•  Prove that $2^n$ divides $(5+\sqrt{17})^n+(5-\sqrt{17})^n$. Use $v_2$
•  Make a recursion for $(5+\sqrt{17})^n+(5-\sqrt{17})^n$
•  If $a_n=(5+\sqrt{17})^n+(5-\sqrt{17})^n$ prove that $a_n=10a_{n-1}-8a_{n-2}$

$(5+\sqrt{17})$ and $5-\sqrt{17}$ are the roots of the equation

\[0=x^2-x(5+\sqrt{17}+5-\sqrt{17})+(5+\sqrt{17})(5-\sqrt{17})=x^2-10x+8\]

So, they both satisfy the equation

$x^2-10x+8=0\implies x^n=10x^{n-1}-8x^{n-2}$

Let $(5+\sqrt{17})^n+(5-\sqrt{17})^n=a_n$

    $$(5+\sqrt{17})^n=10(5+\sqrt{17})^{n-1}-8(5+\sqrt{17})^{n-2}$$

$$ (5-\sqrt{17})^n=10(5-\sqrt{17})^{n-1}-8(5-\sqrt{17})^{n-2}$$

$$a_n=10a_{n-1}-8a_{n-2}$$

Claim: $v_2(a_n)=n$.

Proof:  We will prove using induction.

Base case: 

$a_1=(5+\sqrt{17})+(5-\sqrt{17})=10=2\cdot 5$

$a_2=(5+\sqrt{17})^2+(5-\sqrt{17})^2=2^2\cdot 21$

Inductive step:

Given that $v_2(a_{n-1})=n-1$ and $v_2(a_{n-2})=n-2$, we will prove that $v_2(a_n)=n$.

Let $a_{n-1}=2^{n-1}p$ and $a_{n-2}=2^{n-2}q$

\[a_n=10a_{n-1}-8a_{n-2}=5\cdot2^np-2^{n+1}q\]

$v_2(5\cdot2^np)\neq v_2(2^{n+1}q)\implies v_2(a_n)=\min(v_2(5\cdot2^np), v_2(2^{n+1}q))=\min(n,n+1)=n$.

Now, we need to prove $2^{n+1}$ divides $10^n+(5+\sqrt{17})^n+(5-\sqrt{17})^n=10^n+a_n$

But, $v_2(10^n)=v_2(2^n\cdot5^n)=n$. So, $v_2(a_n)=v_2(10^n)$

So, $v_2(10^n+a_n)>v_2(10^n)=n$

So, $v_2(10^n+(5+\sqrt{17})^n+(5-\sqrt{17})^n)\geq n+1$

So, $2^{n+1}$ divides $10^n+(5+\sqrt{17})^n+(5-\sqrt{17})^n$