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Prove that the next integer greater than $(3+\sqrt{5})^n$ is divisible by $2^n$ where $n$ is a natural number. The problem is given in a chapter of induction. It is actually a part of this whole question:

If $S_n = (3+\sqrt{5})^n + (3-\sqrt{5})^n$, show that $S_n$ is an integer and that $S_{n+1} = 6S_n - 4S_{n-1}$. Deduce that the next integer greater than $(3+\sqrt{5})^n$ is divisible by $2^n$.

I could do the first two parts. The first part I have done by induction and the second part by simply using the given formula. I cannot proceed at all in the third part. I think it may need induction. Please give any solutions regarding the third one and see whether the second one can be proved using induction.

Shubhraneel Pal
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    How about showing that $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is divisible by $2^n$ for all $n\in\mathbb{N}$ using induction? Part I and II are shown here. – StubbornAtom Dec 24 '16 at 12:31
  • How do I prove that and how can I be sure that $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is the next integer greater than $(3+\sqrt{5})^n$. – Shubhraneel Pal Dec 24 '16 at 12:36
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    For the third part: You know that $S_n$ is an integer greater than $\left( 3+\sqrt 5\right)^n$. If $\lceil \left( 3+\sqrt 5 \right)^n \rceil<S_n$ Then show that this implies $\left( 3-\sqrt 5\right)^n>1$ (a contradiction). – lulu Dec 24 '16 at 12:37

1 Answers1

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If we denote $a:=3+\sqrt{5}$ and $b:=3-\sqrt{5}$, then $a$ and $b$ are the roots of the polynomial $$x^2-6x+4$$

Therefore , the number $s_n:=a^n+b^n$ satisfies the recursion :

$$s_1=6$$

$$s_2=28$$

$$s_n=6s_{n-1}-4s_{n-2}$$

Because of $0<b^n<1$ for $n\ge 1$, we have $\lceil a^n \rceil=s_n$

Suppose $s_n$ is divisble by $2^n$ and $s_{n-1}$ is divisible by $2^{n-1}$ (This claim is true for $n=2$) . Then, $s_{n+1}=6s_n-4s_{n-1}$ is divisble by $2^{n+1}$ because $6s_n$ and $4s_{n-1}$ are both divisble by $2^{n+1}$. Therefore, the induction step is completed.

Peter
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  • Both are only divisible by $2^n$. – WimC Dec 24 '16 at 12:51
  • @WimC Fixed. I got the indeces wrong! – Peter Dec 24 '16 at 12:53
  • I did not understand the recursion part. – Shubhraneel Pal Dec 24 '16 at 12:53
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    @user321525 Usually, the calculation is done in the other direction. We know the recursion and calculate the characteristic equation and derive the general formula. To get the characteristic equation, we insert the Ansatz $s_n=x^n$ into the recursion $s_n=us_{n-1}+vs_{n-2}$ and get $x^n-ux^{n-1}-vx^{n-2}=0$ divided by $x^{n-2}$ this gives $x^2-ux-v=0$, the characteristic equation . If the roots $a$ and $b$ are distinct, we can show that there are constants $c_1$ and $c_2$, such that $s_n=c_1a^n+c_2b^n$. The constants can be calculated by using $s_1$ and $s_2$ – Peter Dec 24 '16 at 12:58
  • You can try to show the recursion by induction , a good additional exercise. – Peter Dec 24 '16 at 12:59
  • Now how do I do that? – Shubhraneel Pal Dec 24 '16 at 13:06
  • The induction proof ? Show $s_n=a^n+b^n$ first for $n=1$ and $n=2$. Then, assume $s_{n-1}=a^{n-1}+b^{n-1}$ and $s_{n-2}=a^{n-2}+b^{n-2}$. Plugging into the recursion should give $s_n=a^n+b^n$ – Peter Dec 24 '16 at 13:18