Pat’s doctoral training was in quadratic forms.
… and he was only too happy to say so:
(Translation modified from Satya Prakash, see [2].)
BrShSi 18.64-65:
Put down twice the square-root of a given square multiplied by a multiplier and increased or diminished by an arbitrary number. The product of the first pair multiplied by the multiplier, with the product of the last pair, is the last computed.
The sum of the thunderbolts is the first. The additive is equal to the product of the additives. The two square roots, divided by the additive or the subtractive, are the additive rupas.
Suppose that
Then two solutions of \(Nx^2 + k_1k_2 = y^2\) are:
\[(x,y) = (\alpha_1\beta_2 \pm \alpha_2\beta_1,\beta_1\beta_2 \pm N\alpha_1\alpha_2).\]
Assume that:
Consider “Brahmagupta’s Identity”:
\[(a^2+nb^2)(c^2+nd^2)=(ac+nbd)^2+n(ad-bc)^2.\]
Substitute: \(n = -N, a = \beta_1, b = \alpha_1, c = \beta_2, d = \alpha_2\).
\[(\beta_1-n\alpha_1^2)(\beta_2-N\alpha_2^2)=(\beta_1\beta_2-N\alpha_1\alpha_2)^2-N(\beta_1\alpha_2-\alpha_1\beta_2)^2.\]
\[ \begin{aligned} (\beta_1-n\alpha_1^2)(\beta_2-N\alpha_2^2)=(\beta_1\beta_2-N\alpha_1\alpha_2)^2-N(\beta_1\alpha_2-\alpha_1\beta_2)^2,\\ k_1k_2=(\beta_1\beta_2-N\alpha_1\alpha_2)^2-N(\beta_1\alpha_2-\alpha_1\beta_2)^2,\\ k_1k_2+N(\beta_1\alpha_2-\alpha_1\beta_2)^2=(\beta_1\beta_2-N\alpha_1\alpha_2)^2,\\ N(\alpha_1\beta_2- \beta_1\alpha_2)^2 + k_1k_2=(\beta_1\beta_2-N\alpha_1\alpha_2)^2. \end{aligned} \]
For \(+\) versions, note that \((\alpha_1, -\beta_1)\) also solves \(Nx^2+k_1=y^2\).
\[(a^2+nb^2)(c^2+nd^2)=(ac+nbd)^2+n(ad-bc)^2.\]
Not mentioned by Brahmagupta. Also not mentioned in any extant pre-modern Indian proofs (that I know of).
But it inspired Manjul Bhargava as a child.
\[1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}.\] \[1^3+2^3+\ldots+n^3=\left(\frac{n(n+1)}{2}\right)^2.\]
Part of it has been translated by Kim Plofker (in [1]).
My aim is to fill in the omitted passage.