Let ($x_n$) be a sequence integers defined as such: For all $n>1$ $x_{n+2} =2x_{n+1}x_{n} - x_{n+1} - x_n +1$ , $x_0=a$ , $x_1=2$
Find all integers a such that for all $n>1$ the number $2x_{3n} -1$ is a square of a natural number.
Found this one in an old russian textbook for olympiad practice, couldn't solve it and the author didn't include the solution.
Let $y_n=2x_n-1$. Then $y_0=2a-1$, $y_1=5$, and $$y_{n+2}= y_{n+1}y_n.$$ The power $v_5(y_n)$ with which $5$ occurs in $y_n$ is then governed by a Fibonacci-like recursion $$v_5(y_{n+2})=v_5(y_{n+1})v_5(y_{n}) $$ where we only know $v_5(y_1)=1$ exactly and $v_5(y_0)=v_5(2a-1)$ is some finite integer. We conclude that $$v_5(y_n)=v_5(2a-1)\cdot F_{n-1} +F_n.\tag1$$ For all other primes $p\ne 5$, we find $$v_p(y_n)=v_p(2a-1)\cdot F_{n-1}.\tag2$$ We want that both $(1)$ and $(2)$ are even whenever $n$ is a multiple of $3$ and $>3$. One readily verifies, that for such $n$, $F_n$ is even and $F_{n-1}$ is odd. We conclude that the desired property holds iff $v_(2a-1)$ is even for all $p$, i.e., $$2a-1\text{ is a perfect square},$$ or in other words, $$a=\frac{(2k-1)^2+1}2=2k(k-1)+1$$ for some $k\in\Bbb N$.