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
:$A=\big(\tfrac\big)\sqrt$
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,
:$A = x\sqrt$
Since $\sqrt$ 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 $\$ of the equation can be generated using the following recurrence relations
:$x_=2x_n+3y_n, \quad y_= x_n+2y_n \text n=1,2,\ldots$
or by the following relations
:$\begin x_ & = 4x_-x_\textn=2,3,\ldots \text x_1=2, x_2=7\\ y_ & = 4y_-y_\textn=2,3,\ldots \text y_1=1, y_2=4. \end$
They can also be generated using the following property:
:$x_n+\sqrt y_n=(x_1+\sqrty_1)^n\text n=1,2, \ldots$
The following are the first eight values of $x_n$ and $y_n$ and the corresponding Brahmagupta triangles:
::
The sequence $\$ is entry in the Online Encyclopedia of Integer Sequences (OEIS) and the sequence $\$ is entry 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 $1\le d\le t$. Using Heron's formula, the area $A$ of the triangle can be shown to be
:$A = \big(\tfrac\big)\sqrt$.
For $A$ to be an integer, $t$ must be even and one may take $t=2x$ for some integer. This makes
:$A=x\sqrt$.
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
:$\begin d & = \vert m^2 - 3n^2\vert /g\\ x & = (m^2 + 3n^2)/g\\ y & = 2mn/g \end$
where $m$ and $n$ are relatively prime positive integers and $g = \text(m^2 - 3n^2, 2mn, m^2 + 3n^2)$.

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.

See also

* Brahmagupta polynomials
* Brahmagupta quadrilateral

References

{{reflist

Arithmetic problems of plane geometry
Types of triangles
Eponymous geometric shapes
Elementary mathematics
Elementary number theory
Brahmagupta