1 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 $5$. $\square$