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?

A possible way to the solution.
First, we have the following:
If $$u_n=\frac{a^3}{a^3-1}(a^{3n}-1)-\frac{b^3}{b^3-1}(b^{3n}-1)$$
$$v_n=\frac{a}{a-1}(a^{n}-1)-\frac{b}{b-1}(b^{n}-1)$$
then if $S_n$ is your expression:
$$S_n=A^2\frac{u_n}{v_n}-3A^2$$
In the following, I use at several places that $ab=1$ (and also $a^3b^3=1$). 
Put $P(x)=x^2-(1+\frac{1}{a})x+\frac{1}{a})$. We have 
$$v_n=a^{-n}\frac{a}{a-1}P(a^n)$$
In the same way, if $Q(x)=x^6-x^3(1+\frac{1}{a^3})+\frac{1}{a^3}$, we have
$$u_n=a^{-3n}\frac{a^3}{a^3-1}Q(a^{n})$$
Now, the roots of $P$ are $1$ and $\frac{1}{a}$. As $k>2$, we have $a\not =1$, hence these roots are distincts. We show easily that these are also roots of $Q$. Hence $Q/P$ is a polynomial, and we get that there exist $w_2,w_1,\cdots$ such that 
$$S_n=w_2a^{2n}+w_1 a^{n}+w_0+w_{-1}b^n+w_{-2} b^{2n}$$
Now this show that $S_n$ is a recurrent sequence, of order $\leq 5$. The polynomial of the recurrence is $U(x)=(x-1)(x-a)(x-a^2)(x-b)(x-b^2)$ (or perhaps a divisor of this polynomial, if the coefficient $w_0$ is zero, in this case the factor $(x-1)$ does not appear in the factorisation ) and it has  its coefficients in $\mathbb{Z}$, and to show that $S_n$ is in $\mathbb{Z}$ require only to compute a few first values of $S_n$ ie, $S_0,S_1,S_2,S_3,S_4$.
But of course, there is a lot of not so nice computations to do to finish the proof, if there are not errors above.  
