Bhaskara's Lemma

Lemma

As long as $k \ne 0$:

$\displaystyle N x^2 + k = y^2 \implies N \left({\frac{mx + y}{k}}\right)^2 + \frac{m^2 - N}{k} = \left({\frac{my + Nx}{k}}\right)^2$

for any integer $m$.

Proof

	\(\displaystyle \)	\(\displaystyle N x^2 + k\)	\(=\)	\(\displaystyle y^2\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle N m^2 x^2 - N^2 x^2 + k \left({m^2 - N}\right)\)	\(=\)	\(\displaystyle m^2 y^2 - N y^2\)	\(\displaystyle \)	multiplying both sides by $m^2 - N$
	\(\displaystyle \implies\)	\(\displaystyle N m^2 x^2 + 2 N m x y + N y^2 + k \left({m^2 - N}\right)\)	\(=\)	\(\displaystyle m^2 y^2 + 2 N m x y + N^2 x^2\)	\(\displaystyle \)	adding $N^2 x^2 + 2 N m x y + N y^2$ to both sides
	\(\displaystyle \implies\)	\(\displaystyle N \left({m x + y}\right)^2 + k \left({m^2 - N}\right)\)	\(=\)	\(\displaystyle \left({m y + N x}\right)^2\)	\(\displaystyle \)	factorizing
	\(\displaystyle \implies\)	\(\displaystyle N \left({\frac{m x + y}{k} }\right)^2 + \frac{m^2 - N}{k}\)	\(=\)	\(\displaystyle \left({\frac{m y + N x}{k} }\right)^2\)	\(\displaystyle \)	dividing by $k^2$

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

$\blacksquare$

Notes

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

Source of Name

This entry was named for Bhāskara II Āchārya.