Methods to prove $\frac{21n + 4}{14n + 3}$ is irreducible for every natural number $n$

gcd(21n+4, 14n+3) = gcd(14n+3, 7n+1) = gcd(7n+2, 7n+1) = gcd(7n+1, 1) = 1

Using repeated usage of Euclidean algorithm.

We want to show that $21n+4$ and $14n+3$ are relatively prime. Note that $$(3)(14n+3)-(2)(21n+4)=1.$$ So any common divisor of $21n+4$ and $14n+3$ divides $1$.

Remark: I am not able to construct an argument based on the post. The remainder when $(21n+4)+(14n+3)$ is divided by $7$ is $0$, I would not use DNE, since $0$ is a perfectly respectable remainder. But working modulo $7$ is not enough, we want to rule out all common divisors greater than $1$.