2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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*}

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.