Does $k,l \in\mathbb N\setminus \{1\},\dfrac{k}{l} ,\dfrac{l}{k}\notin\mathbb N$ imply $\dfrac{k}{l}+\dfrac{l}{k}-\dfrac{1}{kl} \notin \mathbb N$?

Question

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 $\dfrac{k}{l} + \dfrac{l}{k} - \dfrac{1}{kl} \notin \mathbb N$?

Answer 1

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 $\notin \mathbb N$.

Answer 2

Let $k=2$ and $l=3$. Then your expression is an integer.

Answer 3

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, $\frac{21}{8} + \frac{8}{21}-\frac{1}{168} = 3$.

Answer 4

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.