$r$ 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 $b^4=16m^4n^4$ 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 https://mathoverflow.net/questions/357444/small-linear-relations-between-primitive-pythagorean-triples-mathsfii.

The relations I found gave following basis for solution space to $a^4u+b^4v+c^2z=0$:
$$v_1=(u,v,z)=(2m^2n^2,m^4+n^4,-2m^2n^2(m^4+2m^2n^2+n^4))=(2m^2n^2,m^4+n^4,-2m^2n^2(m^2+n^2)^2)$$
$$v_2=(u,v,z)=(8m^2n^2,3(m^4+n^4)-2m^2n^2,-8m^2n^2(m^4+n^4))=(8m^2n^2,2(m^4+n^4)+(m^2-n^2)^2,-8m^2n^2(m^4+n^4)).$$


2. In general is there algebraic methods to recover formal relations that guarantee reduced basis for $2$ and $3$ dimensional cases which will help looking for the full integral complement in null space so that lattice methods could be utilized as done in https://mathoverflow.net/questions/357444/small-linear-relations-between-primitive-pythagorean-triples-mathsfii?

[Lenstra-Lenstra-Lovasz](https://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm) suffices for 2.. However I think it will be an overkill here. Perhaps there is an algebraic technique?