Is there a way to know when $(a+l_n)^n+(b+k_n)^n$ is an integer for integer $a$, $b$ and rational $l_n$, $k_n$ with $l_n+k_n=1$? Working with some sequences, I've gotten to something like this:
$$(a+l_n)^n+(b+k_n)^n \tag{1}$$
where $a,b\in \mathbb{N}$ and $l_n,k_n\in\mathbb{Q}\cap(0,1)$ $\forall n \in \mathbb{N}\backslash\{0,1\}$, such that $l_n+k_n=1$. 
My first thought was that it cannot be integer never, but then I remembered the Fibonacci numbers where something with the sum of two irrational numbers that is natural. So,

Is there a way to know when $(1)$ is an integer?

Any help please? Thanks in advance!
Edit: After doing some work on my problem, I got to:
$$l_n=\frac{2^{n-1}-1}{2^n} \qquad k_n=\frac{2^{n-1}+1}{2^n}$$
With this extra information, would this be helpful in order to find a solution?
 A: Let's start with a simple form:
$A=(a+\frac{c}{d})+(b+\frac{e}{d})$
It can be seen that if $d \big|c+e$ then A can be an integer. For example:
$(13+\frac{7}{15})+(17+\frac{8}{15})=\frac{465}{15}=31$
$(7+\frac{3}{5})+(3+\frac{2}{5})=\frac{465}{15}=15$
Where denominators of fraction are equal. We can make denominators equal in the case they are not equal. Let $l_n=\frac{c}{d}$ and $k_n=\frac{e}{f}$, then we may write:
$A=(a+\frac{c}{d})+(b+\frac{e}{f})=(a+\frac{c.f}{d.f})+(b+\frac{e.d}{f.d})$
Then the condition is:
$d.f\big|c.f+e.d$
For example we have:
$c=2$,$d=5$, $e=3$ and we want to find $f$ such that the condition is provided, we must have:
$ed+cf=15+2f=t(5f)$
The only solution is:
$t=1$, $15=5f-2f=3f$ ⇒ $f=5$
Now let $a=8$ and $b=11$ we have:
$(8+\frac{2}{5})+(11+\frac{3}{5})=\frac{100}{5}=20$
For general form $A=(a+\frac{c}{d})^n+(b+\frac{e}{f})^n$
n must be odd ,let $n=2t+1$,such that A can be reduced but the condition will be the same, i.e in following relation:
$A=(a+\frac{c}{d})^{2t+1}+(b+\frac{e}{d})^{2t+1}=\big[(a+\frac{c}{d})+(b+\frac{e}{f})\big]((a+\frac{c}{d})^{2t}(b+\frac{e}{f})+ . . . )$
if $(a+\frac{c}{d})+(b+\frac{e}{f})$ is integer, then A is integer and for that we must have:
$d.f\big|c.f+e.d$
c, d, e and f can be functions of n, for example $c=2^n$, $d=5^n$ etc.
