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.
 A: Let $(n^3 + 2n \:, n^4 + 3n^2 + 1) = a$ 
$$\color {green} {n^3 + 2n = n( n^2 + 2)} \: \:, \: \color {red} {n^4 + 3n^2 + 1 = n^2(n^2+2) + (n^2 + 1)}$$
$$a \mid (n^2 + 2) \implies a \mid (n^2 + 1) \: \: \:, \text{or} \: \: a \mid n \implies a \mid n^2 \implies a \mid (n^2 + 1)$$
Consecutive integers are coprime $\rightarrow a = 1$. 
A: 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}
A: Put  $\,f(x) = x^2\!+3x+1,\ $ and $\ n^k\!-a = n^2\!+2\ $ below
Lemma $\ $ Suppose that  $\,f(x)\,$ is a polynomial with integer coefficients and $f(0)=\pm1 = f(a).\,$ Then $\,f(n^k)\,$
 is coprime to $\, n(n^k-a)$ 
Proof $\,\ {\rm mod}\,\ n\!:\,\ n\equiv 0\,\Rightarrow\, f(n^k)\equiv  f(0)\equiv  \pm1 $
$\ \ \ {\rm mod}\,\ n^k\!-a\!:\ n^k\equiv a\,\Rightarrow\, f(n^k)\equiv f(a)\equiv \pm1 $
Thus $\,f(n^k)\,$ is coprime to $\,n\,$ and $\,n^k\!-a\,$ hence also to their product, by Euclid.
A: You can use the fact that g.c.d of the fraction equals 1 and to do so use Euclide Algorithm.We have$n^4+3n^2+1=(n^3+2n)n+n^2+1$ thus $g.c.d(n^4+3n^2+1,n^3+2n)$= $g.c.d(n^3+2n,n^2+1)$
$n^3+2n=(n^2+1)n+n$ then $g.c.d(n^3+2n,n^2+1)$= $g.c.d(n^2+1,n)$= $g.c.d(n,1)=1$
Hence the fraction in lowest terms.
A: using the Euclidean algorithm multiple times.
(n^4+3n^2+1)=n(n^3+2n)+(n^2+1)
(n^3+2n)=n(n^2+1)+n
(n^2+1)=n(n)+1
n=n(1)+0
Thus gcd =1 so the fraction is in lowest term.
