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Let $n \geqslant 0$ be an integer. Show that $2^n | \lceil (3+\sqrt5)^n \rceil.$

This turned out to be relatively hard. The idea that our lecturer gave was that one should work towards an linear recurrence relation here, but it doesn’t seem to be an trivial task.

Using the binomial theorem on $(3+\sqrt5)^n$ was my first approach to see if there would be any cancellations, but that turned out to be also very messy. What would be the way to approach this?

Later
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  • The linear recurrence approach Is essentially induction. In general these are related to Binet forms (like the ones for Fibonacci numbers). See this for some motivation: https://math.stackexchange.com/questions/674570/prove-that-binets-formula-gives-an-integer-using-the-binomial-theorem – Alex R. Jun 08 '20 at 21:21
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    Show that what? Anyway, can you find a recurrence relation for $a_n=(3+\sqrt5)^n-(3-\sqrt5)^n$ in terms of $a_{n-1}$ and $a_{n-2}$? – J. W. Tanner Jun 08 '20 at 21:22
  • @J.W.Tanner show that $2^n$ divides $\lceil (3+\sqrt{5})^n \rceil$. Apologies for the bad notation. –  Jun 08 '20 at 21:28
  • I see now. So show $ 2^n$ divides $(3+\sqrt5)^n+(3-\sqrt5)^n$ and that equals $\lceil(3+\sqrt5)^n\rceil$ – J. W. Tanner Jun 08 '20 at 21:34
  • Could you elaborate on how did you come up with the expression $(3 + \sqrt5)^n + (3-\sqrt5)^n$? –  Jun 08 '20 at 22:17
  • add the "conjugate" of $(3+\sqrt5)^n$ to get an integer – J. W. Tanner Jun 08 '20 at 22:20

2 Answers2

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

Then can you show $a_0=2$, $a_1=6$, and $a_{n+2}=6a_{n+1}-4a_{n}$?

[$3+\sqrt5$ and $3-\sqrt5$ are roots of $x^2-6x+4=0$.]

It follows that $2^n|(3+\sqrt5)^n+(3-\sqrt5)^n$.

Since $(3+\sqrt5)^n$ is not an integer and $(3-\sqrt5)^n<1$,

it follows that $(3+\sqrt5)^n+(3-\sqrt5)^n=\lceil(3+\sqrt5)^n\rceil$.

J. W. Tanner
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  • Thanks for the answer! I can follow the logic here, but I cannot seem to be able to identify how did you and @OmG come up with the recurrence relation to solve? –  Jun 09 '20 at 08:31
  • Also the characteristic polynomial for $a_{n+2} = 6a_{n-1} - 4a_{n-2}$ seems to be $r^4-6r+4=0$ when divided by the lowest exponent of $r$ e.g $r^{n-2}$? –  Jun 09 '20 at 09:48
  • I'm sorry; I had an error in the recurrence relation; I have corrected it now; it's $a_{n+2}=6a_{n+1}-4a_n$; the roots of the characteristic equation $x^2-6x+4$ are $3\pm\sqrt5$ (their sum is $6$ and their product is $4$) – J. W. Tanner Jun 09 '20 at 10:15
  • Ah alright! Makes sense now. Although I'm still slightly confused on how does one come up with the recurrent relation. It's quite clear that $(3+\sqrt5)^n + (3-\sqrt5)^n = \lceil (3+\sqrt5)^n \rceil$, but the way you expressed it as $a_n$ and managed to find out the terms $a_0, a_1$ is very much unclear still. –  Jun 09 '20 at 10:51
  • @Daniel Take a look at https://en.wikipedia.org/wiki/Linear_difference_equation What this solution does is basically work backwards: We have a sequence that fits the closed form of a linear recurrence, what recurrence is that? – Ingix Jun 09 '20 at 12:58
  • As I commented to the question, I added $(3-\sqrt5)^n$ to make an integer; $a_0=(3+\sqrt5)^0+(3-\sqrt5)^0=2$ and $a_1=(3+\sqrt5)+(3-\sqrt5)=6$ – J. W. Tanner Jun 09 '20 at 13:05
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As hinted in comment, first solve the following recurrent relation:

$$ F_n = 3F_{n-1} - F_{n-2}, F_0 = 1, F_1 = 3 $$

If you can solve the recurrence, you will know the closed form of the above recurrent relation is $F_n = (\frac{3+\sqrt{5}}{2})^n + (\frac{3-\sqrt{5}}{2})^n \Rightarrow 2^n F_n = (3+\sqrt{5})^n + (3-\sqrt{5})^n$.

Now get a ceil from each side of the equation:

$\lceil 2^n F_n\rceil = \lceil (3+\sqrt{5})^n + (3-\sqrt{5})^n\rceil$

As $(3+\sqrt{5})^n + (3-\sqrt{5})^n$ is integer, we can find that $\lceil (3+\sqrt{5})^n + (3-\sqrt{5})^n\rceil = \lceil (3+\sqrt{5})^n \rceil$. Now, we will find:

$$ 2^n F_n = \lceil (3+\sqrt{5})^n \rceil. $$

Because $F_n$ is integer. Therefore $2^n | \lceil (3+\sqrt{5})^n \rceil$.

OmG
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  • What’s the motivation behind finding the correct recuerrence relation to solve? That seems to be my probelm here. –  Jun 08 '20 at 22:35