Pythagorean quadruple generators with a gcd relation

Question

For non-negative integers $m,n,q,p$ with $\gcd(m,n,q,p)=1$, assume we have:

$$\gcd(mq+np,b)=|nq-mp|$$ for some integer $$b<mq+np$$ and that $$8\nmid\,mq+np,$$ $$m+n+p+q\equiv 1\mod 2.$$

Can we show that $b$ always has solutions of either $m^2+n^2$ or $p^2+q^2$?

Answer 1

For $(m,n,q,p)=(5,4,4,3)$ we have $gcd(m,n,q,p)=1$ and $$ gcd(32,b)=gcd(mq+np,b)=nq-mp=1. $$ Hence every odd $b$ is a solution, and not only $m^2+n^2=41$, or $p^2+q^2=25$.