Mongolian Mathematical Olympiad

If we substitute $a_n=2x_n-1$ in the $2x_n-1=2(2x_{n-1}{x}_{n-2}-x_{n-1}-x_{n-2}+1)-1=(2x_{n-1}-1)(2x_{n-2}-1)$ we get $a_n=a_{n-1}{a}_{n-2}$. Initial conditions of the recurrence relation are $a_1=2\cdot 2-1=3$, $a_0=2a-1$. It is required that $a_0=2a-1$ must be a perfect square. Since $a_0=(2k+1{)}^2=4k^2+4k+1=2a-1$, we get $a=2k^2+2k+1$.

Now let us prove that if $\forall k\in \mathbb{Z}:a=2k^2+2k+1$ then $\forall n\in \mathbb{N}$, $2x_{3n}-1$ are perfect squares.

Moreover it is possible to prove for $\forall n\in \mathbb{N}$ terms $a_{3n}$ are perfect squares and terms $a_{3n+1}$, $a_{3n+2}$ are of the form $3s^2$, $s\in \mathbb{Z}$.

Let's proceed by induction. Base of induction is trivial.

If $a_{3n}=k^2$ and $a_{3n+1}=3m^2$ then $a_{3n+2}=k^2\cdot 3m^2=3(km{)}^2$.

If $a_{3n+1}=3k^2$ and $a_{3n+2}=3m^2$ then $a_{3n+3}=(3km{)}^2$.

If $a_{3n+2}=3k^2$ and $a_{3n+3}=m^2$ then $a_{3n+4}=3(km{)}^2$.

Thus we conclude for the numbers $a=2k^2+2k+1$, $\forall k\in \mathbb{Z}$ all terms of the progression with the form $a_{3n}=2x_{3n}-1$ are perfect squares.