2019 Putnam B5

Let $F_m$ be the $m$th Fibonacci number, defined by $F_1 = F_2 = 1$ and $F_m = F_{m-1} + F_{m-2}$ for all $m \geq 3$.

Let $p(x)$ be the polynomial of degree $1008$ such that $p(2n+1) = F_{2n+1}$ for $n=0,1,2,\dots,1008$. Find integers $j$ and $k$ such that $p(2019) = F_j - F_k$.

令 $F_m$ 为第 $m$ 个斐波那契数，对于所有 $m \geq 3$，由 $F_1 = F_2 = 1$ 和 $F_m = F_{m-1} + F_{m-2}$ 定义。

令 $p(x)$ 为 $1008$ 次多项式，使得 $p(2n+1) = F_{2n+1}$ 对于 $n=0,1,2,\dots,1008$。查找整数 $j$ 和 $k$，使得 $p(2019) = F_j - F_k$。