Show that $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n} = \lfloor(1+\sqrt{3})^{2n}+1\rfloor$.

Question

Show that $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n} = \lfloor(1+\sqrt{3})^{2n}+1\rfloor$.

I know that adding the binomial expansions in LHS will cancel out the terms having radical. And we will be left with $2(1+ ^{2n}C_2 3+ \cdots \cdots +^{2n}C_{2n}3^{2n})$. Rest I don't know how to equate these.

Answer 1

Let $a_n+b_n\sqrt{3}=(1+\sqrt{3})^{2n}$ so $a_n-b_n\sqrt{3}=(1-\sqrt{3})^{2n}=(\sqrt{3}-1)^{2n}$. (Edit, to address @diracdeltafunk: since calculating $a_n,\,b_n$ in $a_n+b_n\sqrt{3}=(1+\sqrt{3})^{2n}$ for a specific $n$ only requires using $\sqrt{3}^2=3$, it also works if we globally replace $\sqrt{3}$ with $-\sqrt{3}$, so $a_n-b_n\sqrt{3}=(1-\sqrt{3})^{2n}=(\sqrt{3}-1)^{2n}$.)

So $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n}=2a_n$ is an integer greater than $(1+\sqrt{3})^{2n}$, but only slightly because $|\sqrt{3}-1|<1$. Therefore, it's $\lceil(1+\sqrt{3})^{2n}\rceil=\lfloor(1+\sqrt{3})^{2n}+1\rfloor$.

Answer 2

From that expression that you are describing you can discover that the LHS is indeed an integer. To show that $\lfloor x\rfloor =y$ you have to show that $y\leq x<y+1.$ notice that $\sqrt{3}-1\leq 1$ hence $(\sqrt{3}-1)^n\leq 1\leq 1+(\sqrt{3}-1)^n.$

Answer 3

Let $a = (1 + \sqrt{3})^{2n} + 1$ and $b = (\sqrt{3} - 1)^{2n} - 1$.

First, $$(1 + \sqrt{3})^{2n} + (\sqrt{3} - 1)^{2n} = a + b,$$

and we want to show that this equals $$\lfloor (1 + \sqrt{3})^{2n} + 1 \rfloor = \lfloor a \rfloor,$$ or in other words that $a+b$ is the largest integer less than or equal to $a$. Equivalently, we must show three things:

1. $a+b$ is an integer
2. $b \leq 0$ (so that $a+b \leq a$)
3. $b > -1$ (so that there cannot be any integers strictly between $a+b$ and $a$)

Each of these is not too hard! (see the other excellent answers here for hints)