, 2012

The given relation is equivalent to
$$(b + ac + a + bc)(a + b) = ab + c(a^2 + b^2) + abc^2,$$
so therefore
$$(a+b)^2c + (a+b)^2 = ab(c^2+1) + c(a^2+b^2).$$
It follows
$$\begin{aligned} (a+b)^2 &= ab(c^2+1) + c[a^2 + b^2 - (a+b)^2] \\ &= ab(c^2+1) - 2abc = ab(c-1)^2. \end{aligned} \quad (1)$$
If $c=1$, from the given relation we get
$$\frac{1}{a+b} + \frac{1}{b+a} = \frac{1}{a+b},$$
which is not possible. Therefore, we have $c \ne 1$, and from (1) we obtain
$$ab = \left(\frac{a+b}{c-1}\right)^2. \quad (2)$$
Using (2), it follows
$$\begin{aligned} (c-3)(c+1) &= (c-1)^2 - 4 = \frac{(a+b)^2}{ab} - 4 = \frac{(a-b)^2}{ab} \\ &= \left[ \frac{(a-b)(c-1)}{a+b} \right]^2. \end{aligned}$$
Finally,
$$\sqrt{\frac{c-3}{c+1}} = \frac{\sqrt{(c-3)(c+1)}}{|c+1|} = \frac{|a-b| \cdot |c-1|}{|c+1| \cdot |a+b|} \in \mathbb{Q}.$$

Assume $b=0$. Then we get
$$\frac{1}{a} + \frac{1}{ac} = \frac{1}{a},$$
which is not possible. It follows that $b \neq 0$, so we can divide by $b$ in the given relation and obtain
$$\frac{1}{\frac{a}{b} + c} + \frac{1}{1 + \frac{a}{b}c} = \frac{1}{\frac{a}{b} + 1}, \quad \frac{a}{b} \in \mathbb{Q}. \quad (1)$$
Let $\frac{a}{b} = r \in \mathbb{Q}$. Then (1) is equivalent to
$$\frac{1}{r+c} + \frac{1}{1+rc} = \frac{1}{r+1}. \quad (2)$$
From (2) we obtain $(1+r+rc+c)(r+1) = (r+c)(rc+1)$, hence $r^2 - [(c-1)^2 - 2]r + 1 = 0$. This is a quadratic equation with rational roots, hence the discriminant $\Delta$ must be a perfect square of a rational number. We have
$$\begin{aligned} \Delta &= [(c-1)^2 - 2]^2 - 4 = [(c-1)^2 - 4](c-1)^2 \\ &= (c-3)(c+1)(c-1)^2 = \frac{c-3}{c+1}(c+1)^2(c-1)^2, \end{aligned}$$
implying that $\frac{c-3}{c+1}$ is a perfect square of a rational number, and we are done.