A Pell's equation hidden in Problem 29 052

 In translation:

                     "29052.   Let $(x_n)_{n\geqslant 1}$ be the string with $x_n=[(3+2\sqrt{2})^n],\;n\geqslant 1$.

                   Prove that for any $n\in \mathbb{N}^*$, the number  $\frac{(x_n-1)(x_n+3)}{8}$  is a perfect square.

                      ($[\alpha]$ represents the integer part of the real number $\alpha$.)

Mihai Pîrvulescu, Calafat"

ANSWER CiP

$$\frac{(x_n-1)(x_n+3)}{8}=B_n^2 \tag{1}$$

                         where $B_n$ is defined by  

$B_1=2,\;B_2=12,\;\;B_{n+2}=6\cdot B_{n+1}-B_n,\;n\geqslant 1. \tag{2}$

                    Solution CiP

               Let

$(3+2\sqrt{2})^n=A_n+B_n\cdot \sqrt{2},\;n\geqslant 1. \tag{3}$

We have $A_1=3,\;B_1=2$, and since $(3+2\sqrt{2})^2=17+12\sqrt{2}$, we get $A_2=17,\;B_2=12.$ Examining the expansion with Binomial formula of $(3\pm2\sqrt{2})^n$, it results that

$(3-2\sqrt{2})^n=A_n-B_n\sqrt {2}. \tag{4}$

Moreover

$(3+2\sqrt{2})^n+(3-2\sqrt{2})^n\;\;\overset{(3),(4)}{=}\;2\cdot A_n. \tag{5}$

          Writing $(3+2\sqrt{2})^{n+1}=(3+2\sqrt{2})^n\cdot(3+2\sqrt{2})$, we have from (3)

$A_{n+1}+\sqrt{2}B_{n+1}=(A_n+\sqrt{2}B_n) \cdot (3+2\sqrt{2})=(3A_n+4B_n)+\sqrt{2}(2A_n+3B_n),$ so

\begin{cases} A_{n+1}=3A_n+4B_n\;\;\;\;\;\;\;\;\;\;\;\;(6) \\B_{n+1}=2A_n+3B_n\;\;\;\;\;\;\;\;\;\;\;\;(7)\end{cases}

Using relations (6), (7) several times we have

$B_{n+2}\;\overset{(7)}{\underset{n\to n+1}{=}}\;2A_{n+1}+3B_{n+1}\overset{(6)}{=}2(3A_n+4B_n)+3B_{n+1}=3B_{n+1}+8B_n+6A_n=$

$\overset{(7)}{\underset{\overbrace{2A_n=B_{n+1}-3B_n}}{=}}\;\;3B_{n+1}+8B_n+3(B_{n+1}-3B_n)=6B_{n+1}-B_n.$

So the sequence $(B_n)_{n \geqslant 1}$ checks the relations (2).

          Let us now prove that $(x,y)=(A_n,B_n)$ verifies Pell's equation

$x^2-2y^2=1. \tag{8}$

 Obviously $(A_1,B_1)=(3,2)$ verifies (8). Then assuming

$A_n^2-2B_n^2=1 \tag{9}$

we have $A_{n+1}^2-2B_{n+1}^2\overset{(6)}{\underset{(7)}{=}}(3A_n+4B_n)^2-2(2A_n+3B_n)^2=(9A_n^2+24A_nB_n+16B_n^2)-$

$-2(4A_n^2+12A_nB_n+9B_n^2)=A_n^2-2B_n^2\overset{(9)}{=}1,$

so (9) is also true for  $n+1$, so by the Principle of Mathematical Induction, (9) is valid for all $n\in \mathbb{N}^*$.

          Finally $(3+2\sqrt{2})^n\overset{(3)}{=}A_n+(\sqrt{2}B_n)\overset{(4)}{=}A_n+\left ( A_n-(3-2\sqrt{2})^n \right )=2A_n-(3-2\sqrt{2})^n=$

$$=(2A_n-1)+\left ( 1-(3-2\sqrt{2})^n \right ),$$

and since $0<1-(3-2\sqrt{2})^n<1$, for $n=1,\;2,\dots$, it results

$$x_n=[(3+2\sqrt{2})^n]=2A_n-1. \tag{10}$$

Then  $\frac{(x_n-1)(x_n+3)}{8}\overset{(10)}{=}\frac{(2A_n-2)(2A_n+2)}{8}=\frac{A_n^2-1}{2}\overset{(9)}{=}\frac{(2B_n^2+1)-1}{2}=B_n^2.$

QED $\blacksquare$