Proving an expression is an integer

Question

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?

Note that $ab=1$, which can be used to simplify the term by writing the numerator in terms of the denominator — a25bedc5-3d09-41b8-82fb-ea6c353d75ae, Jul 28 '16 at 04:12
you can observe that $a, b$ are solutions to $x^2 - kx +1 = 0$, which can let you deduce $ab = 1$ through $x^2-kx+1 = (x-a)(x-b)$. I noticed since $a, b$ look suspiciously similar to quadratic formula. Admittedly I have no clue where to go from there though — Hwai-Ray Tung, Jul 28 '16 at 04:35
This may be related to or answered by this answer. The ratios in this question should satisfy equation $(9)$ in the cited answer. — robjohn, Jul 28 '16 at 14:18
I agree with @robjohn: the question is essentially a duplicate. The reduction I gave below reduces the question to exactly the question that robjohn cited. — Willie Wong, Jul 28 '16 at 14:25
Possible duplicate of Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$ — Willie Wong, Jul 28 '16 at 14:27

Kelenner · Answer 1 · 2016-07-28T14:28:12.770

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.

(+1) Nice one. It might be better to use different letters for the coefficients of $S_n$ since you already defined $u_n$. The calculation of first few terms (along with full solution) is given in robjohn's answer (as commented above). — Sungjin Kim, Jul 28 '16 at 14:26

score 1 · Answer 2 · answered Jul 28 '16 at 07:11

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.

Proving an expression is an integer

2 Answers2