Take any $n\in\mathbb N$ and take $k=n$, $l=n+1$ and you will get $$\frac kl + \frac lk -\frac{1}{kl} =\frac{n+1}{n} + \frac{n}{n+1} - \frac{1}{n(n+1)} = \\ = \frac{n^2 + (n+1)^2 - 1}{n(n+1)} = \frac{2n^2 + 2n + 1 - 1}{n(n+1)} =  2\in\mathbb N$$, so no, you don't have $\notin \mathbb N$.
A: There are several options.  The meaning is the same everywhere. To represent the equation.
$$\frac{k}{l}+\frac{l}{k}-\frac{1}{kl}=q$$
$$k^2-qkl+l^2=1$$
Then use different replacement. Such.  $k=x+y$  $:$ $l=x-y$
$$(q+2)y^2-(q-2)x^2=1$$
Or such.   $k=x+al$  We get this equation.
$$x^2+(2a-q)xl+(a^2-qa+1)l^2=1$$
And the decision will be reduced to a Pell equation.  You need to use this formula.  http://www.artofproblemsolving.com/community/c3046h1048219
The only condition is to choose the ratio of $q$ such that the root was a rational number.
