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

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The discussion focuses on proving that the expression $\left\lfloor{(2+\sqrt3)^n}\right\rfloor$ is odd for all positive integers n. A key point is the use of binomial expansion, where the sum $(2+\sqrt3)^n + (2-\sqrt3)^n$ results in an even integer. Since $(2-\sqrt3)^n$ is less than 1, it implies that $\left\lfloor(2+\sqrt3)^n\right\rfloor$ must be odd. The mathematical reasoning hinges on the properties of even and odd integers in relation to the floor function. This establishes a clear pattern for the oddness of the expression across all natural numbers n.
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show that $\left\lfloor{(2+\sqrt3)^n}\right\rfloor$ is odd for every positive integer n
 
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My solution:

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

$$x^2-4x+1$$

And by inspecting initial conditions, we may then state:

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

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

$$W_{n+1}=4W_{n}-W_{n-1}$$

and:

$$U_0=1,\,U_1=2$$

$$V_0=0,\,U_1=1$$

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

$$\left\lfloor U_n+V_n\sqrt{3} \right\rfloor=U_n+V_n\left\lfloor \sqrt{3} \right\rfloor=U_n+V_n$$

We may then conclude that for all natural numbers $n$ that $$\left\lfloor \left(2+\sqrt{3}\right)^n \right\rfloor$$ must be odd, since the sum of an odd and an even natural number will always be odd.
 
A slight variation on the same idea.
[sp] $$(2+\sqrt3)^n + (2-\sqrt3)^n = 2\sum_{k=0}^{\lfloor k/2\rfloor}{n\choose 2k}2^{n-2k}3^k$$ (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]
 
MarkFL said:
My solution:

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

$$x^2-4x+1$$

And by inspecting initial conditions, we may then state:

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

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

$$W_{n+1}=4W_{n}-W_{n-1}$$

and:

$$U_0=1,\,U_1=2$$

$$V_0=0,\,U_1=1$$

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

$$\left\lfloor U_n+V_n\sqrt{3} \right\rfloor=U_n+V_n\left\lfloor \sqrt{3} \right\rfloor=U_n+V_n$$

We may then conclude that for all natural numbers $n$ that $$\left\lfloor \left(2+\sqrt{3}\right)^n \right\rfloor$$ must be odd, since the sum of an odd and an even natural number will always be odd.
$\left\lfloor{V_n\sqrt{3}}\right\rfloor\ne V_n\left\lfloor{\sqrt3}\right\rfloor$
or am I mssing some thing
 
kaliprasad said:
$\left\lfloor{V_n\sqrt{3}}\right\rfloor\ne V_n\left\lfloor{\sqrt3}\right\rfloor$
or am I mssing some thing

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