Autumn Tournament

Since $a$ and $b$ are co-prime, so are $b$ and $a^{n+1000}$, respectively, what is requested is equivalent to $b$ dividing $(n + 2)a^2 - (n + 1)a - n$ for each $n$.

From $n = 1$ and $n = 2$ we get that necessarily $b$ divides $3a^2 - 2a - 1$ and $4a^2 - 3a - 2$. Hence $b$ divides
$$4(3a^2 - 2a - 1) - 3(4a^2 - 3a - 2) = a + 2$$
and since $3a^2 - 2a - 1 = 3(a - 2)(a + 2) - 2(a + 2) + 15$, then necessarily $b$ divides $15$.

Clearly $b > a \ge 1$, i.e. $b \ge 2$. If $b = 15$, then $b \mid a + 2$ gives $a = 13$, but $4 \cdot 13^2 - 3 \cdot 13 - 2 = 635$ is not divisible by $3$. If $b = 3$, then $a = 1$, but $4 \cdot 1^2 - 3 \cdot 1 - 2 = -1$ is not divisible by $3$.

It remains $b = 5$, respectively $a = 3$. Indeed, $(n + 2) \cdot 3^2 - (n + 1) \cdot 3^1 - n = 5(n + 3)$ is divisible by $5$. $\square$