proving that given fraction is irreducible.

Question

Prove that for every natural number n, fraction $\frac{21n+4}{14n+3}$ is irreducible. I deduced that if we can prove that numerator and denominator have 1 as their GCD, we can get the result, but I cannot get it from thereon.

Answer 1

Going by Euclidean algorithm: $$ GCD(21n+4, 14n+3) = GCD(7n+1, 14n+3) = GCD(7n+1, 1) = 1$$

Answer 2

Since $$3(14n+3)-2(21n+4)=1$$ we have that GCD$(14n+3,21n+4)$ divides $1$, then GCD$(14n+3,21n+4)=1$

Answer 3

$21n+4=(14n+3)+7n+1$

$14n+3=2(7n+1)+1$