to $n$, gives a Type I solution (note that $abd$ is automatically coprime to $n$, thanks to ). In the converse direction, all Type I solutions arise in this manner:

\[type-1\] Let $n \in {\mathbb{N}}$, and let $(x,y,z)$ be a Type I solution. Then there exists a unique $(a,b,c,d,e,f) \in {\mathbb{N}}^6 \cap \Sigma^{{\operatorname{I}}}_n$ with $abcd$ coprime to $n$ and $a,b,c$ having no common factor, such that $\pi^{{\operatorname{I}}}_n(a,b,c,d,e,f) = (x,y,z)$.

The uniqueness follows since $\pi^{{\operatorname{I}}}_n$ is injective after quotienting out by dilations. To show existence, we factor $x=ndx', y = dy', z=dz'$, where $x',y',z'$ are coprime, then after multiplying by $ndx'y'z'$ we have $$\label{dxy}
4dx'y'z' = y'z' + nx'y' + nx'z'.$$ As $y',z'$ are coprime to $n$, we conclude that $x'$ divides $y'z'$, $y'$ divides $x'z'$, and $z'$ divides $x'y'$. Splitting into prime factors, we conclude that $$\label{xabc}
 x' = ab, y' = ac, z' = bc$$ for some natural numbers $a,b,c$; since $x',y',z'$ have no common factor, $a,b,c$ have no common factor also. As $y,z$ were coprime to $n$, $abcd$ is coprime to $n$ also.

Substituting into we obtain , which in particular implies (as $c$ is coprime to $n$) that $c$ divides $a+b$. If we then set $e := 