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

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.

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.

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.

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