Formula for $\binom{n}{2}/\binom{m}{2}=1/2$ or $2n(n-1) = m(m-1)$?

Question

Where could I find a formula that produces integers $n$ and $m$ such that $\binom{n}{2}/\binom{m}{2}=1/2$?

Of course, this questions can be reformulated as: How to find all the integer values of n and m such that $$2n(n-1) = m(m-1)\quad ?$$

Answer 1

Multiply by $4$ to get

$$ 2(2n-1)^2 - 1 = (2m-1)^2$$

This is a form of Pell's equation ($\displaystyle 2x^2 - y^2 = 1$) and has infinite solutions in which both $x$ and $y$ are odd. These can be read off from the continued fraction of $\sqrt{2}$

The solutions to $\displaystyle 2x^2 - y^2 = 1$ with $\displaystyle x$ and $\displaystyle y$ odd, can be read off from the recurrence

$\displaystyle x_{n+1} = 3x_n + 2y_n$
$\displaystyle y_{n+1} = 4x_n + 3y_n$

starting with $\displaystyle x_1 = 1,\ y_1 = 1$.

Answer 2

In my answer here you will find Lagrange's method for reducing the general bivariate quadratic Diophantine equation to a Pell equation, along with some interesting links, including a Java applet that will show you the complete solution, including complete step-by-step details of the solution. According to said applet the solutions are given by

$$\rm N_{i+1}\ =\ 3\ N_i + 2\ M_i -2 $$

$$\rm M_{i+1}\ =\ 4\ N_i + 3\ M_i -3 $$

with $\rm\ N_0,M_0\ =\ 0,1\:;\ \ 1,1\:.\ $ To see complete details of the solution, including the relationship with the continued fraction of $\rm\:\sqrt{2}\:,\:$ choose the "step by step" solution method.