main.tex

	- (\aa r+2\bb)\aa m
	\equiv k
	\quad \mod 2n^2
	\quad \text{for some } k \in \ZZ_{>0}
\\ &\iff
	- \aa^2 m r - 2\aa\bb m
	\equiv k
	\quad \mod 2n^2
	\quad \text{for some } k \in \ZZ_{>0}
\\ &\Longrightarrow
  \aa^2 \aa^{'}\bb m - 2\aa\bb m
	\equiv k
	\quad \mod \gcd(2n^2, \aa^2 mn)
	\quad \text{for some } k \in \ZZ_{>0}
	\label{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
\\ &\Longrightarrow
  -\aa\bb m
	\equiv k
	\quad \mod n
	\quad \text{for some } k \in \ZZ_{>0}
	\label{eqn:better_eps_problem_k_mod_n}
\end{align}

In our situation, we want to find the gap between the right-hand-side of eqn
\ref{eqn:positive_rad_condition_in_terms_of_q_beta},
and the least element of $\frac{1}{m}\ZZ$ which is strictly greater.
This amounts to finding the least $k \in \ZZ_{>0}$ for which 
eqn \ref{eqn:finding_better_eps_problem} holds.
Since such a $k$ satisfies eqn \ref{eqn:better_eps_problem_k_mod_n},
we can pick the smallest $k \in \ZZ_{>0}$ which satisfies this new condition
(a computation only depending on $q$ and $\beta$, but not $r$)
and we are then guaranteed that the gap is at least $\frac{k}{2mn^2}$.

A potentially larger gap can also be guaranteed if we choose the least
$k \in \ZZ_{>0}$ satisfying eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
instead, but at the cost of computing several $\gcd$'s and modulo reductions
for each $q$ considered.





\minorheading{Irrational $\beta$}

\egroup % end scope where beta redefined to beta_{-}

\section{Conclusion}

\newpage
\section{Appendix - SageMath code}

\usemintedstyle{tango}

\begin{footnotesize}
\inputminted[
	obeytabs=true,
	tabsize=2,
	breaklines=true,
	breakbefore=./
]{python}{filtered_sage.txt}
\end{footnotesize}

\end{document}