Prove that the elements of these two sequences are not null Let $x_{n+1}=x_n+2y_n$ and $y_{n+1}=y_n-x_n$, where $x_1=1$ and $y_1=-1$.
I tried proving by contradiction, I tried by induction, I got nothing.
This is a question I had on an exam, I didn't manage to solve it, and afterwards I spent one day thinking about it and still came up with nothing.
Please note that a full solution is not necessary, if you could just provide a hint, that would be awesome.
The origin of the question was to show that $\left( \begin{array}{ccc}
1 & 2 \\
-1 & 1 \\
\end{array} \right)^n$ has no zero elements for any positive integer $n$.
 A: The pair $(x_n, y_n)^T$ is equal to 
$$
\begin{pmatrix}
1 & 2 \\
-1 & 1
\end{pmatrix}^{n-1}
\begin{pmatrix}
1 \\ -1
\end{pmatrix}.
$$
If you want to prove that there is no $n$ s.t. $x_n=y_n=0$, then you just use that fact that a regular matrix never maps a nonzero vector to zero. If you want to prove that for any $n$ neither $x_n$ nor $y_n$ is zero, do the eigenvalue decomposition and compute $x_n, y_n$ exactly.
A: Let $A=
\left( {\begin{array}{*{20}c}
   1 & 2  \\
   { - 1} & 1  \\
\end{array}} \right)
$. Let try us to diagonalize this matirx, we have to find first the eigenvalue of $A$.
\begin{align}
\left| {\lambda I - A} \right| = \left| {\begin{array}{*{20}c}
   {\lambda  - 1} & { - 2}  \\
   1 & {\lambda  - 1}  \\
\end{array}} \right| = \left( {\lambda  - 1} \right)^2  + 2 = \lambda ^2  - 2\lambda  + 3 \ne 0\,\,\left( {\text{for}\,\,\text{all}\,\,\text{real}\,\,\lambda } \right)
\end{align}
so that no "real" eigenvalues and thus the matrix $A$ is not diagonalizable; i.e., there is NO a non-singular matirx $P$ such that $PAP^{-1}=D$ (D:=diagonal matirx), or we write $A=P^{-1}DP$
A: Hint: look at their remainders $\pmod p$ for some small primes $p$.
