China Hong Kong Mathematical Olympiad

a.
Observe that $f(k) = k^2 - (k+1)^2 - (k+2)^2 + (k+3)^2 = 4$ for any integer $k$.

* If $c = b + 4m$ for some $m \in \mathbb{Z}$, then $c = b + \sum_{k=1}^{m} f(4k - 3)$.
* If $c = b + 4m + 1$ for some $m \in \mathbb{Z}$, then $c = b + 1^2 + \sum_{k=1}^{m} f(4k - 2)$.
* If $c = b + 4m + 2$ for some $m \in \mathbb{Z}$, then $c = b - 1^2 - 2^2 - 3^2 + 4^2 + \sum_{k=1}^{m} f(4k + 1)$.
* If $c = b + 4m + 3$ for some $m \in \mathbb{Z}$, then $c = b - 1^2 + 2^2 + \sum_{k=1}^{m} f(4k - 1)$.

Therefore, in any case we can write $c$ as $b \pm 1^2 \pm 2^2 \pm \cdots \pm n^2$ for some $n$ and some choices of the signs. This clearly solves the problem.

b.
The answer is $19$.

Firstly, since $1^2 + 2^2 + \cdots + 17^2 = \frac{17 \cdot 18 \cdot 35}{6} = 1785 < 2012$, we have $n \ge 18$.

Secondly, when $n = 18$, $\pm 1^2 \pm 2^2 \pm \cdots \pm 18^2 \equiv 1 + 0 + \cdots + 1 + 0 \equiv 1 \pmod{2}$. This shows $b$ and $c$ cannot have the same parity, which is a contradiction.

Lastly, it remains to provide a construction for $n = 19$. Since
$$1^2 + 2^2 + \cdots + 19^2 = \frac{19 \cdot 20 \cdot 39}{6} = 2470 = 2012 + 2 \times 229 = 2012 + 2(2^2 + 15^2),$$
we can change the signs of $2^2$ and $15^2$ to negative, so that the new sum becomes $2012$. This gives a corresponding quadratic sequence that satisfies the requirement.