Prove that $\frac{(2+{\sqrt 3})^{2k-1}+(2-{\sqrt 3})^{2k-1}}{2}-1$ always results in a square number

Question

Given the expression

$$\frac{(2+{\sqrt 3})^{2k-1}+(2-{\sqrt 3})^{2k-1}}{2}-1$$

prove that for positive integer k, this expression results in a perfect square.

My attempt: I tried to prove this by induction. The base step is easy to check, assuming the expression is true for $k=n$, I have no idea how to proceed with $k=n+1$. Any help would be appreciated.

Answer 1

Notice that $$(2+\sqrt{3})^{2k-1}=a+b\sqrt{3}\tag{1}$$ for suitable integers $a,b$ (for any $x,y \in \mathbb{Z}[\sqrt{3}]$ we have $xy \in \mathbb{Z}[\sqrt{3}]$, i.e. it is closed under multiplication). Similarly for conjugate, we get $$(2-\sqrt{3})^{2k-1}=a-b\sqrt{3}\tag{2}.$$ Adding the two, we can verify that $a-1$ is the number we need to show to be a perfect square.

Applying binomial theorem on $(2+\sqrt{3})^{2k-1}$, we get $$ a=\sum_{i=0}^{k-1}\binom{2k-1}{2i}2^{2k-2i-1}3^i. $$ Thus $a \equiv 0 \bmod 2$, and so $a-1$ is odd. Similarly, $a \equiv 2^{2k-1} \equiv 2 \bmod 3$, and so $3 \mid a+1$.

Finally, multiplying out $(1)$ and $(2)$, we see $1=a^2-3b^2$. But then $$ (a-1)\left(\frac{a+1}{3}\right)=b^2 \tag{3}, $$ and since $a-1$ is odd, $a-1$ and $a+1$ are coprime. Then $(3)$ implies $a-1$ is a perfect square.

Answer 2

Note $$I=\frac{(2+{\sqrt 3})^{2k-1}+(2-{\sqrt 3})^{2k-1}}{2}-1 =\left(\frac{(2+{\sqrt 3})^{k}}{\sqrt3+1}- \frac{(2-{\sqrt 3})^{k}}{\sqrt3-1} \right)^2 $$

So, it suffices to prove

\begin{align} & S_k = \frac{(2+{\sqrt 3})^{k}}{\sqrt3+1}- \frac{(2-{\sqrt 3})^{k}}{\sqrt3-1}= n\\ & C_k = \frac{(2+{\sqrt 3})^{k}}{\sqrt3+1}+ \frac{(2-{\sqrt 3})^{k}}{\sqrt3-1}=m\sqrt3 \end{align}

where $m$ and $n$ are integers. Then, by induction with $S_1=1$ and $C_1=\sqrt3$

\begin{align} &S_{k+1} = 2S_k +\sqrt3C_k = 2n+ 3m\\ &C_{k+1} = 2C_k +\sqrt3S_k = (2m+n)\sqrt3 \end{align}

Thus, $I$ is a perfect square.

Answer 3

$$\dfrac{(2+{\sqrt 3})^{2k-1}+(2-{\sqrt 3})^{2k-1}}{2}-1$$ $$=\dfrac{(2+{\sqrt 3})^{2k-1}+\left(\frac{1}{2+{\sqrt 3}}\right)^{2k-1}-2}{2}$$ $$=\dfrac{\left((2+{\sqrt 3})^{2k-1}\right)^2-2(2+{\sqrt 3})^{2k-1}+1}{2(2+{\sqrt 3})^{2k-1}}$$ $$=\dfrac{\left((2+{\sqrt 3})^{2k-1}-1\right)^2}{2(2+{\sqrt 3})^{2k-1}}$$ $$=\left(\dfrac{(2+{\sqrt 3})^{2k-1}-1}{\sqrt2(2+{\sqrt 3})^{k-\frac12}}\right)^2$$ $$=\left(\dfrac{(2+{\sqrt 3})^{2k-1}-1}{\frac{2(2+{\sqrt 3})^{k}}{\large \sqrt2\cdot \sqrt{2+\sqrt3}}}\right)^2$$ $$=\left(\dfrac{(\sqrt{4+2\sqrt3})((2+{\sqrt 3})^{2k-1}-1)}{2(2+{\sqrt 3})^{k}}\right)^2$$ $$=\left(\dfrac{(\sqrt3+1)\left((2+{\sqrt 3})^{k-1}-(2-\sqrt3)^k\right)}{2}\right)^2$$ Above is always a square number.