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Proof that for all $n\in \mathbb{N}$ : $$ (3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n] $$

Hamza
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    Have you tried writing out the binomial expansion? There are clearly going o be a lot of cancellations – Simon S Nov 28 '14 at 05:56
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    The binomial is good to show that the quantity is an integer, but not for factors. Recursion plus induction is the simplest way to go. – Winther Nov 28 '14 at 06:02

3 Answers3

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HINT:

If $a_n=(3+\sqrt5)^n+(3-\sqrt5)^n$

$$a_{n+2}-6a_{n+1}+4a_n=0$$

Now use Strong induction like $2^m\mid a_m$ for $1\le m\le n$

Community
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lab bhattacharjee
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Consider the recurrence relation $$a_n=3a_{n-1}-a_{n-2}$$ with initial conditions $a_0=2$ and $a_1=3$. Use this to find that $$a_n=\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n.$$

  • Effectively the same as lab bhattacharjee's answer, but doing it this way makes the induction much easier, in fact, trivial. +1 – David Nov 28 '14 at 06:13
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Another view on this:

$\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$

Is the trace of the $n$-th power of

$\left(\begin{array}{cc} 2 & 1 \\ 1 & 1\end{array}\right)$

And another:

$\dfrac{3+\sqrt 5}{2}$ and $\dfrac{3-\sqrt 5}{2}$ are algebraic integers (they're roots of $x^2 - 3x + 1$). Then since $\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$ is rational, it's an integer.

All of these methods are closely related.

aes
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  • all the methods are identical ! It only differs at the point of how the recurssion is established ! .. using $\alpha^{n+1}+\beta^{n+1} = (\alpha+\beta)(\alpha^{n}+\beta^{n}) - \alpha\beta(\alpha^{n-1}+\beta^{n-1})$ is another one for example ! – r9m Nov 28 '14 at 06:21