Generating triangles with consecutive-integer sides and rational area

Question

I'm stuck on this problem for quite some time:

Call a triangle a Special Rational triangle if it's area is rational, and the side lengths are consecutive positive integers, Can we find a closed form which generates all Special Rational triangles?

I have tried this one for quite some time, I was able to find a nice closed form in terms of a Diophantine equation, but I'm totally not satisfied with it. Your insight would be very helpful.

Thanks in advance.

Answer 1

Almost-equilateral Heronian triangles can be generated using expression for the length of the middle side

\begin{align} a_n&=(2+\sqrt3)^n+(2-\sqrt3)^n \quad\text{for }\quad n=1,2,\dots . \end{align}

The two other side lengths are $a_n-1$ and $a_n+1$, see also A003500.

Alternatively,

\begin{align} a_1&=4,\quad a_2=14 ,\\ a_n&=4a_{n-1}-a_{n-2} . \end{align}

Answer 2

Let the sides of the triangle be $a,a+1,a+2$ where $a$ is a positive integer. The perimeter of the triangle is an integer, so we need only be concerned with the area. The semiperimeter is $\frac{3a+3}2$, so by Heron's formula, the area is $$\sqrt{\frac{3a+3}2\cdot\frac{a+3}2\cdot\frac{a+1}2\cdot\frac{a-1}2} =\frac{a+1}4\sqrt{3a^2+6a-9}$$ For this to be rational, it is necessary and sufficient that $3a^2+6a-9$ be a perfect square.

Suppose $3a^2+6a-9=x^2$ where $x$ is a positive integer. Let $y=a+1$. Then $$3y^2-x^2=3a^2+6a+3-(3a^2+6a-9)=12$$ so we need the solutions to $3y^2-x^2=9$ in positive integers. This is a Pell-type equation, and methods of solution are well-known. I simply typed it into Wolfram Alpha. The positive solutions for $y$ are $$y=\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n$$ so that the solutions for $a$ are $$a=\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n-1$$

The first few values of $a$ are $$3,13,51,193,723$$ Plugging these into OEIS, we find A016064, which gives some other formulas and recurrences.

Answer 3

Let the sides be $n-1$, $n$, and $n+1$ for $n \in \Bbb{Z}$ with $n > 0$. (You don't explicitly reject degenerate triangles with $n - 1 = 0$...)

By Heron's formula, the area is $$ A = \sqrt{s(s-(n-1))(s-n)(s-(n+1))} \text{,} $$ where $s = (1/2)((n-1) + n + (n+1)) = (3n/2)$, so \begin{align*} A &= \sqrt{s(s-(n-1))(s-n)(s-(n+1))} \\ &= \sqrt{(3n/2)((3n/2)-n+1)((3n/2)-n)((3n/2)-n-1)} \\ &= \sqrt{(3n/2)((n/2)+1)(n/2)((n/2)-1)} \\ &= \sqrt{\frac{3}{16}n^2(n^2 - 4)} \\ \end{align*} By plugging in $0$, $1$, and $2$, we quickly see that $3 \mid n^2(n^2 - 4)$. With a similar computation, we find $n^2(n^2 - 4)$ is divisible by $16$ exactly when $n$ is even, so $n = 2m$. $$ A = 3\sqrt{\frac{4m^2(4m^2 - 4)}{3 \cdot 16}} = 3\sqrt{\frac{m^2(m^2 - 1)}{3}} $$ By plugging in $0$, $1$, and $2$, we quickly see that $3 \mid m^2(m^2 - 1)$, so $\frac{m^2(m^2 - 1)}{3}$ is an integer for any choice of $m$.

So we want to solve $$ \frac{m^2(m^2 - 1)}{3} = u^2 $$ in integers. That is, $$ (m^2)^2 - 3u^2 = m^2 \text{.} $$ Notice $m^2$ divides two terms, so it divides all three. Divide through by $m^2$, obtaining $$ m^2 - 3(u/m)^2 = 1 \text{,} $$ a Pell's equation, where we seek solutions in the integers $m > 0$ and $u/m \geq 0$.

By the usual application of continued fractions, we obtain a general solution. We don't actually need the form of $u/m$, but, $$ m = \frac{1}{2} \left( (2-\sqrt{3})^c + (2+\sqrt{3})^c \right) $$ for integer $c \geq 1$. Then suitable triangles are obtained from all $$ n = \left( (2-\sqrt{3})^c + (2+\sqrt{3})^c \right) $$ for integer $c \geq 1$. (This includes the degenerate $1$-$2$-$3$ triangle when $c = 0$.)