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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 $a^4u+b^4v+c^2z=0$: $$v_1=(u,v,z)=(2m^2n^2,m^4+n^4,-2m^2n^2(m^4+2m^2n^2+n^4))=(2m^2n^2,m^4+n^4,-2m^2n^2(m^2+n^2)^2)$$ $$v_2=(u,v,z)=(8m^2n^2,3(m^4+n^4)-2m^2n^2,-8m^2n^2(m^4+n^4))=(8m^2n^2,2(m^4+n^4)+(m^2-n^2)^2,-8m^2n^2(m^4+n^4)).$$

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.

  1. 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?

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  • $\begingroup$ Whatever the reduced basis should be is not clear to me. $\endgroup$
    – VS.
    Apr 17 '20 at 4:15
  • $\begingroup$ $(3v_1-v_2)/(2m^2n^2)$ gives a shorter basis $(-1,1,(m^4+n^4-6m^2n^2))$. $\endgroup$
    – VS.
    Apr 17 '20 at 23:00
  • $\begingroup$ $(3v_1-v_2)/(2m^2n^2)$ is $(-1,1,a^2-b^2)$. $\endgroup$
    – VS.
    Apr 18 '20 at 3:49
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Consider the Linear Diophantine Equation $$a^{2t}u+b^{2t}v+c^2z=0$$ where $t\geq2$.

There is always a $(u,v,z)\neq(0,0,0)$ solution with $\|(u,v,z)\|_\infty=O(r^{4(t-1)})$ since $c^2|(a^{2t}-b^{2t})$ and we can take $(u,v,z)=(-1,1,\frac{a^{2t}-b^{2t}}{c^2})=(-1,1,\frac{a^{2t}-b^{2t}}{a^2+b^2})$.

Thus for $a^{4}u+b^{4}v+c^2z=0$ there is a solution with $|u|,|v|=1$ and $|z|\leq 64r^4$ if $m,n\in\mathbb Z$ are coprime in $[r,2r]$ with $mn$ even and $a=m^2-n^2$, $b=2mn$ and $c=m^2+n^2$.

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