Oddity of $\left\lfloor{(2+\sqrt3)^n}\right\rfloor$ for All $n \in \mathbb{N}$

[kaliprasad]

show that $\left\lfloor{(2+\sqrt3)^n}\right\rfloor$ is odd for every positive integer n

 

[MarkFL]

My solution:

Spoiler

First, we find that the minimal polynomial for $2+\sqrt{3}$ is:

\(\displaystyle x^2-4x+1\)

And by inspecting initial conditions, we may then state:

\(\displaystyle \left(2+\sqrt{3}\right)^n=U_n+V_n\sqrt{3}\)

where both $U$ and $V$ are defined recursively by:

\(\displaystyle W_{n+1}=4W_{n}-W_{n-1}\)

and:

\(\displaystyle U_0=1,\,U_1=2\)

\(\displaystyle V_0=0,\,U_1=1\)

Now, we can see that $U_n$ and $V_n$ have opposing parities, and given:

\(\displaystyle \left\lfloor U_n+V_n\sqrt{3} \right\rfloor=U_n+V_n\left\lfloor \sqrt{3} \right\rfloor=U_n+V_n\)

We may then conclude that for all natural numbers $n$ that \(\displaystyle \left\lfloor \left(2+\sqrt{3}\right)^n \right\rfloor\) must be odd, since the sum of an odd and an even natural number will always be odd.

 

[Opalg]

A slight variation on the same idea.
[sp] \(\displaystyle (2+\sqrt3)^n + (2-\sqrt3)^n = 2\sum_{k=0}^{\lfloor k/2\rfloor}{n\choose 2k}2^{n-2k}3^k\) (binomial expansion). The right side is clearly an even integer. On the left side, $(2-\sqrt3)^n < 1$, so it follows that $\left\lfloor(2+\sqrt3)^n\right\rfloor$ is odd.[/sp]

 

[kaliprasad]

My solution:

Spoiler

First, we find that the minimal polynomial for $2+\sqrt{3}$ is:

\(\displaystyle x^2-4x+1\)

And by inspecting initial conditions, we may then state:

\(\displaystyle \left(2+\sqrt{3}\right)^n=U_n+V_n\sqrt{3}\)

where both $U$ and $V$ are defined recursively by:

\(\displaystyle W_{n+1}=4W_{n}-W_{n-1}\)

and:

\(\displaystyle U_0=1,\,U_1=2\)

\(\displaystyle V_0=0,\,U_1=1\)

Now, we can see that $U_n$ and $V_n$ have opposing parities, and given:

\(\displaystyle \left\lfloor U_n+V_n\sqrt{3} \right\rfloor=U_n+V_n\left\lfloor \sqrt{3} \right\rfloor=U_n+V_n\)

We may then conclude that for all natural numbers $n$ that \(\displaystyle \left\lfloor \left(2+\sqrt{3}\right)^n \right\rfloor\) must be odd, since the sum of an odd and an even natural number will always be odd.

Spoiler

$\left\lfloor{V_n\sqrt{3}}\right\rfloor\ne V_n\left\lfloor{\sqrt3}\right\rfloor$
or am I mssing some thing

 

[MarkFL]

Spoiler

$\left\lfloor{V_n\sqrt{3}}\right\rfloor\ne V_n\left\lfloor{\sqrt3}\right\rfloor$
or am I mssing some thing

Spoiler

Yes, my mistake, that's only true for $V_n\in\{0,1\}$. :D