How Many 120° Triples?

We want a triangle with sides $a,a+1,b$ where $b > a+1$ , and $b$ is the side opposite the angle of $120^\circ$ , so that $b^2 \; = \; a^2 + (a+1)^2 - 2a(a+1)\cos120^\circ \; = \; 3a^2 + 3a + 1$ so that $4b^2 - 3(2a+1)^2 \; = \; 1$ The positive integer solutions of the equation $x^2 - 3y^2 = 1$ are given by the formula $x_n + y_n\sqrt{3} = (2 + \sqrt{3})^n$ for $n \ge 0$ . It is easy to show that we can ensure that $x_n$ is even and $y_n$ is odd by choosing $n$ to be odd. If we define $a_n \; = \; \tfrac12(y_{2n+1} - 1) \hspace{1cm} b_n \; = \; \tfrac12x_{2n+1} \hspace{2cm} n \ge 0$ we obtain sequences of integers defined by the recurrence relations $\begin{aligned} a_n & = \; 7a_{n-1} + 4b_{n-1} + 3 \\ b_n & = \; 12a_{n-1} + 7b_{n-1} + 6 \end{aligned}$ with $a_0=0$ and $b_0=1$ , and we have $b_n^2 \; = \; 3a_n^2 + 3a_n + 1 \hspace{2cm} n \ge 0$ while it is clear that $b_n > a_n+1$ for all $n \ge 1$ . Thus we have $\boxed{\mathrm{infinitely}}$ many triangles.