10

Let $n$ be a nonnegative integer. Show that $\lfloor (2+\sqrt{3})^n \rfloor $ is odd and that $2^{n+1}$ divides $\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $.

My attempt:

$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$

$$ 0\leq (2-\sqrt{3})^n \leq1$$

$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$

$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$

$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$

darij grinberg
  • 17,777
Chon
  • 6,002
  • Related to http://math.stackexchange.com/questions/48508/how-are-the-integral-parts-of-9-4-sqrt5n-and-9-4-sqrt5n-relate – lhf Dec 23 '11 at 11:30
  • 3
    What do you mean by "integer part"? The floor function, or the first coordinate with respect to the basis $(1,\sqrt 3)$ of the $\Bbb Z$-module $\Bbb Z[\sqrt3]$? –  Aug 12 '19 at 21:20
  • Because the latter is even for $n=1$. On the other hand, I don't see what having some $2$ on the RHS has to do with evaluating the floor function of that sum of irrational numbers (for instance, the last term of the sum on the RHS would be $\binom nn2^{n-n}3^{n/2}=1\cdot 1\cdot \sqrt{3^n}$ ) –  Aug 12 '19 at 21:24
  • The reason GaeS is asking is because your proof seems to be either evaluating or ignoring the “evenness” of terms that are clearly irrational. If you were concerned with the floor function this would be unjustified. – Erick Wong Aug 12 '19 at 21:25
  • If the interpretation were the first coordinate with respect to the basis $(1, \sqrt{3})$ then the statement would be false: for example, $(2+\sqrt{3})^1 = 2+\sqrt{3}$ and $(2+\sqrt{3})^3 = 26 + 15\sqrt{3}$. – Daniel Schepler Aug 12 '19 at 21:27
  • 1
    Hint (with the interpretation of "integer part" being the floor function): examine the sequence $a_n = (2+\sqrt{3})^n + (2-\sqrt{3})^n$. – Daniel Schepler Aug 12 '19 at 21:30
  • Cf. this question – J. W. Tanner Aug 12 '19 at 21:32
  • 1
    Use \binom nk or \binom{n}{k} or {n \choose k} instead of C^k_n – gen-ℤ ready to perish Aug 12 '19 at 22:13
  • for example, in $a + \sqrt {b}$, the question wants to prove that a is odd – the cat Aug 12 '19 at 22:15
  • FWIW I'd intepret the "integer part" to mean if $(2+\sqrt 3)^k -1= a + b\sqrt 3$ then the "integer part" is $a$. As $(2+\sqrt 3)^2 - 1= (4 + 4\sqrt 3 + 3) -1 = 6+4\sqrt 3$ I'd say it is false. – fleablood Aug 12 '19 at 22:47
  • But it is the integer part of $(2+\sqrt{3})^k$, you add "1" to the end – the cat Aug 13 '19 at 00:08
  • But it is the integer part of $(2+\sqrt{3})^k$, you add "1" to the end – the cat Aug 13 '19 at 00:08

4 Answers4

15

Hint for the first part: Consider $u_n = (2+\sqrt{3})^{n} + (2-\sqrt{3})^{n}$. Prove that $u_n$ is always an even integer and that $u_n = \lceil (2+\sqrt{3})^n \rceil$. Use that $(2-\sqrt{3})^{n}\to 0$.

(This has now been incorporated into the edited question.)

Hint for the second part: Consider $v_n = (1+\sqrt{3})^{n} + (1-\sqrt{3})^{n}$. Find a second-order recursion for $v_n$ based on the quadratic equation that defines $1\pm\sqrt{3}$.

lhf
  • 216,483
  • You don't really need $(2-\sqrt 3)^n\to 0$, just that $0<(2-\sqrt 3)^n<1$, right? And you've confused $u_n$ and $u_2n$ in the second part. – Thomas Andrews Dec 22 '11 at 15:18
  • @Thomas, right, just that $0<(2-\sqrt 3)^n<1$. I've fixed the typo, thanks. – lhf Dec 22 '11 at 15:23
  • Could someone detail the second part? – Chon Dec 22 '11 at 17:20
  • @PlaneChon-Ju, actually, I don't think that the second part is true. The numbers in the sequence are 2,4,14,52,194,724,2702,10084,37634,140452,524174,1956244,7300802,27246964 and they are either 2 or 4 times an odd number. – lhf Dec 22 '11 at 18:40
  • Sorry, the quantity is actually $\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $ – Chon Dec 22 '11 at 21:19
  • @PlaneChon-Ju, after so many edits I think it's best if you asked a separate question. But the answers here should give you enough to start. – lhf Dec 22 '11 at 21:36
11

It is not hard to prove that$$(\forall n\in\mathbb N):\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n\in2\mathbb N.$$This, together with the fact that $2-\sqrt3\in(0,1),$ is enough to prove that $\left\lfloor\left(2+\sqrt3\right)^n\right\rfloor$ is an odd integer.

José Carlos Santos
  • 427,504
6

You can use recurrences such as $$f(n)=4f(n-1)-f(n-2)+2$$ or $$f(n)=5f(n-1)-5f(n-2)+f(n-3)$$ starting at $f(0)=1, f(1)=3$.

Then show the various results by induction.

Henry
  • 157,058
2

For the first claim:

Here is a more general discussion. As is often the case, none of this is original, and I did work it out on my own for fun.

To make one root of $f(x) =x^2-2ax+b =0$ with $a, b > 0$ in $(0, 1)$.

The roots are $x_{\pm} =\dfrac{2a\pm\sqrt{4a^2-4b}}{2} =a\pm\sqrt{a^2-b} $.

$\begin{array}\\ x_{-} &=a-\sqrt{a^2-b}\\ &=(a-\sqrt{a^2-b})\dfrac{a+\sqrt{a^2-b}}{a+\sqrt{a^2-b}}\\ &=\dfrac{a^2-(a^2-b)}{a+\sqrt{a^2-b}}\\ &=\dfrac{b}{a+\sqrt{a^2-b}}\\ \end{array} $

So we want $0 < b < a+\sqrt{a^2-b} $ so that $ b < a^2$ and $(b-a)^2 < a^2-b $ or $b^2-2ab+a^2 < a^2-b $ or $b^2+b < 2ab $ or $b< 2a-1$.

If $a=2$ then $b < 3$ so, if $b$ is an integer, $b \in \{1, 2\} $ so $a^2-b \in \{3, 2\} $ so $\{x_-, x_+\} =2\pm\sqrt{3}, 2\pm\sqrt{2} $.

Let $c = a^2-b$.

$\begin{array}\\ x_+^n+x_-^n &=(a+\sqrt{c})^n+(a-\sqrt{c})^n\\ &=\sum_{k=0}^n \binom{n}{k}a^kc^{(n-k)/2}+\sum_{k=0}^n \binom{n}{k}a^k(-1)^{n-k}c^{(n-k)/2}\\ &=\sum_{k=0}^n \binom{n}{k}a^kc^{(n-k)/2}(1+(-1)^{n-k})\\ &=\sum_{k=0}^n \binom{n}{k}a^{n-k}c^{k/2}(1+(-1)^{k})\\ &=\sum_{k=0}^{\lfloor n/2 \rfloor}2\binom{n}{2k}a^{n-2k}c^{k}\\ &=2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}\\ \end{array} $

Since $0 < x_-^n < 1$. $\lfloor x_+^n-1 \rfloor $ is odd and $\lfloor x_+^n\rfloor =2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}-1 $ so

$\begin{array}\\ x_+^n-\lfloor x_+^n \rfloor &=x_+^n-2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}+1\\ &=1-x_-^n\\ &\to_- 1\\ \end{array} $

If $a=2, b=1$ then $x_+^n-\lfloor x_+^n \rfloor =1-(2-\sqrt{3})^n =1-\dfrac{1}{(2+\sqrt{3})^n} $.

darij grinberg
  • 17,777
marty cohen
  • 107,799