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Small linear relations in unbalanced diophantine equations from primitive Pythagorean triples

$r$ is parameter. Pick coprime $m,n\in[r,2r]$ with $mn$ even. Consider the Linear Diophantine Equation $$a^4u+b^4v+c^2z=0$$ where $a=m^2-n^2$, $b=2mn$ and $c=m^2+n^2$.

  1. Is it true that there are constants $$\alpha,\beta,\gamma,\delta>0$$ such that $$|u|,|v|<\alpha r^2\implies|z|>\beta r^6$$ $$|z|<\gamma r^6\implies|u|+|v|>\delta r^2$$ holds?

I think above is true for the following reason:

$a^4u+b^4v\bmod c^2$ seems to admit enough room to get $|u|,|v|>c>r^2$. Then since $a^4|u|,b^4|v|>r^{10}$ then it seems $r^6$ should be the lower bound for $|z|$. How to show this formally is unclear to me.

I tried playing with $a^4=m^8-4m^6n^2+6m^4n^4-4m^2n^6+n^8$ and $b^4=16m^2n^2$ and $c^2=m^4+2m^2n^2+n^4$. I can't seem to nail down enough relations to make a formal proof as done in Small linear relations between primitive Pythagorean triples $\mathsf{II}$.

One relation I found was $u=1$, $2v=m^4+n^4$ and $z=-c^2$ but is not very illuminating.

  1. In general is there algebraic methods to recover formal relations which might help looking at basis for the integral complement in null space so that lattice methods could be utilized as done in Small linear relations between primitive Pythagorean triples $\mathsf{II}$?
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