, 2010

Consider the relations $p^2 + p q + q^2 = u^2$ and $p^2 + m p q + q^2 = v^2$. Thus $(m - 1) p q = (v - u)(v + u)$. Suppose we choose $m - 1 = r^2$ where $r$ is a positive integer and $v - u = r p$ and $v + u = r q$. Then $u = (r q - r p)/2$ and $4(p^2 + p q + q^2) = (r q - r p)^2$. This leads to
$$(r^2 - 4) p^2 - (2 r^2 + 4) p q + (r^2 - 4) q^2 = 0.$$
Solving for $p/q$, we get
$$\frac{p}{q} = \frac{2 r^2 + 4 \pm \sqrt{4(r^2 + 2)^2 - 4(r^2 - 4)^2}}{2(r^2 - 4)}.$$
We want the discriminant to be a perfect square. This forces $3(r^2 - 1)$ to be a perfect square, which leads to the Pell's equation $r^2 - 3 t^2 = 1$. The equation has infinitely many solutions $(r_n, t_n)$ given by
$$r_n + t_n \sqrt{3} = (2 + \sqrt{3})^n.$$
We also have recurrence relations:
$$r_{n+1} = 2 r_n + 3 t_n, \quad t_{n+1} = r_n + 2 t_n,$$
where $r_1 = 2$ and $t_1 = 1$. Induction shows that $t_{2l}$ is even for all $l \ge 1$. We can express $p/q$ in terms of $t$:
$$\frac{p}{q} = \frac{(t \pm 1)^2}{t^2 - 1} = \frac{t + 1}{t - 1} \text{ or } \frac{t - 1}{t + 1}.$$
Take $m = r_{2l}^2 + 1$, $p_m = t_{2l} - 1$, $q_m = t_{2l} + 1$; we see that $p_l, q_l$ are consecutive odd integers. Moreover,
$$\begin{align*} p_m^2 + p_m q_m + q_m^2 &= r_{2l}^2, \\ p_m^2 + m p_m q_m + q_m^2 &= (t_{2l} r_{2l})^2. \end{align*}$$
This proves our claim.