PR review comments

Bhaskar1312 · Bhaskar1312 · commit eea90be52cbc · 2025-10-20T18:30:05.000+05:30
diff --git a/scripts/install-mkdocs.sh b/scripts/install-mkdocs.sh
@@ -9,4 +9,4 @@ pip install \
     mkdocs-git-revision-date-localized-plugin \
     mkdocs-simple-hooks \
     mkdocs-rss-plugin \
-    mkdocs-git-committers-plugin-2
+    plugins/mkdocs-git-committers-plugin-2
diff --git a/src/others/pells_equation.md b/src/others/pells_equation.md
@@ -22,16 +22,21 @@ This factorization is important because it shows the connection to quadratic irr
 
 The norm of an expression $u + v \sqrt{d}$ is defined as $N(u + v \sqrt{d}) = (u + v \sqrt{d})(u - v \sqrt{d}) = u^2 - d v^2$. The norm is multiplicative: $N(ab) = N(a)N(b)$. This property is crucial in the descent argument below.
 
-We will prove that all solutions to Pell's equation are given by powers of the smallest positive solution. Let's assume it to be $x_0 + y_0 \sqrt{d}$, where $x_0 > 1$ is the smallest possible value for $x$.
+We will prove that all solutions to Pell's equation are given by powers of the smallest non-trivial solution. Let's assume it to be the minimum possible $x_0 + y_0 \sqrt{d} > 1$. For such a solution, it also holds that $y_0 > 0$, and its $x_0 > 1$ is the smallest possible value for $x$.
 
 ## Method of Descent
 Suppose there is a solution $u + v \cdot \sqrt{d}$ such that $u^{2} - d \cdot v^{2} = 1$ and is not a power of  $( x_{0} + \sqrt{d} \cdot y_{0} )$
 Then it must lie between two powers of $( x_{0} + \sqrt{d} \cdot y_{0} )$. 
-i.e, For some n, $( x_{0} + \sqrt{d} \cdot y_{0} )^{n}  < u + v \cdot \sqrt{d} < ( x_{0} + \sqrt{d} \cdot y_{0} )^{n+1}$
+i.e, For some $n$,
+$$
+( x_{0} + \sqrt{d} \cdot y_{0} )^{n}  < u + v \cdot \sqrt{d} < ( x_{0} + \sqrt{d} \cdot y_{0} )^{n+1}
+$$
 
-Multiplying the above inequality by $( x_{0} - \sqrt{d} \cdot y_{0} )^{n}$,(which is > 0 and < 1) we get
+Multiplying the above inequality by $( x_{0} - \sqrt{d} \cdot y_{0} )^{n}$,(which is $> 0$ and $< 1$) we get
 
-$1 < (u + v \cdot \sqrt{d})( x_{0} - \sqrt{d} \cdot y_{0} )^{n} < ( x_{0} + \sqrt{d} \cdot y_{0} )$
+$$
+1 < (u + v \cdot \sqrt{d})( x_{0} - \sqrt{d} \cdot y_{0} )^{n} < ( x_{0} + \sqrt{d} \cdot y_{0} )
+$$
 Because both $(u + v \cdot \sqrt{d})$ and $( x_{0} - \sqrt{d} \cdot y_{0} )^{n}$ have norm 1, their product is also a solution.
 But this contradicts our assumption that $( x_{0} + \sqrt{d} \cdot y_{0} )$ is the smallest solution. Therefore, there is no solution between $( x_{0} + \sqrt{d} \cdot y_{0} )^{n}$ and $( x_{0} + \sqrt{d} \cdot y_{0} )^{n+1}$.
 
