Brahmagupta's identity

In algebra, Brahmagupta's identity says that, for given integer N, the product of two numbers of the form ${\displaystyle x^{2}-Ny^{2}}$ is again a number of the form. More precisely, we have

${\displaystyle (x_{1}^{2}-Ny_{1}^{2})(x_{2}^{2}-Ny_{2}^{2})=(x_{1}x_{2}+Ny_{1}y_{2})^{2}-N(x_{1}y_{2}+x_{2}y_{1})^{2}.}$

This identity can be used to generate infinitely many solutions to the Pell's equation. It can also be used to generate successively better rational approximations to square roots of arbitrary integers.

Brahmagupta matrix

If, for an arbitrary real number ${\displaystyle t}$, we define the matrix

${\displaystyle B(x,y)={\begin{bmatrix}x&y\\ty&x\end{bmatrix}}}$

then, Brahmagupta's identity can be expressed in the following form:

${\displaystyle \det B(x_{1},y_{1})\det B(x_{2},y_{2})=\det(B(x_{1},y_{1})B(x_{2},y_{2}))}$

The matrix ${\displaystyle B(x,y)}$ is called the Brahmagupta matrix.

Brahmagupta polynomials

Let ${\displaystyle B=B(x,y)}$ be as above. Then, it can be seen by induction that the matrix ${\displaystyle B^{n}}$ can be written in the form

${\displaystyle B^{n}={\begin{bmatrix}x_{n}&y_{n}\\ty_{n}&x_{n}\end{bmatrix}}}$

Here, ${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$ are polynomials in ${\displaystyle x,y,t}$. These polynomials are called the Brahmagupta polynomials. The first few of the polynomials are listed below:

${\displaystyle {\begin{alignedat}{2}x_{1}&=x&y_{1}&=y\\x_{2}&=x^{2}+ty^{2}&y_{2}&=2xy\\x_{3}&=x^{3}+3txy^{2}&y_{3}&=3x^{2}y+ty^{3}\\x_{4}&=x^{4}+6t^{2}x^{2}y^{2}+t^{2}y^{4}\qquad &y_{4}&=4x^{3}y+4txy^{3}\end{alignedat}}}$

Recurrence relations

The polynomials ${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$ satisfy the following recurrence relations:

• ${\displaystyle x_{n+1}=xx_{n}+tyy_{n}}$
• ${\displaystyle y_{n+1}=xy_{n}+yx_{n}}$
• ${\displaystyle x_{n+1}=2xx_{n}-(x^{2}-ty^{2})x_{n-1}}$
• ${\displaystyle y_{n+1}=2xy_{n}-(x^{2}-ty^{2})y_{n-1}}$
• ${\displaystyle x_{2n}=x_{n}^{2}+ty_{n}^{2}}$
• ${\displaystyle y_{2n}=2x_{n}y_{n}}$

Exact expressions

The eigenvalues of ${\displaystyle B(x,y)}$ are ${\displaystyle x\pm y{\sqrt {t}}}$ and the corresponding eigenvectors are ${\displaystyle [1,\pm {\sqrt {t}}]^{T}}$. Hence

${\displaystyle B[1,\pm {\sqrt {t}}]^{T}=(x\pm y{\sqrt {t}})[1,\pm {\sqrt {t}}]^{T}}$.

It follows that

${\displaystyle B^{n}[1,\pm {\sqrt {t}}]^{T}=(x\pm y{\sqrt {t}})^{n}[1,\pm {\sqrt {t}}]^{T}}$.

This yields the following exact expressions for ${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$:

• ${\displaystyle x_{n}={\tfrac {1}{2}}{\big [}(x+y{\sqrt {t}})^{n}+(x-y{\sqrt {t}})^{n}{\big ]}}$
• ${\displaystyle y_{n}={\tfrac {1}{2{\sqrt {t}}}}{\big [}(x+y{\sqrt {t}})^{n}-(x-y{\sqrt {t}})^{n}{\big ]}}$

Expanding the powers in the above exact expressions using the binomial theorem and simplifying one gets the following expressions for ${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$:

• ${\displaystyle x_{n}=x^{n}+t{n \choose 2}x^{n-2}y^{2}+t^{2}{n \choose 4}x^{n-4}y^{4}+\cdots }$
• ${\displaystyle y_{n}=nx^{n-1}y+t{n \choose 3}x^{n-3}y^{3}+t^{2}{n \choose 5}x^{n-5}y^{5}+\cdots }$

Special cases

1. If ${\displaystyle x=y={\tfrac {1}{2}}}$ and ${\displaystyle t=5}$ then, for ${\displaystyle n>0}$:

${\displaystyle 2y_{n}=F_{n}}$ is the Fibonacci sequence ${\displaystyle 1,1,2,3,5,8,13,21,34,55,\ldots }$.

${\displaystyle 2x_{n}=L_{n}}$ is the Lucas sequence ${\displaystyle 2,1,3,4,7,11,18,29,47,76,123,\ldots }$.

1. If we set ${\displaystyle x=y=1}$ and ${\displaystyle t=2}$, then:

${\displaystyle x_{n}=1,1,3,7,17,41,99,239,577,\ldots }$ which are the numerators of continued fraction convergents to ${\displaystyle {\sqrt {2}}}$.^[6] This is also the sequence of half Pell-Lucas numbers.

${\displaystyle y_{n}=0,1,2,5,12,29,70,169,408,\ldots }$ which is the sequence of Pell numbers.

A differential equation

${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$ are polynomial solutions of the following partial differential equation:

${\displaystyle \left({\frac {\partial ^{2}}{\partial x^{2}}}-{\frac {1}{t}}{\frac {\partial ^{2}}{\partial y^{2}}}\right)U=0}$