VN-TST 2003 Problem 3

Let $f(0, 0) = 5^{2003}, f(0, n) = 0$ for every integer $n \neq 0$ and $$\begin{array}{c}\ f(m, n) = f(m-1, n) - 2 \cdot \Bigg\lfloor \frac{f(m-1, n)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n-1)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n+1)}{2}\Bigg\rfloor \end{array}$$ for every natural number $m > 0$ and for every integer $n$ .

Prove that there exists a positive integer $M$ such that $f(M, n) = 1$ for all integers $n$ such that $|n| \leq \frac{(5^{2003}-1)}{2}$ and $f(M, n) = 0$ for all integers n such that $|n| > \frac{5^{2003}-1}{2}.$