Prove $2^{n+1} \mid 10^n-(5+\sqrt{17})^n-(5-\sqrt{17})^n$ for all positive integers $n$.

Question

Problem:

$f(n) = 10^n-(5+\sqrt{17})^n-(5-\sqrt{17})^n$ is a function which is valid for all integers $n\geq1$. Prove that $f(n)$ is always perfectly divisible by $2^{n+1}$.

My approach:

I have used the theory of recurrence relations and induction.

The general form of third order homogeneous recurrence relation is $$c_1a_n+c_2a_{n-1}+c_3a_{n-2}+c_4a_{n-3}=0$$

If we change $a_n$ to $x^n$, we get $$c_1x^n+c_2x^{n-1}+c_3x^{n-2}+c_4x^{n-3}=0$$

If we want to get the characteristic equation of it, the equation will be $c_1x^3+c_2x_2+c_3x+c_4=0$.

If we solve the equation, we will get three solutions $k_1$, $k_2$, $k_3$.

Then $a_n=Ak_1^n+Bk_2^n+Ck_3^n$ where $A$, $B$ and $C$ are constants according to the equation.

$$\begin{equation}\tag{i}a_n=Ak_1^n+Bk_2^n+Ck_3^n\end{equation}$$

Let $$\begin{equation}\tag{ii}a_n=10^n-(5+\sqrt{17})^n-(5-\sqrt{17})^n\end{equation}$$

If we compare $(\mathrm{i})$ and $(\mathrm{ii})$, we get $A=1$, $B=-1$, $C=-1$. If we calculate $k_1=10$, $k_2=5+\sqrt{17}$, $k_3=5-\sqrt{17}$

So, the polynomial will be (in terms of $b$) $$(b-10)[b-(5+\sqrt{17})][b-(5-\sqrt{17})]=0$$ or $b^3-20b^2+108b-80=0$.

If we multiply the equation by $b^{n-3}$ and replace $b^n$ by $a_n$, it will give the result of $$a_n=20a_{n-1}-108a_{n-2}+80a_{n-3}$$ or $$a_{n+1}=20a_{n}-108a_{n-1}+80a_{n-2}$$

We will use strong induction now.

Base case:

$2^{2+1}$ divides $f(2)$.

Induction step:

Suppose it is true for all the way up to $k+1$. Then, $2^{k+1} \mid a_n$, $2^{k} \mid a_{n-1}$ and $2^{k-1} \mid a_{n-2}$. Therefore $a_{n}=2^{k+1}m$, $a_{n-1}=2^{k}n$ and $a_{n-2}=2^{k-1}p$ where $m$, $n$ and $p$ are integers.

If we input the values of $a_{n}$, $a_{n-1}$, $a_{n-2}$ in $a_{n+1}=20a_{n}-108a_{n-1}+80a_{n-2}$, we will get $a_{n+1}=2^{k+2}(10m-27n+10p)$, and so $2^{k+2} \mid a_{n+1}$. Therefore $2^{n+1} \mid f(n)$ for all positive integers. $\square$

Conclusion: I don't like this proof because it includes recurrence relation.I don't want to solve it using recurrence relation. Feel free to help me.

Source: Problem 8, BDMO (Bangladesh Mathematical Olympiad) 2024, junior, national

Answer 1

A Freshman's $\rm\color{#0af}{Dream}$ Binomial Theorem yields a simple proof: cancelling $\:\!2^n$ reduces it to

$\qquad\quad\ \ \exists\:\!j\in\Bbb Z\!:\ a^n\! + b^n\ \ \, =\,\ \ \color{#0a0}5^n\ +\ \color{#e80}2\,j\in\Bbb Z,\,$ for $\,a,b = \frac{5\pm\sqrt{17}}2,\,$ $\small \begin{align}\color{#0a0}{a\!+\!b = 5}\\ \color{#e80}{ab = 2}\end{align}\:\!;\ $ it's true by ${\rm\color{#0af}{Dream}}\ \ \exists\:\!j\in\Bbb Z^{\phantom{|^{|}}}\!\!\!\!:\ a^n\! + b^n\! = (\color{#0a0}{a\!+\!b})^n \!+\!\color{#e80}{ab}j\in \Bbb Z,\,$ if $\ \color{#0a0}{a\!+\!b},\color{#e80}{ab} \in \Bbb Z,\,$ by $\rm\color{#c00}{induction},$ with step: $\qquad\quad\ \ a^{n+1}\!+b^{n+1} = (a\!+\!b)\:\!\overbrace{(a^n\!+b^n)}^{\large \!\!\!\color{#c00}{(a+b)^n\:\!\:\!+\:\color{#e80}{ab}^{\phantom{|^|}}\!\!j}\!\!\!} $ ${-}\:\color{#e80}{ab}\:\!(\overbrace{a^{n-1}\!+b^{n-1}}^{\color{#c00}{\large k\ \in\ \Bbb Z}}) \,=\, \overbrace{(\color{#0a0}{a\!+\!b})^{n+1}\!+\color{#e80}{ab}\:\!\color{#90f}J}^{ (a+b)\ \color{#c00}{j\,\ -\,\ k}\,\ =\:\!:\,\ \color{#90f}{J}\ \in\ \Bbb Z\!\!\!\!\!\!\!\!\!}\ \ \ \ \bf\small QED$

Answer 2

Have you seen these formulae?: $$p_n=p_{n-1}e_1-p_{n-2}e_2+p_{n-3}e_3-...\tag1$$ Let $x_1=w=5+\sqrt{17}$ and $x_2=\bar w=5-\sqrt{17}.$ Then we have two elementary symmetric polynomials: $e_1=x_1+x_2=10$, $e_2=x_1x_2=8.$ Power sum symmetric polynomials are $$p_n=x_1^n+x_2^n,\,\,n\geq0.$$ First examples are $p_0=2,$ $p_1=e_1=10,$ $p_2=84.$

By $(1)$, we have the recursive equation $p_n=2(5p_{n-1}-4p_{n-2}).$ İt is not difficult to show that $p_n=2^n\times \text{odd}$ for $n\geq1$ by induction. Hence, $2^{n+1}$ will divide $10^n-p_n$ for positive integer $n$.