Showing that $\frac{21n+4}{14n+3}$ is irreducible for every natural number n.

Question

I wanted to prove that the fraction is irreducible using induction and have written the following proof:

Let's take $n=1$, then $\frac{21\cdot1+4}{14\cdot1+3}=\frac{24}{17}$ which is not divisible. assuming that for $n=k$ it is also not divisible. So now we will show that it is not divisible for $n=k+1$. $$\frac{21(k+1)+4}{14\cdot(k+1)+3}=\frac{21k+21+4}{14k+14+3}=\frac{21k+25}{14k+17}$$ we have assumed that the fraction is not true for n=k and it is also not true for $n=k+1$.

Is this a concrete proof of induction?

P.s I know the Euclidean algorithm proof but wanted to try it with induction.

Answer 1

Hint : $$3 \times (14n+3)-2 \times (21n+4)=1$$

Now, Bezout is the man.