Brazilian Mathematical Olympiad

Let $p(x) = q(x)(x-a)(x-b) + mx + n$. So putting $x = a, b$ we get $p(a) = ma + n$, $p(b) = mb + n$. Solving, $m = \frac{p(a)-p(b)}{a-b}$, $n = \frac{p(b)a-p(a)b}{a-b}$.

In the case given $a = 2$, $b = -1$, so $m = \frac{2^{200}-1}{3}$, $n = \frac{2^{200}+2}{3}$. Note that $2 \equiv -1 \pmod{3}$, so $2^{200} \equiv 1 \pmod{3}$ and hence $2^{200} - 1$ is a multiple of $3$, so $m$ is an integer. $n = m + 1$, so $n$ is also an integer.