main.tex

	\sage{positive_radius_condition.subs([q_value_expr,beta_value_expr]).factor()}
\end{equation*}

\noindent
Then we have:

\begin{equation*}
	d - \frac{(\aa r + 2\bb)\aa}{2n^2}
	\geq \epsilon_q^2 \geq \epsilon_q^1 > 0
\end{equation*}

Where $\epsilon_q^1$ and $\epsilon_q^2$ are defined as follows:

\begin{equation*}
	\epsilon_q^1 :=
	\frac{k_q^1}{2mn^2}
	\qquad
	\epsilon_q^2 :=
	\frac{k_q^2}{2mn^2}
\end{equation*}
\begin{align*}
	\text{where }
	&k_q^1 \text{ is the least }
	k\in\ZZ_{>0}\: s.t.:\:
	k \equiv -\aa\bb m \mod n
\\
	&k_q^2 \text{ is the least }
	k\in\ZZ_{>0}\: s.t.:\:
	k \equiv \aa\bb m (\aa\aa^{'}-2)
	\mod n\gcd(2n,\aa^2 m)
\end{align*}
	
\end{lemmadfn}

It's worth noting that $\epsilon_q^2$ is potentially larger than $\epsilon_q^2$
but calculating it involves a $\gcd$, a modulo reduction, and a modulo $n$
inverse, for each $q$ considered.

\begin{proof}

Consider the following tautology:

\begin{align}
	&\frac{ x }{ m }
	- \frac{
		(\aa r+2\bb)\aa
	}{
		2n^2
	}
	= \frac{ k }{ 2mn^2 }
	\quad \text{for some } x \in \ZZ, k \in \ZZ_{>0}
	\label{eqn:finding_better_eps_problem}
\\ &\iff
	- (\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 least $k$ satisfying 
eqn \ref{eqn:finding_better_eps_problem}.
Since such a $k$ must also satisfy eqn \ref{eqn:better_eps_problem_k_mod_n},
we can pick the smallest $k_q^1 \in \ZZ_{>0}$ which satisfies this new condition
(a computation only depending on $q$ and $\beta$, but not $r$).
We are then guaranteed that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_q^1$.
Furthermore, $k$ also satisfies
eqn \ref{eqn:better_eps_problem_k_mod_gcd2n2_a2mn}
so we can also pick the smallest $k_q^2 \in \ZZ_{>0}$ satisfying this condition,
which also guarantees that the gap $\frac{k}{2mn^2}$ is at least $\epsilon_q^2$.

\end{proof}




\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}