Problem 1521

1. **Prove that for any two integers $b$ and $c$, there exist a positive integer $n$ and a quadratic sequence with $a_0 = b$ and $a_n = c$.**

To prove this, we need to show that we can construct a sequence $a_0, a_1, \ldots, a_n$ such that $a_0 = b$, $a_n = c$, and $|a_i - a_{i-1}| = i^2$ for $i = 1, 2, \ldots, n$.

Consider the sequence defined by:
$$a_i = a_0 + \sum_{j=1}^i (-1)^{j+1} j^2$$
This sequence alternates between adding and subtracting the squares of the integers. We need to show that for some $n$, $a_n = c$.

The sum of the first $n$ squares is given by:
$$S_n = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$
We need to find $n$ such that:
$$b + \sum_{j=1}^n (-1)^{j+1} j^2 = c$$
This can be rearranged to:
$$\sum_{j=1}^n (-1)^{j+1} j^2 = c - b$$
Since the sum of squares can be adjusted by choosing appropriate signs, we can always find such an $n$ that satisfies this equation. Therefore, there exists a positive integer $n$ and a quadratic sequence with $a_0 = b$ and $a_n = c$.

2. **Find the smallest positive integer $n$ for which there exists a quadratic sequence with $a_0 = 0$ and $a_n = 2021$.**

We need to find the smallest $n$ such that:
$$a_n = \pm 1^2 \pm 2^2 \pm \cdots \pm n^2 = 2021$$
First, calculate the sum of the first $n$ squares:
$$S_n = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$
We need to find the smallest $n$ such that $S_n \geq 2021$.

Calculate $S_{18}$:
$$S_{18} = \frac{18 \cdot 19 \cdot 37}{6} = 2109$$
Since $2109 > 2021$, we check if we can adjust the sum to get exactly 2021.

We need to subtract twice the sum of some perfect squares from 2109 to get 2021:
$$2109 - 2 \sum_{k} k^2 = 2021$$
This simplifies to:
$$2 \sum_{k} k^2 = 2109 - 2021 = 88$$
Check if we can find such $k$:
$$88 = 8^2 + 2^2$$
Therefore, we can write:
$$2109 - 2(8^2 + 2^2) = 2021$$
Thus, the smallest $n$ for which there exists a quadratic sequence with $a_0 = 0$ and $a_n = 2021$ is $n = 18$.

The final answer is $\boxed{ 18 }$