, 2008

Let $a$, $b$, $c$ be a solution of the given equation. Putting
$$\frac{a}{b} = X, \quad \frac{b}{c} = Y,$$
we have $c/a = (1/XY)$. Note that $X$ and $Y$ are rational numbers. The equation takes the form
$$X^2Y + XY^2 + 1 = 3XY. \quad (1)$$
This is a cubic curve. We observe that $(1, 1)$ is a point on the curve (1). If we know the rational points on (1), then we can get integer solutions of the given equation.
Suppose $(x, y)$ is a rational point on the curve (1). Then the line joining $(x, y)$ to $(1, 1)$ has rational slope. Conversely, any line through $(1, 1)$ having rational slope intersects the curve (1) in a point having rational coordinates.
Consider the line
$$\frac{y-1}{x-1} = m,$$
where $m \ne -1$ is a non-zero rational number. Let its intersection with (1) be $(X_0, Y_0) \ne (1, 1)$. Then we have $Y_0 = m(X_0 - 1) + 1$. Putting this in (1), we get
$$X_0^2(m(X_0 - 1) + 1) + X_0(m(X_0 - 1) + 1)^2 + 1 = 3X_0(m(X_0 - 1) + 1).$$
This reduces to
$$(X_0 - 1)^2(m(1 + m)X_0 + 1) = 0.$$
Since $X_0 \ne 1$, we have
$$X_0 = -\frac{1}{m(1+m)}, \quad Y_0 = m(X_0 - 1) + 1 = -\frac{m^2}{(1+m)}.$$
Thus
$$\frac{a}{b} = -\frac{1}{m(1+m)}, \quad \frac{b}{c} = -\frac{m^2}{(1+m)}, \quad \frac{a}{c} = \frac{m}{(1+m)^2}.$$
Suppose $m = v/u$, where $v$, $u$ are integers and $\gcd(v, u) = 1$. We get
$$\frac{a}{c} = \frac{uv^2}{v(v+u)^2}.$$
This gives
$$\frac{a}{uv^2} = \frac{c}{v(v+u)^2}.$$
Similarly, the expression for $a/b$ gives
$$\frac{a}{v^2} = -\frac{b}{u(u+v)}.$$
We thus obtain
$$\frac{a}{uv^2} = -\frac{b}{u^2(u+v)} = \frac{c}{v(v+u)^2}.$$
Let each of these ratios be $p/q$ for some integers $p$, $q$. We obtain
$$qa = puv^2, \quad qb = -pu^2(u+v), \quad qc = pv(u+v)^2.$$
This shows that $\gcd(qb, qc) = \pm p(u + v)$. Thus
$$\pm q(b, c) = \pm p(u + v).$$
But $qa = puv^2$. It follows that $\pm q \gcd(a, b, c) = p$. Hence
$$\frac{p}{q} = \pm \gcd(a, b, c) = k,$$
say. Finally, we have
$$a = kuv^2, \quad b = -ku^2(u+v), \quad c = kv(u+v)^2.$$
It is easy to check that $\{a, b, c\}$ satisfies the given relation.