Difference between revisions of "Chakravala method"

Revision as of 15:08, 3 March 2023

The chakravala method is an algorithm for solving the Pell equation $x^2 - Dy^2 = 1.$

Method of composition

We let $a$ and $b$ be integers such that $\gcd(a,b) = 1$ , and we notate $a^2 - Db^2 = q$ .

We then choose an integer $c$ and let $\begin{align*} \alpha &= \frac{ac+Db}{q}, \\ \beta &= \frac{a+bc}{q}.\\ \end{align*}$

Existence of suitable choice

We claim that it is always possible to choose $c$ such that $\beta$ is an integer.

Because $\gcd(a,b) = 1$ , we have $\gcd(a^2,b) = 1$ , so $\gcd(q,b) = \gcd(a^2-Db^2,b) = 1.$

Suppose $a + bc \equiv a + bc' \pmod{|q|}$ . Then $q \mid b(c - c')$ . Because $\gcd(q,b) = 1$ , $q$ also divides $c - c'$ , so $c \equiv c' \pmod{|q|}$ .

We can therefore construct a set of $q$ possible integer values of $c$ , none congruent to another $\mathrm{mod} \; |q|$ ; the corresponding values of $a + bc$ take all $|q|$ distinct values $\mathrm{mod} \; |q|$ , so there must be one element $c_0$ in the set such that $a + bc_0 \equiv 0 \pmod q$ ; that is, $\frac{a + bc_0}{q}$ is an integer.

Recovery of initial conditions

We further claim that if $\beta$ is an integer, then

$\alpha$ is also an integer, and
$\gcd(\alpha, \beta) = 1$ .

For the first claim, we use the fact that $\beta$ is an integer to conclude that $a \equiv -bc \pmod{|q|}$ . Therefore, $a(-bc) - Db^2 \equiv a^2 - Db^2 \pmod{|q|}.$ The right-hand side of the above congruence is $q$ ; the left side is $-b(ac+Db) = -bq\alpha$ . Because $-bq\alpha$ is a multiple of $q$ and $\gcd(q,b) = 1$ , $-q\alpha$ is also a multiple of $q$ . Thus, $\alpha$ is an integer.

For the second claim, we prove that $\gcd(q\alpha,q\beta) = |q|$ . Suppose that a positive integer $k$ divides both $q\alpha$ and $q\beta$ . Similarly to before, we consider $-bq\alpha = -b(ac+Db) = a(-bc) - Db^2$ and use the assumption that $q\beta$ is a multiple of $k$ to make the substitution $a \equiv -bc \pmod k$ , obtaining $-bq\alpha \equiv a^2 - Db^2 \pmod k.$ But $-bq\alpha$ is a multiple of $k$ , so $a^2 - Db^2 = q$ is also a multiple of $k$ . Thus, $k$ is a divisor of $|q|$ .

Evaluation

We now claim that $\alpha^2 - D\beta^2 = \frac{c^2-D}{q}$ .

From Brahmagupta's Identity (with $n = -D$ and $d = 1$ ) we have $(ac+Db)^2 - D(a+bc)^2 = (a^2-Db^2)(c^2-D).$ That is, $(q\alpha)^2 - D(q\beta)^2 = q(c^2-D).$ Dividing both sides by $q^2$ gives the desired result.

Algorithm

We begin by choosing initial relatively prime integers $a$ and $b$ . At each step, we choose the value of $c$ that minimizes $|c^2 - D|$ (among the values of $c$ for which $\beta$ is an integer) and replace the values of $a$ and $b$ with the resulting values of $\alpha$ and $\beta$ . Repeating this step, the value of $q$ eventually reaches $1$ , yielding a solution to the Pell equation.

@@ Line 34: / Line 34: @@
 From [[Brahmagupta's Identity]] (with <math>n = -D</math> and <math>d = 1</math>) we have <cmath>(ac+Db)^2 - D(a+bc)^2 = (a^2-Db^2)(c^2-D).</cmath>
 That is, <cmath>(q\alpha)^2 - D(q\beta)^2 = q(c^2-D).</cmath> Dividing both sides by <math>q^2</math> gives the desired result.
+==Algorithm==
+We begin by choosing initial [[relatively prime]] integers <math>a</math> and <math>b</math>. At each step, we choose the value of <math>c</math> that minimizes <math>|c^2 - D|</math> (among the values of <math>c</math> for which <math>\beta</math> is an integer) and replace the values of <math>a</math> and <math>b</math> with the resulting values of <math>\alpha</math> and <math>\beta</math>. Repeating this step, the value of <math>q</math> eventually reaches <math>1</math>, yielding a solution to the Pell equation.

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