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$r$ 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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$.

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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$?

Lenstra-Lenstra-Lovasz suffices for 2.. However I think it will be an overkill here. Perhaps there is an algebraic technique?

$r$ 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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$.

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

It is unclear that if these are the shortest basis. It is not clear from this how to prove 1. even though these basis satisfy 1.

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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$?

Lenstra-Lenstra-Lovasz suffices for 2.. However I think it will be an overkill here. Perhaps there is an algebraic technique?

$r$ 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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$.

The relations I found gave following basis for solution space to $a^4u+b^4v+c^2z=0$: $$v_1=(u,v,z)=(2m^2n^2,m^4+n^4,-(2m^2n^2)(m^4+2m^2n^2+n^4))$$ $$v_2=(u,v,z)=(8m^2n^2,3(m^4+n^4)-2m^2n^2,- 8m^2n^2).$$ These are not very illuminating. It is not clear if these constitute the reduced basis.

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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$?

Lenstra-Lenstra-Lovasz suffices for 2.. However I think it will be an overkill here. Perhaps there is an algebraic technique?

$r$ 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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$.

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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$?

Lenstra-Lenstra-Lovasz suffices for 2.. However I think it will be an overkill here. Perhaps there is an algebraic technique?

$r$ 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 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.

2. 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 $\mathsf{II}$?

$r$ 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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$.

The relations I found gave following basis for solution space to $a^4u+b^4v+c^2z=0$: $$v_1=(u,v,z)=(2m^2n^2,m^4+n^4,-(2m^2n^2)(m^4+2m^2n^2+n^4))$$ $$v_2=(u,v,z)=(8m^2n^2,3(m^4+n^4)-2m^2n^2,- 8m^2n^2).$$ These are not very illuminating. It is not clear if these constitute the reduced basis.

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 Small linear relations between primitive Pythagorean triples $\mathsf{II}$?

Lenstra-Lenstra-Lovasz suffices for 2.. However I think it will be an overkill here. Perhaps there is an algebraic technique?