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.
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$\begingroup$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 $\rightarrow a = 1$.
$\endgroup$ 2 $\begingroup$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}
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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