remove comments in markdown

Bhaskar1312 · Bhaskar1312 · commit 55f35b4ce412 · 2025-04-18T22:22:11.000+05:30
diff --git a/src/others/pells_equation.md b/src/others/pells_equation.md
@@ -21,14 +21,6 @@ The first part $( x + \sqrt{d} \cdot y )$ is always greater than 1. And the seco
 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} \cdot \sqrt{d} $
 
-[//]: # (We claim that every solution to the equation is given by $ &#40;x_{0} +  y_{0} \cdot \sqrt{d}&#41;^{n} $ for some integer $ n $. The idea is that any other solution)
-[//]: # (  $ &#40; x + \sqrt{d} \cdot y &#41;$ must be of this form, meaning that all solutions are generated by repeated multiplication of the smallest solution. Here we use Brahmagupta's identity of quadratic decomposition.)
-[//]: # (   $ &#40;a^{2} - n \cdot b^{2}&#41; \cdot &#40;c^{2} - n \cdot d^{2}&#41; = &#40;a \cdot c + n \cdot b \cdot d&#41;^{2} - n \cdot &#40;a \cdot d + b \cdot c&#41;^{2}$)
-[//]: # (   )
-[//]: # (Here, $a = x_{1}$, $b = y_{1}$, $c = x_{2}$, $d = y_{2}$)
-[//]: # (   So if &#40;x_{1}, y_{1}&#41; and &#40;x_{2}, y_{2}&#41; are solutions, then $&#40;x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2}, x_{1} \cdot y_{2} + y_{1} \cdot x_{2}&#41;$ is also a solution.)
-[//]: # (First, we prove that product of two solutions is also a solution.)
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 We use method of descent to prove it. suppose there is a solution $u + v \cdot \sqrt{d}$ such that $u^{2} - d \cdot v^{2} = 1 $
 Therefore, $ ( x_{0} + \sqrt{d} \cdot y_{0} )^{n}  < u + v \cdot \sqrt{d} < ( x_{0} + \sqrt{d} \cdot y_{0} )^{n+1} $
 
