<s>[INST] Proving the fraction $\frac{n^3 + 2n}{n^4 + 3n^2 + 1}$ is irreducible for $n \in \mathbb{N}$ 
Prove that the fraction $\dfrac{n^3 + 2n}{n^4 + 3n^2 + 1}$ is in lowest terms for every positive integer $n$.

I just don't know how to solve this. I tried polynomial division, expressing the gdc of the two terms as a linear combination, and factorizing the sum of the two terms but nothing really leads anywhere. I'd really appreciate some help.
 [/INST] You can do it with linear combinations:
$$
(n^4+3n^2+1)-n(n^3+2n)=n^2+1 \tag{1}
$$
Then
$$
(n^3+2n)-n(n^2+1)=n \tag{2}
$$
and
$$
(n^2+1)-n\cdot n=1 \tag{3}
$$
Just work backwards if you need the linear combination, but this is already sufficient.
\begin{align}
1
&=\color{red}{(n^2+1)}-n\cdot\color{red}{n}
&&\text{by (3)}\\
&=\color{red}{(n^2+1)}-
  n\cdot\bigl(\color{green}{(n^3+2n)}-n\color{red}{(n^2+1)}\bigr) 
&&\text{by (2)} \\
&=(-n)\color{green}{(n^3+2n)}+(n^2+1)\color{red}{(n^2+1)} 
&&\text{reorder}\\
&=(-n)\color{green}{(n^3+2n)}+
  (n^2+1)\bigl(\color{green}{(n^4+3n^2+1)}-n\color{green}{(n^3+2n)}\bigr) 
&&\text{by (1)}\\
&=(n^2+1)\color{green}{(n^4+3n^2+1)}+(-n^3-2n)\color{green}{(n^3+2n)}
&&\text{reorder}
\end{align}
</s>