Idea behind proof that $\frac{21n+4}{14n+3}$ is irreducible for all $n$

I have a proof of $\frac{21n+4}{14n+3}$ being an irreducible fraction for any positive integer $n$. The proof starts with the following: Indeed, from the equality: $$2(21n+4)-3(14n+3)=-1$$ and so on..But how does this equality follow from the fraction?

Just work out the formula, and see it holds. It does not follow from the fraction.

It just shows that $\gcd(21n+4, 14n+3) = 1$, so the fraction is irreducible.

Quick proof: suppose that $p$ is a prime that divides $21n+4$ and $14n+3$. Then $p$ also divides $2(21n+4) - 3(14n+3) = -1$ which cannot be.

The diophantine equation $$2p-3q=-1$$ has solutions exactly when $-1$ is a multiple of $\gcd(p,q)$, that is $\gcd(21n+4,14n+3)=1$. But as you can see, the equation holds for all integers $n$, since $$2(21n+4)-3(14n+3)=42n+8-42n-9=-1.$$ Thus, we must have $\gcd(21n+4,14n+3)=1$.

To prove the fraction is irreducible we need to show $\,\gcd(3\cdot 7n+4,\,2\cdot 7n+3) = 1.\,$ By Bezout it suffices to find $\,a,b\,$ such that $\, a(\color{#c00}3\cdot 7n+4) + b(\color{#0a0}2\cdot 7n+3) = 1.\,$ A natural first try is to choose $\,a=\color{#0a0}{-2},\,b=\color{#c00}3\,$ so to eliminate $n,\,$ which yields $\,4a+3b = -8+9 = 1,\,$ as sought

You can do euclidean algorithm... I'm starting off with bigger $n$...

So

$21n+4=1(14n+3)+(7n+1) $ since only one 14 in 21.

$14n+3=2(7n+1)+1 $ since two 7's in 14.

So now we work backwards...

$(14n+3)-2(7n+1)=1$ Then replace $(7n+1)$ with $(21n+4)-1(14n+3)$

$(14n+3)-2[(21n+4)-(14n+3)]=1$

$3(14n+3)-2(21n+4)=1$ Then you could multipy both sides by -1 giving you $2(21n+4)-3(14n+3)=-1$

But this might be overkill.