Working with some sequences, I’ve gotten to something like this:

$$(a+l_n)^n+(b+k_n)^n tag{1}$$

where $a,bin mathbb{N}$ and $l_n,k_ninmathbb{Q}cap(0,1)$ $forall n in mathbb{N}backslash{0,1}$, 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?

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:

$d.fbig|c.f+e.d$

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:

$d.fbig|c.f+e.d$

c, d, e and f can be functions of n, for example $c=2^n$, $d=5^n$ etc.