From 1a26030f239714f4c3bf3ead418688661d725f61 Mon Sep 17 00:00:00 2001
From: Pietro Monticone <38562595+pitmonticone@users.noreply.github.com>
Date: Mon, 15 Apr 2024 15:06:20 +0200
Subject: [PATCH] update

---
 blueprint/src/chapters/0-introduction.tex |  2 +-
 blueprint/src/chapters/2-case2.tex        | 10 +++++-----
 2 files changed, 6 insertions(+), 6 deletions(-)

diff --git a/blueprint/src/chapters/0-introduction.tex b/blueprint/src/chapters/0-introduction.tex
index 30fb22c..ab75b4c 100644
--- a/blueprint/src/chapters/0-introduction.tex
+++ b/blueprint/src/chapters/0-introduction.tex
@@ -68,7 +68,7 @@ \chapter{Introduction}
     Since $\eta$ corresponds to a third root of unity, we have that $\eta^3 = 1$,
     which implies that $\eta^2 = \eta^{-1}$.
     Since $\eta$ corresponds to a root of the equation $x^2 + x + 1 = 0$, then $\eta^2 = -1 - \eta$.
-    Substituting back, we have that $\lambda^2 = (-1 - \eta) - 2\eta + 1 = -3\eta$.
+    Substituting back, we have that $\lambda^2 = (-1 - \eta) - 2\eta + 1 = s-3\eta$.
 \end{proof}
 
 \begin{theorem}
diff --git a/blueprint/src/chapters/2-case2.tex b/blueprint/src/chapters/2-case2.tex
index 1859a62..9bfdfdb 100644
--- a/blueprint/src/chapters/2-case2.tex
+++ b/blueprint/src/chapters/2-case2.tex
@@ -13,7 +13,7 @@ \chapter{Case 2}
 \end{lemma}
 \begin{proof}
     \leanok
-    Since $3 \divides a^3 + b^3 = c^3$, then $3 \divides c$. \\
+    Since $3 \divides a^3 + b^3 = c^3$, then $3 \divides c$.
     Then $3 \divides gcd(a,b,c)$.
 \end{proof}
 
@@ -29,7 +29,7 @@ \chapter{Case 2}
 \end{lemma}
 \begin{proof}
     \leanok
-    Since $3 \divides c^3 - a^3 = b^3$, then $3 \divides b$. \\
+    Since $3 \divides c^3 - a^3 = b^3$, then $3 \divides b$.
     Then $3 \divides gcd(a,b,c)$.
 \end{proof}
 
@@ -44,7 +44,7 @@ \chapter{Case 2}
 \end{lemma}
 \begin{proof}
     \leanok
-    Since $3 \divides c^3 - b^3 = a^3$, then $3 \divides a$. \\
+    Since $3 \divides c^3 - b^3 = a^3$, then $3 \divides a$.
     Then $3 \divides gcd(a,b,c)$.
 \end{proof}
 
@@ -184,10 +184,10 @@ \chapter{Case 2}
     lmm:lambda_not_dvd_two}
     Since $\lambda \notdivides a$, then
     $\lambda^4 \divides a^3 - 1 \lor \lambda^4 \divides a^3 + 1$ by
-    \Cref{lmm:lambda_pow_four_dvd_cube_sub_one_or_add_one_of_lambda_not_dvd}. \\
+    \Cref{lmm:lambda_pow_four_dvd_cube_sub_one_or_add_one_of_lambda_not_dvd}.
     Since $\lambda \notdivides b$, then
     $\lambda^4 \divides b^3 - 1 \lor \lambda^4 \divides b^3 + 1$ by
-    \Cref{lmm:lambda_pow_four_dvd_cube_sub_one_or_add_one_of_lambda_not_dvd}. \\
+    \Cref{lmm:lambda_pow_four_dvd_cube_sub_one_or_add_one_of_lambda_not_dvd}.
     We proceed by analysing each case:
     \begin{itemize}
         \item Case $\lambda^4 \divides a^3 - 1 \land \lambda^4 \divides b^3 - 1$.