Proving irreducibility for $\frac{a^2+b^2}{ac+bd}$

Question

For positive integers $a,b,c,d$ satisfying $ad-bc=1,$ prove that $\frac{a^2+b^2}{ac+bd}$ is irreducible. That is, $\gcd(a^2+b^2, ac+bd)=1.$

I know how to prove $\frac{a+b}{c+d}$ is irreducible if $ad-bc$ with bezouts theorem, but am not sure how to prove $\frac{a^2+b^2}{ac+bd}$ is irreducible. I tried using proof by contradiction, writing $km=a^2+b^2$ and $kn=ac+bd$ and $\gcd(m,n)=1$ for $k>1.$ Then from the first equation I got $b=\sqrt{km-a^2}.$ Then plugging it into the second equation I got $kn=ac+(\sqrt{km-a^2})d.$ Then, $c=\tfrac{kn-d\sqrt{km-a^2}}{a}.$ So then $$ad-bc=ad-(\sqrt{km-a^2})\left(\frac{kn-d\sqrt{km-a^2}}{a}\right)=ad-\left(\frac{kn\sqrt{km-a^2}}{a}-\frac{dkm}{a}+ad\right).$$ Thus, $$ad-bc=\frac{dkm}{a}-\frac{kn\sqrt{km-a^2}}{a}=1.$$ This doesn't seem to lead anywhere useful, though. There are two many variables and there seems like no good way to prove that $k=1.$ Is there a better way then this? Possibly a way to prove irreducibility like proving $\frac{a+b}{c+d}$ is irreducible using bezouts theorem? Thanks in advance.

score 9 · Accepted Answer · answered Jul 02 '22 at 02:31

9

Prove that if $ad-bc=1$, $(a^2+b^2, ac+bd)=1.$

First, we can recall the Brahmagupta-Fibonacci identity. $$ (ad-bc)^2+(ac+bd)^2=(a^2+b^2)(c^2+d^2). $$

Then, it's easy to find that: $$ (ac+bd)^2+1=(a^2+b^2)(c^2+d^2). $$

So, $$ \Big((ac+bd)^2, a^2+b^2\Big)=1. $$

Therefore, $$ (a^2+b^2, ac+bd)=1. \blacksquare$$

answered Jul 02 '22 at 02:31

RDK

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Oh my, I can't believe I forgot about such a well known identity! Nice solution :) – mathisfun Jul 02 '22 at 02:35
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@mathisfun Thanks, it was lucky that I just saw this question while I was getting interested in the Brahmagupta-Fibonacci identity, as you can see from my profile. – RDK Jul 02 '22 at 02:40
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@mathisfun FYI, regarding that identity, you may wish to check this OP's fairly recent question & couple of answers at Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.. – John Omielan Jul 02 '22 at 02:40
@mathisfun This has a more intuitive form in terms of Gaussian integer arithmetic - see my answer. The above proof using the sum of squares corresponds to the proof taking norms in my answer (in fact a simple natural way to prove the sum of squares identity is via such norms). – Bill Dubuque Jul 02 '22 at 08:52

Bill Dubuque · Answer 2 · 2022-07-03T15:39:28.443

This has a natural proof using Gaussian integers $\,a+b\:\!i,\ a,b\in \Bbb Z,\,$ where it becomes a Gaussian integer analog of the integer case: $\ \,\bbox[5px,border:1px solid #c00]{\color{#0a0}ab = \color{#c00}n-1\ \,\Rightarrow\,\ (\color{#0a0}a,\color{#c00}n) \,=\, \color{darkorange}1}^{\phantom{|^|}}$

Lemma $\,\alpha = a\!+\!b\:\!i,\ \beta = c\!-\!d\:\!i,\ \bbox[5px,border:1px solid #c00]{\color{#0a0}\alpha\beta = \color{#c00}n-i\,\Rightarrow\, (\color{#0a0}{\alpha\bar \alpha},\,\color{#c00}n)=\color{darkorange}1}\,,\ \ \begin{align} \color{#0a0}{\alpha\bar\alpha} &= a^2+b^2\\ \color{#c00}n &= ac+bd\end{align}$

Proof $\, $ by taking the norm of: $\ \ \beta\alpha = n-i\: \Rightarrow\: \beta\bar\beta\color{#0a0}{\alpha\bar\alpha} = \color{#c00}n^2+\color{darkorange}1.\ \ \bf\small QED $

Or $\ \color{#c00}n\!-\!\color{#0a0}\alpha\beta = i\,$ is a unit, so $\,(\color{#0a0}\alpha,\color{#c00}n)\!=\!1.\,$ Conjugating $\Rightarrow (\bar\alpha,n)\! =\! 1,\,$ so $\,(\alpha\bar\alpha,n)\!=\!1\,$ by Euclid, which is a prototypical Gaussian example of coprimes (to $\color{#c00}n$) are closed under products.

We used (as in Bezout) $\ b\color{#0a0}a!-!c\color{#c00}n = u\Rightarrow (\color{#0a0}a,\color{#c00}n)=1,,$ if $,u,$ is a unit (i.e. $u\mid 1,,$ e.g. $, u = \pm1,,\pm i),$ since $,d:=( \color{#0a0}a,\color{#c00}n)\mid \color{#0a0}a,\color{#c00}n,\Rightarrow, d\mid u!=!b\color{#0a0}a!-!c\color{#c00}n,,$ so $, d\mid u\mid 1\ \ $ — Bill Dubuque, Jul 03 '22 at 15:43

Proving irreducibility for $\frac{a^2+b^2}{ac+bd}$

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