<s>[INST] Proving irreducibility for $\frac{a^2+b^2}{ac+bd}$ 
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.
 [/INST] 
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,
$$ (a^2+b^2, ac+bd)=1. \blacksquare$$
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