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

Question

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?

Answer 1

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.

Answer 2

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 $\gcd(21n+4,14n+3)=1$.

Answer 3

To prove the fraction is irreducible we need to show $\,d:= \gcd(3\cdot 7n+4,\,2\cdot 7n+3) = 1.\,$ By Bezout it suffices to find $\,a,b\,$ such that $\, a(\color{#0a0}3\cdot 7n+4) + b(\color{#c00}2\cdot 7n+3) = 1.\:$ A natural choice is to eliminate $\,n\,$ by choosing $\,a=\color{#c00}{-2},\,b=\color{#0a0}3\,$ yielding $\,4\color{c00}a\!+\!3\:\!\color{0a0}b = 4(\color{#c00}{-2})\!+\!2(\color{#0a0}3) = 1,\,$ as sought.

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Below are hints/sketches for a few more proofs, by elimination and Euclid's algorithm, and linear algebra. First we repeat the above more concisely, then recast it into modular language.

$(1)\,\ \ d\mid 21n\!+\!4,14n\!+\!3\,\Rightarrow\,d\mid 3(14n\!+\!3)\!-\!2(21n\!+\!4) = 1\ $ by eliminating $\,n$

$\qquad\qquad\quad\!\! \begin{align} \llap{(2)\ \bmod d\!:\ \ \ \ }\color{#c00}2\,[\,21n&\equiv -4\,]\\[.1em] -\,\color{#0a0}3\,[\,14n&\equiv -3\,]\\[.2em] \hline \Longrightarrow\ \ \ \ 0&\equiv 1\\[.1em] \Longrightarrow\ \ \ \ \ \ &\!\!\!\:\!d\,\mid\, 1\end{align},\ \ $ get $\,\color{#c00}2,\color{#0a0}3\,$ by $\ \color{#c00}a[21] = \color{#0a0}b[14]\iff \dfrac{\color{#c00}a}{\color{#0a0}b} = \dfrac{14}{21} = \dfrac{\color{#c00}2}{\color{#0a0}3}$

$\qquad\qquad\quad\!\begin{align} \llap{(3)\ \bmod d\!:\qquad }[-4&\,\equiv\, 21n\,]\\[.1em] \div\, \ [\,-3&\,\equiv\, 14n\,]\\[.2em] \hline \Longrightarrow\ \ \frac{\color{#c00}4}{\color{#0a0}3}&\,\equiv\, \frac{21}{14}\equiv \frac{\color{#0a0}3}{\color{#c00}2}\end{align}\ \ \ $ so $\ \ \color{#c00}{4\cdot 2}\equiv \color{#0a0}{3\cdot 3}\,\Rightarrow\,0\equiv 1$

$(4)\ $ By Euclidean algorithm $\:(21n\!+\!4,14n\!+\!3)=(14n\!+\!3,7n\!+\!1)=(7n\!+\!1,1) = 1\ $

$(5)\ $ The linear map $\,(n,1)\mapsto (14n\!+\!3,21n\!+\!4)\,$ has $\, \det = 14\cdot 4-21\cdot 3=\color{#c00}{\bf 1},\,$ therefore, by this very simple theorem $\ \gcd(14n\!+\!3,21n\!+\!4)\mid \color{#c00}{\bf 1}\cdot \gcd(n,1)= 1$

Answer 4

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.