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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.

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Lê Thành Đạt
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  • Thanh Dat: What have you done so far? Sound like an olympic type problem that is a particular case of a more advanced theory. Try to find the topic and book or article that the problem belongs to before going into the tedious calculation. – DeepSea Sep 03 '19 at 17:39
  • I agree with DeepSea. Where does the problem come from? If we write $x_n = a_n + b_n \sqrt 2$, $y_n = c_n + d_n \sqrt 2$, $z_n = e_n + f_n \sqrt 2$, then it looks like $a_n, b_n, c_n, d_n, e_n, f_n$ grow very fast for $n \ge 3$ (and therefore the answer to the question is indeed no). – Luca Bressan Sep 04 '19 at 18:35

1 Answers1

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$x_n,y_n$ 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.

Helmut
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