From ff1b012b41867a453de076301b2a0b814cb5dffb Mon Sep 17 00:00:00 2001
From: Luke Naylor <l.naylor@sms.ed.ac.uk>
Date: Fri, 12 May 2023 18:22:40 +0100
Subject: [PATCH] Rearrange and correct work from last 4 commits
---
main.tex | 24 ++++++++++++++++--------
1 file changed, 16 insertions(+), 8 deletions(-)
diff --git a/main.tex b/main.tex
index 93741a6..70ab47f 100644
--- a/main.tex
+++ b/main.tex
@@ -807,22 +807,25 @@ beta_value_expr = (beta == a/n)
\noindent
That is, $r \equiv -a^{-1}b$ mod $n$ ($a$ is coprime to $n$, and so invertible mod $n$).
+
Substituting the current values of $q$ and $\beta$ into the condition for the
radius of the pseudo-wall being positive
(eqn \ref{eqn:positive_rad_d_bound_betamin}) we get:
\begin{equation}
- \frac{1}{\lcm(m,2n^2)}\ZZ
+\label{eqn:positive_rad_condition_in_terms_of_q_beta}
+ \frac{1}{m}\ZZ
\ni
\sage{positive_radius_condition.subs([q_value_expr,beta_value_expr]).factor()}
\in
\frac{1}{2n^2}\ZZ
\end{equation}
-For each $r$, the smallest element of $\frac{1}{\lcm(m,2n^2)}\ZZ$ strictly larger
-than the lower bound here is exactly $\frac{1}{\lcm(m,2n^2)}$ greater.
-Therefore, if any of the two upper bounds come to within
-$\frac{1}{\lcm(m,2n^2)}$ of this lower bound, then there are no solutions for $d$.
+\noindent
+Both $d$ and the lower bound are elements of $\frac{1}{\lcm(m,2n^2)}\ZZ$.
+So, if any of the two upper bounds on $d$ come to within
+$\frac{1}{\lcm(m,2n^2)}$ of this lower bound, then there are no solutions for
+$d$.
Considering equations
\ref{eqn:bgmlv2_d_bound_betamin},
@@ -835,9 +838,11 @@ this happens when:
\sage{bgmlv2_d_upperbound_exp_term},
\sage{bgmlv3_d_upperbound_exp_term_alt.subs(chbv==0)},
\right)
- < \frac{1}{\lcm(m,2n^2)}
+ < \epsilon := \frac{1}{\lcm(m,2n^2)}
\end{equation}
+%% refinements using specific values of q and beta
+
\begin{sagesilent}
rhs_numerator = (
positive_radius_condition
@@ -849,7 +854,8 @@ rhs_numerator = (
\end{sagesilent}
\noindent
-Considering the numerator of the right-hand-side:
+Considering the numerator of the right-hand-side of
+(eqn \ref{eqn:positive_rad_condition_in_terms_of_q_beta}):
\begin{align}
\sage{rhs_numerator}
@@ -858,7 +864,9 @@ Considering the numerator of the right-hand-side:
&\equiv ab &\mod n
\end{align}
-And so, we also have $a(ar+2b) \equiv ab$ (mod $\lcm(m,2n^2)$).
+\noindent
+And so, we also have $a(ar+2b) \equiv ab$ (mod $2n^2$).
+
\minorheading{Irrational $\beta$}
--
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