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.
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 $\rightarrow a = 1$.