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We have three mutually coprime integers $r,t,M$ where $M\asymp K^{\frac12-2\epsilon}$ and $r,t\asymp K^{\frac14+\epsilon}$ holds with some fixed $\epsilon>0$ and $K>0$ is a large parameter. Assume $M^2-4rt$ is a prime if need be.

  1. What is the largest integer $T$ (preferably prime) with $$2\sqrt{rt}+M\equiv0\bmod T?$$

  2. Is there a fast method to find the largest $T$ allowable and can it be much larger than $K$ (preferably order $\Omega(K^\alpha)$ at arbitrary $\alpha\geq1$)?

Preferably the method should be in $e^{o(\log K)}$ time. Ideally it should be $O(polylog(K))$.

  1. Also for every possible can $\zeta_r,\zeta_t\asymp T^{\frac12+\beta}$ hold where $\zeta_r^2\equiv r\bmod T$ and $\zeta_t^2\equiv t\bmod T$ while $\beta>0$ is very small and is fixed?
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  • $\begingroup$ Is there $T\gg M^2+4rt$ here? $\endgroup$
    – Turbo
    Commented Aug 24, 2019 at 15:36
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    $\begingroup$ Is your first sentence a question or a statement? It sounds like a statement (a condition) but it ends with a question mark. If it is a question, it should read "Do we have three mutually etc." In Question 2 which integer do you ask about: $r$ or $t$ or $M$ or $T$? Please try to ask more clearly. $\endgroup$
    – GH from MO
    Commented Aug 24, 2019 at 21:09
  • $\begingroup$ Corrected. In 2. I am trying to find largest $T$ possible and it appears $M^2-4rt$ will work however I want something much larger of order $K^2$. $\endgroup$
    – Turbo
    Commented Aug 25, 2019 at 0:00
  • $\begingroup$ It is not clear what you mean by the congruence, since $\sqrt{rt}$ is not an integer. Note also that, in analytic number theory, the symbol $\gg$ means "larger than a positive constant times". So $K\gg 0$ makes little sense (it is equivalent to $K>0$); you probably meant that $K$ is a large parameter. Finally, I inserted $T$ in the first question. $\endgroup$
    – GH from MO
    Commented Aug 25, 2019 at 2:11
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    $\begingroup$ Ninth version today, not to mention four self-deletions & undeletions. $\endgroup$
    – Gerry Myerson
    Commented Aug 25, 2019 at 12:17

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