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# Understand the problem

Define a sequence $< a_n > _{n\geq0}$ by $a_0 = 0$, $a_1 = 1$ and
$$a_n = 2a_{n - 1} + a_{n - 2},$$
for $n\geq2.$

$(a)$ For every $m > 0$ and $0\leq j\leq m,$ prove that $2a_m$ divides
$a_{m + j} + ( - 1)^ja_{m - j}$.

$(b)$ Suppose $2^k$ divides $n$ for some natural numbers $n$ and $k$. Prove that $2^k$ divides $a_n.$

##### Source of the problem

Indian National Mathematical Olympiad 2010

Number Theory
Medium
##### Suggested Book
Problem Solving Strategies by Arthur Engel

Do you really need a hint? Try it first!

For (a), use strong induction on $j$.
The recurrence has the characteristic equation $x^2-2x-1=0$ (see this post for details). This has two distinct roots, $1\pm\sqrt{2}$. Hence the general solution is of the form $y_n=a(1+\sqrt{2})^n+b(1-\sqrt{2})^n$.
Solving for $a,b$, we get $$a_n=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$$. Suppose that $2^k\;|\; n$, i.e. $n=2^kj$ for some $j$. Now it suffices to show that $\frac{a_n}{2^k}=$ $$\frac{(1+\sqrt{2})^{2^kj}-(1-\sqrt{2})^{2^kj}}{2^{k+1}\sqrt{2}}$$ is an integer. Use induction on $k$ to prove this.
Note that the given expression is equivalent to $\frac{(1+\sqrt{2})^{2^{k-1}j}+(1-\sqrt{2})^{2^{k-1}j}}{2}\cdot \frac{(1+\sqrt{2})^{2^{k-1}j}-(1-\sqrt{2})^{2^{k-1}j}}{2^k\sqrt{2}}$. By induction, it suffices to show that $\frac{(1+\sqrt{2})^{2^{k-1}j}+(1-\sqrt{2})^{2^{k-1}j}}{2}$ is an integer. Write $b_n=\frac{(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}}{2}$. Note that $b_n$ satisfies $b_n=2b_{n-1}+b_{n-2}$ hence it suffices to prove that $b_0$ and $b_1$ are both integers. As $b_0=1=b_1$, we are done.

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