<s>[INST] Find the quotient and the remainder of $(n^6-7)/(n^2+1)$ 
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.

So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
 [/INST] We have $n^6-7 = (n^2+1)q(n)+an+b$, with $a,b \in \mathbb Z$, because $n^2+1$ is monic.
Plugging $n=\pm i$, we get
$$
-8 = \pm ai+b
$$
which implies $a=0$ and $b=-8$. Therefore,
$$
q(n) = \frac{n^6-7+8}{n^2+1} = \frac{n^6+1}{n^2+1} = n^4 - n^2+ 1
$$
Thus,
$$
n^6-7 = (n^2+1)(n^4 - n^2+ 1)-8
$$
</s>