# Bhaskara's Lemma

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## Contents

## Lemma

Let $m \in \Z$ be an integer.

For $k \ne 0$:

- $N x^2 + k = y^2 \implies N \paren {\dfrac {m x + y} k}^2 + \dfrac {m^2 - N} k = \paren {\dfrac {m y + N x} k}^2$

## Proof

\(\displaystyle N x^2 + k\) | \(=\) | \(\displaystyle y^2\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle N m^2 x^2 - N^2 x^2 + k \paren {m^2 - N}\) | \(=\) | \(\displaystyle m^2 y^2 - N y^2\) | multiplying both sides by $m^2 - N$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle N m^2 x^2 + 2 N m x y + N y^2 + k \paren {m^2 - N}\) | \(=\) | \(\displaystyle m^2 y^2 + 2 N m x y + N^2 x^2\) | adding $N^2 x^2 + 2 N m x y + N y^2$ to both sides | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle N \paren {m x + y}^2 + k \paren {m^2 - N}\) | \(=\) | \(\displaystyle \paren {m y + N x}^2\) | factorizing | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle N \paren {\frac {m x + y} k}^2 + \frac {m^2 - N} k\) | \(=\) | \(\displaystyle \paren {\frac {m y + N x} k}^2\) | dividing by $k^2$ |

The implication goes the other way if $m^2 - N \ne 0$.

$\blacksquare$

## Also see

This lemma is used when deriving the Chakravala Method of solving indeterminate quadratic equations.

## Source of Name

This entry was named for Bhaskara II Acharya.