# Proving an expression is perfect square

I have this expression I got in one larger exercise: $$\frac{(2+\sqrt3)^{2n+1}+(2-\sqrt3)^{2n+1}-4}{6}\frac{(2+\sqrt3)^{2(n+1)+1}+(2-\sqrt3)^{2(n+1)+1}-4}{6}+1$$ and i need to prove it is perfect square. I tried many different approaches but couldn't find way to show it is square. Interesting fact is $(2+\sqrt3)(2-\sqrt3)=1$ so I tried replacing $(2+\sqrt3)=x$ and $(2-\sqrt3)=1/x$ to see if I would get an idea.

Alternative form I got after some steps and using equality giving $1$ I got: $$\frac{(2+\sqrt3)^{4n+4}(1-16((2- \sqrt3)^{2n+2})+66((2- \sqrt3)^{2n+2})^2-16((2- \sqrt3)^{2n+2})^3+((2- \sqrt3)^{2n+2})^4)}{36}$$ which is interesting as I have "rising exponent" but this coefficients doesn't make sense to me. Any ideas?

EDIT: I actually need to prove $$(y_{n+1}^2-1)(y_{n+2}^2-1)+1$$ is perfect square where $y_0=1$, $y_1=3$ and $y_{n}=4y_{n-1}-y_{n-2}$. Expression from above I got solving this recursion and using exact expression for $y_n$. I also tried solving directly using induction and this recursion but didn't get result.

Let $y_n=4y_{n-1}-y_{n-2}$ with $y_0=1$ and $y_1=3$.

First we show by induction that $$y_n^2+y_{n+1}^2-4y_ny_{n+1}+2=0 \; \forall n \in \mathbb{N}. \label{*}\tag{*}$$ Clearly it holds for $n=0$ and \begin{align*} & y_{n+1}^2+y_{n+2}^2-4y_{n+1}y_{n+2}+2 =\\ & (y_{n+2}-y_{n+1})^2-2y_{n+1}y_{n+2}+2 =\\ & (3y_{n+1}-y_n)^2-2y_{n+1}y_{n+2}+2 =\\ & y_n^2+y_{n+1}^2-4y_ny_{n+1}+2+8y_{n+1}^2-2y_ny_{n+1}-2y_{n+1}y_{n+2} =\\ & 8y_{n+1}^2-2y_{n+1}(y_n+y_{n+2}) =\\ & 8y_{n+1}^2-2y_{n+1}4y_{n+1} = 0. \end{align*}

Finally $\eqref{*}$ implies the desired result: \begin{align*} -y_n^2-y_{n+1}^2 & = -4y_ny_{n+1} + 2 \\ y_n^2y_{n+1}^2-y_n^2-y_{n+1}^2+2 &= y_n^2y_{n+1}^2-4y_ny_{n+1} + 4 \\ (y_n^2-1)(y_{n+1}^2-1)+1 & = (y_ny_{n+1}-2)^2. \end{align*}

Let $a_n=(2+\sqrt 3)^n+(2-\sqrt 3)^n$.

One can prove the followings (see here for the details):

• $a_n=4a_{n-1}-a_{n-2}$.

• $a_{2n+1}a_{2n+3}=a_{4n+4}+14.$

• $a_{4n+4}={a_{n+1}}^4-4{a_{n+1}}^2+2.$

• $a_{2n+1}+a_{2n+3}=4{a_{n+1}}^2-8$.

• ${a_n}^2=12b_n+4$ where $b_n$ is an integer.

Then, we have \begin{align}\frac{a_{2n+1}-4}{6}\cdot\frac{a_{2n+3}-4}{6}+1&=\frac{a_{2n+1}a_{2n+3}-4(a_{2n+1}+a_{2n+3})+16}{36}+1\\&=\frac{(a_{4n+4}+14)-4(4{a_{n+1}}^2-8)+16}{36}+1\\&=\frac{({a_{n+1}}^4-4{a_{n+1}}^2+16)-4(4{a_{n+1}}^2-8)+16}{36}+1\\&=\frac{{a_{n+1}}^4-20{a_{n+1}}^2+64}{36}+1\\&=\frac{(12b_{n+1}+4)^2-20(12b_{n+1}+4)+100}{36}\\&=\frac{144{b_{n+1}}^2-144b_{n+1}+36}{36}\\&=(2b_{n+1}-1)^2.\end{align}