Bijection between $\mathbb{Z}^2$ and bounded sequences - Miklos Schweitzer 
Let $\alpha \leq-2 $  be an integer. Prove that for every pair $\beta_{0},\beta_{1}$
   of integers there exists a uniquely determined sequence 
  $0 \leq q_{0},...,q_{k}<\alpha^2-\alpha$ of integers, such that $q_{k}\neq0$ if
  $(\beta_{0},\beta_{1})\neq(0,0)$ and 
$\beta_{i}=\sum\limits_{j=0}^k q_{j}(\alpha-i)^j$ for $i=0,1$.



This question is from Miklos Schweitzer 2001.
My idea was to show that there are unique polynomials $P_1,P_2 \in \mathbb{Z}[x]$ such that:
$P(x)=-\beta_{1}(x-\alpha) \displaystyle \frac{P_1(x)}{P_1(\alpha-1)}-\beta_{0}(x-\alpha-1)\displaystyle \frac{P_2(x)}{P_2(\alpha)}$  
and all coefficients of $P(x)$ are in the interval $[0,\alpha^2-\alpha).$  
However I couldn't go any further.
 A: A reformulation of the problem is :
Given a pair $(\beta_1,\beta_2)$, we have to show there is a unique polynomial $P$ with coefficients in $0 \ldots \alpha^2-\alpha-1$ sych that $P(\alpha) = \beta_0$ and $P(\alpha-1) = \beta_1$.
Consider the constant coefficient $q_0$ of such a polynomial.
We have $\beta_0 = q_0 \pmod \alpha$ and $\beta_1 = q_0 \pmod {\alpha-1}$.
Since $\alpha$ and $\alpha-1$ are coprime, by the Chinese Remainder Theorem,
the value of $q_0$ modulo $\alpha(\alpha-1)$ is uniquely determined by $\beta_0 \pmod \alpha$ and $\beta_1 \pmod {\alpha-1}$. Since the integers from $0$ to $\alpha^2-\alpha-1$ form a complete set of representatives, there is indeed only one possible choice for $q_0$. 
Let $q_0$ be that integer. Then we need $(P(\alpha-i)-q_i)/(\alpha-i) = (\beta_i-q_i)/(\alpha-i) = \lceil \beta_i/(\alpha-i) \rceil$. So we are left with having to prove that there is a unique polynomial having new values $\lceil \beta_i/(\alpha-i) \rceil$ at $\alpha-i$.
Let $f : \Bbb Z \to \Bbb Z$ defined by $f(x,y) = (\lceil x/\alpha \rceil, \lceil y/(\alpha-1) \rceil)$.
If $a \le -2$, it is easy to check that $|\lceil b/a \rceil| < |b|$ except when $b=0$ or $b=-1$ in which case we get to $b= 1$ and then to $b=0$ on nthe next iteration.
Therefore, by repeatedly applying $f$ to any $(\beta_0,\beta_1) \in \Bbb Z^2$, we will reach $(0,0)$ in a finite number of steps.
So, we know that there is a unique polynomial getting values $\beta_i$ at $\alpha-i$ if and only if there is a unique polynomial having values $0$ at $\alpha-i$.
But from $(0,0)$ we get that $q_0$ has to be $0$ and so on : every $q_i$ has to be $0$. 
Thus the zero polynomial is the only polynomial having a value of $0$ at $\alpha-i$, which concludes the proof.
