On the board, there are three numbers $$\large x_0 = -\sqrt 2 – 1, y_0 = \sqrt 2, z_0 = -\sqrt 2 + 1$$
. Every minute, a computer is programmed to erase the numbers previously written and write on the board numbers $$\large \begin{align} x_n &= y_{n – 1}^2 + y_{n-1}z_{n-1} + z_{n-1}^2\\ y_n &= z_{n – 1}^2 – \sqrt 2z_{n-1}x_{n-1} + x_{n-1}^2\\ z_n &= x_{n – 1}^2 – x_{n-1}y_{n-1} + y_{n-1}^2 \end{align}$$
Is there a moment where $x_n, y_n, z_n \in \mathbb Q$?
If there is, calculate the minimum value of $n$.
If there isn't, explain why.
The answer is probably no. (I feel sorry for the computer, it got overheated.) But I don't really know how to establish that.
Best Answer
$x_n,y_n$ and $z_n$ are never rational as is shown below.
Obviously, all $x_n,y_n,z_n$ are in $\mathbb Z[\sqrt2]$. We simply calculate modulo $2\mathbb Z[\sqrt2]$. Then we claim
Lemma For all $n$, $x_n\equiv z_n\equiv 1+\sqrt2$ and $y_n\equiv \sqrt2$ modulo 2.
Proof: We proceed by induction. For $n=0$, this is true. Now assume that it is true for some $n$, $n\geq0$. Then we calculate $$x_n^2\equiv z_n^2\equiv x_nz_n\equiv 1, y_n^2\equiv0, x_ny_n\equiv z_ny_n\equiv \sqrt2.$$ This yields $$\begin{matrix}x_{n+1}\equiv0+\sqrt2+1\equiv1+\sqrt2,\\ y_{n+1}\equiv1+\sqrt2+1\equiv\sqrt2\end{matrix},\\ z_{n+1}\equiv 1+\sqrt2+0\equiv 1+\sqrt2.$$ Hence the statement is also true for $n+1$.
The principle of mathematical induction now gives the Lemma.