# Brahmagupta triangle

A Brahmagupta triangle is a triangle whose side lengths are consecutive positive integers and area is a positive integer. The triangle whose side lengths are 3, 4, 5 is a Brahmagupta triangle and so also is the triangle whose side lengths are 13, 14, 15. The Brahmagupta triangle is a special case of the Heronian triangle which is a triangle whose side lengths and area are all positive integers but the side lengths need not necessarily be consecutive integers. A Brahmagupta triangle is called as such in honor of the Indian astronomer and mathematician Brahmagupta (c. 598 – c. 668 CE) who gave a list of the first eight such triangles without explaining the method by which he computed that list.
A Brahmagupta triangle is also called a Fleenor-Heronian triangle in honor of Charles R. Fleenor who discussed the concept in a paper published in 1996. Some of the other names by which Brahmagupta triangles are known are super-Heronian triangle and almost-equilateral Heronian triangle.
The problem of finding all Brahmagupta triangles is an old problem. A closed form solution of the problem was found by Reinhold Hoppe in 1880.

## Generating Brahmagupta triangles

Let the side lengths of a Brahmagupta triangle be $t-1$, $t$ and $t+1$ where $t$ is an integer greater than 1. Using Heron's formula, the area $A$ of the triangle can be shown to be
    $A={\big{\tfrac {t}{2}}{\big )}{\sqrt {3{\big{\big{\tfrac {t}{2}}{\big )}^{2}-1{\big ]}}}$
Since $A$ has to be an integer, $t$ must be even and so it can be taken as $t=2x$ where $x$ is an integer. Thus,
    $A=x{\sqrt {3(x^{2}-1)}}$
Since ${\sqrt {3(x^{2}-1)}}$ has to be an integer, one must have $x^{2}-1=3y^{2}$ for some integer $y$. Hence, $x$ must satisfy the following Diophantine equation:
    $x^{2}-3y^{2}=1$.
This is an example of the so-called Pell's equation $x^{2}-Ny^{2}=1$ with $N=3$. The methods for solving the Pell's equation can be applied to find values of the integers $x$ and $y$.
Obviously $x=2$, $y=1$ is a solution of the equation $x^{2}-3y^{2}=1$. Taking this as an initial solution $x_{1}=2,y_{1}=1$ the set of all solutions $\{(x_{n},y_{n})\}$ of the equation can be generated using the following recurrence relations
    $x_{n+1}=2x_{n}+3y_{n},\quad y_{n+1}=x_{n}+2y_{n}{\text{ for }}n=1,2,\ldots$
or by the following relations
    ${\begin{aligned}x_{n+1}&=4x_{n}-x_{n-1}{\text{ for }}n=2,3,\ldots {\text{ with }}x_{1}=2,x_{2}=7\\y_{n+1}&=4y_{n}-y_{n-1}{\text{ for }}n=2,3,\ldots {\text{ with }}y_{1}=1,y_{2}=4.\end{aligned}}$
They can also be generated using the following property:
    $x_{n}+{\sqrt {3}}y_{n}=(x_{1}+{\sqrt {3}}y_{1})^{n}{\text{ for }}n=1,2,\ldots$
The following are the first eight values of $x_{n}$ and $y_{n}$ and the corresponding Brahmagupta triangles:
       $n$12345678 $x_{n}$2726973621351504218817 $y_{n}$141556209780291110864 Brahmagupta triangle3,4,513,14,1551,52,53193,194,195723,724,7252701,2702,270310083,10084,1008537633,37634,37635
The sequence $\{x_{n}\}$ is entry A001075 in the Online Encyclopedia of Integer Sequences (OEIS) and the sequence $\{y_{n}\}$ is entry A001353 in OEIS.

## Generalized Brahmagupta triangles

In a Brahmagupta triangle the side lengths form an integer arithmetic progression with a common difference 1. A generalized Brahmagupta triangle is a Heronian triangle in which the side lengths form an arithmetic progression of positive integers. Generalized Brahmagupta triangles can be easily constructed from Brahmagupta triangles. If $t-1,t,t+1$ are the side lengths of a Brahmagupta triangle then, for any positive integer $k$, the integers $k(t-1),kt,k(t+1)$ are the side lengths of a generalized Brahmagupta triangle which form an arithmetic progression with common difference $k$. There are generalized Brahmagupta triangles which are not generated this way. A primitive generalized Brahmagupta triangle is a generalized Brahmagupta triangle in which the side lengths have no common factor other than 1.
To find the side lengths of such triangles, let the side lengths be $t-d,t,t+d$ where $t,d$ are integers satisfying $1\leq d\leq t$. Using Heron's formula, the area $A$ of the triangle can be shown to be
    $A={\big{\tfrac {b}{4}}{\big )}{\sqrt {3(t^{2}-4d^{2})}}$.
For $A$ to be an integer, $t$ must be even and one may take $t=2x$ for some integer. This makes
    $A=x{\sqrt {3(x^{2}-d^{2})}}$.
Since, again, $A$ has to be an integer, $x^{2}-d^{2}$ has to be in the form $3y^{2}$ for some integer $y$. Thus, to find the side lengths of generalized Brahmagupta triangles, one has to find solutions to the following homogeneous quadratic Diophantine equation:
    $x^{2}-3y^{2}=d^{2}$.
It can be shown that all primitive solutions of this equation are given by
    ${\begin{aligned}d&=\vert m^{2}-3n^{2}\vert /g\\x&=(m^{2}+3n^{2})/g\\y&=2mn/g\end{aligned}}$
where $m$ and $n$ are relatively prime positive integers and $g={\text{gcd}}(m^{2}-3n^{2},2mn,m^{2}+3n^{2})$.
If we take $m=n=1$ we get the Brahmagupta triangle $(3,4,5)$. If we take $m=2,n=1$ we get the Brahmagupta triangle $(13,14,15)$. But if we take $m=1,n=2$ we get the generalized Brahmagupta triangle $(15,26,37)$ which cannot be reduced to a Brahmagupta triangle.