Brahmagupta triangle

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

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.^[4][5][6][7] Some of the other names by which Brahmagupta triangles are known are super-Heronian triangle^[8] and almost-equilateral Heronian triangle.^[9]

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.^[10]

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 $${\displaystyle A={\frac {t}{2}}{\sqrt {3\left(\left({\tfrac {t}{2}}\right)^{2}-1\right)}}}$$ Since A has to be an integer, t must be even and so it can be taken as ${\displaystyle t=2x}$ where x is an integer. Thus, $${\displaystyle A=x{\sqrt {3(x^{2}-1)}}}$$ Since ${\textstyle {\sqrt {3(x^{2}-1)}}}$ has to be an integer, one must have ${\displaystyle x^{2}-1=3y^{2}}$ for some integer y. Hence, x must satisfy the following Diophantine equation: $${\displaystyle x^{2}-3y^{2}=1.}$$ This is an example of the so-called Pell's equation x² − Ny² = 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 x² − 3y² = 1. Taking this as an initial solution x₁ = 2, y₁ = 1, the set of all solutions {(x_n, y_n)} of the equation can be generated using the following recurrence relations^[1] $${\displaystyle {\begin{aligned}&x_{n+1}=2x_{n}+3y_{n},&&y_{n+1}=x_{n}+2y_{n},&&n=1,2,\ldots \\[2pt]&\quad x_{1}=2.&&\quad y_{1}=1.\end{aligned}}}$$ or by the following relations with additional initial solutions x₂ = 7, y₂ = 4: $${\displaystyle {\begin{aligned}&x_{n+1}=4x_{n}-x_{n-1},&&y_{n+1}=4y_{n}-y_{n-1},&&n=2,3,\ldots \\[2pt]&\quad x_{1}=2,\ x_{2}=7.&&\quad y_{1}=1,\ y_{2}=4.\end{aligned}}}$$

They can also be generated using the following property: $${\displaystyle x_{n}+{\sqrt {3}}y_{n}=\left(x_{1}+{\sqrt {3}}y_{1}\right)^{n},\quad n=1,2,\ldots }$$ The following are the first eight values of x_n and y_n and the corresponding Brahmagupta triangles:

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.

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.^[11]

To find the side lengths of such triangles, let the side lengths be ${\displaystyle t-d,t,t+d}$ where ${\displaystyle t,d}$ 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^[11]

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

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.

• Brahmagupta polynomials
• Brahmagupta quadrilateral