Brahmagupta triangle

A Brahmagupta triangle is a triangle whose side lengths are consecutive positive integers and area is a positive integer.^[1][2][3] The triangle whose side lengths are 3, 4, 5 is a Brahagupta 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.^[1][4]

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.^[5][6][7][8] Some of the other names by which Brahmagupta triangles are known are super-Heronian triangle^[9] and almost-equilateral Heronian triangle.^[10]

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.^[11]

Generating Brahmagupta triangles

Let the side lengths of a Brahmagupta triangle be ${\displaystyle t-1}$, ${\displaystyle t}$ and ${\displaystyle t+1}$ where ${\displaystyle t}$ is an integer greater than 1. Using Heron's formula, the area ${\displaystyle A}$ of the triangle can be shown to be

${\displaystyle A={\big (}{\tfrac {t}{2}}{\big )}{\sqrt {3{\big [}{\big (}{\tfrac {t}{2}}{\big )}^{2}-1{\big ]}}}}$

Since ${\displaystyle A}$ has to be an integer, ${\displaystyle t}$ must be even and so it can be taken as ${\displaystyle t=2x}$ where ${\displaystyle x}$ is an integer. Thus,

${\displaystyle A=x{\sqrt {3(x^{2}-1)}}}$

Since ${\displaystyle {\sqrt {3(x^{2}-1)}}}$ has to be an integer, one must have ${\displaystyle x^{2}-1=3y^{2}}$ for some integer ${\displaystyle y}$. Hence, ${\displaystyle x}$ must satisfy the following Diophantine equation:

${\displaystyle x^{2}-3y^{2}=1}$.

This is an example of the so-called Pell's equation ${\displaystyle x^{2}-Ny^{2}=1}$ with ${\displaystyle N=3}$. The methods for solving the Pell's equation can be applied to find values of the integers ${\displaystyle x}$ and ${\displaystyle y}$.

Obviously ${\displaystyle x=2}$, ${\displaystyle y=1}$ is a solution of the equation ${\displaystyle x^{2}-3y^{2}=1}$. Taking this as an initial solution ${\displaystyle x_{1}=2,y_{1}=1}$ the set of all solutions ${\displaystyle \{(x_{n},y_{n})\}}$ of the equation can be generated using the following recurrence relations^[1]

${\displaystyle x_{n+1}=2x_{n}+3y_{n},\quad y_{n+1}=x_{n}+2y_{n}{\text{ for }}n=1,2,\ldots }$

or by the following relations

${\displaystyle {\begin{aligned}x_{n+1}&=4x_{n}-x_{n-1}{\text{ for }}n=2,3,\ldots {\text{ with }}x_{1}=2,x_{2}=7\\y_{n+1}&=4y_{n}-y_{n-1}{\text{ for }}n=2,3,\ldots {\text{ with }}y_{1}=1,y_{2}=4.\end{aligned}}}$

They can also be generated using the following property:

${\displaystyle x_{n}+{\sqrt {3}}y_{n}=(x_{1}+{\sqrt {3}}y_{1})^{n}{\text{ for }}n=1,2,\ldots }$

The following are the first eight values of ${\displaystyle x_{n}}$ and ${\displaystyle y_{n}}$ and the corresponding Brahmagupta triangles:

${\displaystyle n}$	1	2	3	4	5	6	7	8
${\displaystyle x_{n}}$	2	7	26	97	362	1351	5042	18817
${\displaystyle y_{n}}$	1	4	15	56	209	780	2911	10864
Brahmagupta triangle	3,4,5	13,14,15	51,52,53	193,194,195	723,724,725	2701,2702,2703	10083,10084,10085	37633,37634,337635

The sequence ${\displaystyle \{x_{n}\}}$ is entry A001075 in the Online Encyclopedia of Integer Sequences (OEIS) and the sequence ${\displaystyle \{y_{n}\}}$ 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 ${\displaystyle t-1,t,t+1}$ are the side lengths of a Brahmagupta triangle then, for any positive integer ${\displaystyle k}$, the integers ${\displaystyle k(t-1),kt,k(t+1)}$ are the side lengths of a generalized Brahmagupta triangle which form an arithmetic progression with common difference ${\displaystyle k}$. 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.^[12]

To find the side lengths of such triangles, let the side lengths be ${\displaystyle t-d,t,t+d}$ where ${\displaystyle b,d}$ are integers satisfying ${\displaystyle 1\leq d\leq t}$. Using Heron's formula, the area ${\displaystyle A}$ of the triangle can be shown to be

${\displaystyle A={\big (}{\tfrac {b}{4}}{\big )}{\sqrt {3(t^{2}-4d^{2})}}}$.

For ${\displaystyle A}$ to be an integer, ${\displaystyle t}$ must be even and one may take ${\displaystyle t=2x}$ for some integer. This makes

${\displaystyle A=x{\sqrt {3(x^{2}-d^{2})}}}$.

Since, again, ${\displaystyle A}$ has to be an integer, ${\displaystyle x^{2}-d^{2}}$ has to be in the form ${\displaystyle 3y^{2}}$ for some integer ${\displaystyle y}$. Thus, to find the side lengths of generalized Brahmagupta triangles, one has to find solutions to the following homogeneous quadratic Diophantine equation:

${\displaystyle x^{2}-3y^{2}=d^{2}}$.

It can be shown that all primitive solutions of this equation are given by^[12]

${\displaystyle {\begin{aligned}d&=\vert m^{2}-3n^{2}\vert /g\\x&=(m^{2}+3n^{2})/g\\y&=2mn/g\end{aligned}}}$

where ${\displaystyle m}$ and ${\displaystyle n}$ are relatively prime positive integers and ${\displaystyle g={\text{gcd}}(m^{2}-3n^{2},2mn,m^{2}+3n^{2})}$.

If we take ${\displaystyle m=n=1}$ we get the Brahmagupta triangle ${\displaystyle (3,4,5)}$. If we take ${\displaystyle m=2,n=1}$ we get the Brahmagupta triangle ${\displaystyle (13,14,15)}$. But if we take ${\displaystyle m=1,n=2}$ we get the generalized Brahmagupta triangle ${\displaystyle (15,26,37)}$ which cannot be reduced to a Brahmagupta triangle.

See also

• Brahmagupta polynomials
• Brahmagupta quadrilateral

References