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Copy file name to clipboardExpand all lines: src/others/pells_equation.md
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@@ -20,18 +20,19 @@ $x^{2} - d \cdot y^{2} = ( x + \sqrt{d} \cdot y ) ( x - \sqrt{d} \cdot y ) = 1$
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The first part $( x + \sqrt{d} \cdot y )$ is always greater than 1. And the second part $( x - \sqrt{d} \cdot y )$ is always less than 1.
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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}$.
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We will prove that all solutions to Pell's equation are given by powers of the smallest positive solution. Let's assume it to be
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$ x_{0} + y_{0} \cdot \sqrt{d} $
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[//]: #(We claim that every solution to the equation is given by $ (x_{0} + y_{0} \cdot \sqrt{d})^{n}$ for some integer $n$. The idea is that any other solution)
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[//]: #( $( x + \sqrt{d} \cdot y )$ 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.)
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[//]: #( $(a^{2} - n \cdot b^{2}) \cdot (c^{2} - n \cdot d^{2}) = (a \cdot c + n \cdot b \cdot d)^{2} - n \cdot (a \cdot d + b \cdot c)^{2}$)
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[//]: #(We claim that every solution to the equation is given by $ (x_{0} + y_{0} \cdot \sqrt{d})^{n}$ for some integer $ n $. The idea is that any other solution)
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[//]: #($ ( x + \sqrt{d} \cdot y )$ 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.)
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[//]: #( $(a^{2} - n \cdot b^{2}) \cdot (c^{2} - n \cdot d^{2}) = (a \cdot c + n \cdot b \cdot d)^{2} - n \cdot (a \cdot d + b \cdot c)^{2}$)
[//]: #( So if (x_{1}, y_{1}) and (x_{2}, y_{2}) are solutions, then $(x_{1}
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\cdot x_{2} + n \cdot y_{1} \cdot y_{2}, x_{1} \cdot y_{2} + y_{1} \cdot x_{2})$ is also a solution.)
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[//]: #(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 $u^{2} - d \cdot v^{2} = 1$
we get the continued fraction of $\sf$ as $[2; 1, 1, 4, 1, 1, 4, \ldots]$.
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It can also be calculated using only [integer based calculation](https://cp-algorithms.com/algebra/continued-fractions.html)
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### Finding the solution using Chakravala method
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The Chakravala method is an ancient Indian algorithm to solve Pell's equation. It is based on the Brahmagupta's identity of quadratic decomposition.
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The Chakravala method is an ancient Indian algorithm to solve Pell's equation. It is based on the Brahmagupta's identity of quadratic decomposition
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$(x_{1}^{2} - n \cdot y_{1}^{2}) \cdot (x_{2}^{2} - n \cdot y_{2}^{2}) = (x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2})^{2} - n \cdot (x_{1} \cdot y_{2} + x_{2} \cdot y_{1})^{2}$
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$(x_{1}^{2} - n \cdot y_{1}^{2}) \cdot (x_{2}^{2} - n \cdot y_{2}^{2}) = (x_{1} \cdot x_{2} - n \cdot y_{1} \cdot y_{2})^{2} - n \cdot (x_{1} \cdot y_{2} - x_{2} \cdot y_{1})^{2}$
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And Bhaskara's Lemma:
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If $x^{2} - n \cdot y^{2} = k$, then $ ( \frac{ m \cdot x + n \cdot y }{k})^{2} - n \cdot ( \frac{ x + m \cdot y }{k})^{2} = \frac{m^2 - n}{k}$
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Using above Brahmagupta's identity, If $(x_{1}, y_{1}, k_{1})$ and $(x_{2}, y_{2}, k_{2})$ satisfy $(x_{1}^{2} - y_1^{2}) \cdot (x_{2}^{2} - y_2^{2}) = k_{1} \cdot k_{2} $, then $(x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2}, x_{1} \cdot y_{2} + y_{1} \cdot x_{2}, k_{1} \cdot k_{2})$ is also a solution of $(x_{1} \cdot x_{2} + n \cdot y_{1} \cdot y_{2})^{2} - n \cdot (x_{1} \cdot y_{2} + x_{2} \cdot y_{1})^{2} = k_{1} \cdot k_{2}$
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[//]: #(First, we choose $y_{1} = 1$ and choose $x_{1} = \lfloor \sqrt n \rfloor$ such that $k_{1}$ is a small integer.)
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#### Steps
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1. Initialization:Choose an initial solution $(p_{0}, q_{0}, m_{0})$ where $p_{0}$ and $q_{0}$ are co-prime such that $p_{0}^{2} - N \cdot q_{0}^{2} = m_{0}$. Typically, start with $p_{0} = \lfloor \sqrt N \rfloor $, $q_{0} = 1$, $m_{0} = p_0^2 - N $.
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2. Key step: Find $x_{1}$ such that: $q_{0} \cdot x_{1} \equiv -p_{0} \pmod {\lvert m_{0}\rvert}$ and $\lvert x_{1}^2 - N \rvert$ is minimized.
3. Termination: When $m_{k}=1$, the values of $p_{k}$ and $q_{k}$ are the smallest positive solution of the Pell's equation.
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##### Example
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Let's solve the equation $x^{2} - 13 \cdot y^{2} = 1$ using Chakravala method.
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1. Start with $(p_{0}, q_{0}, m_{0}) = (3, 1, -4)$ because $3^2 - 13 \cdot1^2 = -4$.
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[//]: #()
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[//]: #(Then, Iteratively we adjust $x_{1}$ and $y_{1}$ so that $k_{1} = 1$. The solution is given by $(x_{1}, y_{1})$ to original Pell's equation.)
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[//]: #()
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[//]: #( Choosing m. At e)
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2. Find $x_{1}$ such that $x_{1} \equiv -3 \pmod {4}$ and $\lvert x_{1}^2 - 13 \rvert$ is
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minimized.
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