Bhaskara's Lemma

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

\(\ds N x^2 + k\)

\(=\)

\(\ds y^2\)

\(\ds \leadsto \ \ \)

\(\ds N m^2 x^2 - N^2 x^2 + k \paren {m^2 - N}\)

\(=\)

\(\ds m^2 y^2 - N y^2\)

multiplying both sides by $m^2 - N$

\(\ds \leadsto \ \ \)

\(\ds N m^2 x^2 + 2 N m x y + N y^2 + k \paren {m^2 - N}\)

\(=\)

\(\ds m^2 y^2 + 2 N m x y + N^2 x^2\)

adding $N^2 x^2 + 2 N m x y + N y^2$ to both sides

\(\ds \leadsto \ \ \)

\(\ds N \paren {m x + y}^2 + k \paren {m^2 - N}\)

\(=\)

\(\ds \paren {m y + N x}^2\)

factorizing

\(\ds \leadsto \ \ \)

\(\ds N \paren {\frac {m x + y} k}^2 + \frac {m^2 - N} k\)

\(=\)

\(\ds \paren {\frac {m y + N x} k}^2\)

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.