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Given small parameters $0<\epsilon<\epsilon'$ is there an $n_\epsilon>0$ such that at every $n>n_\epsilon$ if we are given a prime $n^2<p<2n^2$ then can we always find integers $a,b,c,d,e,f$ with $a,b$ coprime and $n^{1+\epsilon}<a,b<2n^{1+\epsilon}$ with $a$ even and $b,e,f$ odd, $p/2<c,d<p$ and $p^{\epsilon'}<e,f<2p^{\epsilon'}$ such that $$p|(ad-bc)\mbox{ and }p|(cf-de)$$ holds?

$$\mbox{Note }ad\equiv bc\bmod p,\quad cf\equiv de\bmod p\implies cd^{-1}\equiv ab^{-1}\equiv ef^{-1}\bmod p\mbox{ holds}.$$

If $e,f$ odd are fixed with $p^{\epsilon'}<e,f<2p^{\epsilon'}$ then there are $\frac{\frac{n^{1+\epsilon}}2n^{1+\epsilon}}{\zeta(2)}=\frac{n^{2(1+\epsilon)}}{2\zeta(2)}$ choices of coprime $a,b$ with $a$ even and so at least one of these choices should give right $a,b$ with $ab^{-1}\equiv ef^{-1}\bmod p$ (we should expect $\frac{n^{2\epsilon}}{2\zeta(2)}$ choices of coprime $a,b$ with $a$ even since $p$ is of size $O(n^2)$). A similar argument holds for $c,d$.

Do such sextuples really exist? The argument indicates each of the $O(p^{2\epsilon'})$ different pairs of $e,f$ have at least one $a,b,c,d$ associated with them.

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    $\begingroup$ Exponential sum arguments can detect whether $cd^{-1}$ is congruent to any fixed reduced residue class modulo $p$, such as $ef^{-1}$ for fixed $e,f$; and the number of choices for the pair $(c,d)$ is $\gg p^2$, so you should get an asymptotic formula. ... $\endgroup$ Feb 23, 2018 at 16:45
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    $\begingroup$ ... Then, for fixed $c,d$, use the same method to detect pairs $(a,b)$ such that $ab^{-1}$ is congruent to the (now fixed) $cd^{-1}$ modulo $p$; the number of potential pairs $ab^{-1}$ is $\gg p^{1+\varepsilon}$, so I believe the method still works. Restricting $a$ to be even is easy with exponential sums, and one can add the additional condition $(a,b)=1$ by including an extra sum of $\mu(d)$ over $d\mid(a,b)$. $\endgroup$ Feb 23, 2018 at 16:46
  • $\begingroup$ @GregMartin could you post a full answer particularly for existence of $a,b$? existence for $c,d$ should follow. $\endgroup$
    – Turbo
    Feb 23, 2018 at 18:53
  • $\begingroup$ @GregMartin it does not exist. I think I should close this problem. $\endgroup$
    – Turbo
    Feb 23, 2018 at 20:18

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