Proving an expression is an integer 
Prove that $$A^2 \cdot \dfrac{\sum_{m=1}^{n}(a^{3m}-b^{3m})}{\sum_{m=1}^n(a^m-b^m)}-3A^2$$ is an integer if $a=\frac{k+\sqrt{k^2-4}}{2}, b=\frac{k-\sqrt{k^2-4}}{2}$, where $k>2$ is a positive integer, and $A =  \dfrac{1}{\sqrt{k^2-4}}$ for all positive integers $n$.

I thought about expanding the numerator and denominator. For the denominator we get $\sum_{m=1}^n(a^m-b^m) = \left(a \cdot \dfrac{a^n-1}{a-1}-b \cdot \dfrac{b^n-1}{b-1}\right)$ and we can similarly do the same thing with the numerator. How do we use all this to solve the question?

Partial answer:
Observe that $a,b$ are the roots of the quadratic equation $x^2 - kx + 1 = 0$, and hence $ab = 1$. 
This implies that 
$$ \frac{\sum (a^{3m}- b^{3m})}{\sum (a^m - b^m)} - 3 = \frac{\sum (a^m - b^m)^3}{\sum (a^m - b^m)} $$
Next, we also have the recurrence relation 
$$ k(a^m - b^m) - (a^{m-1} - b^{m-1}) = a^{m+1} - b^{m+1} $$
so defining $c_m$ by $c_1 = 1$, $c_2 = k$ and 
$$ c_{m+2} = k c_{m+1} - c_m $$
we have that it suffices to show that
$$ \frac{\sum_{m = 1}^n (c_m)^3}{\sum_{m = 1}^n c_m} $$
is an integer. This last step however I don't know how to treat. 
