Mathematical Reflections 2011, Issue 5 - Problem S207

Problem:
Let $a,b,c$ be distinct nonzero real numbers such that $ab + bc + ca = 3$ and $ab+bc+ca \neq abc+\dfrac{2}{abc}$. Prove that
$$\left(\sum_{cyc}\dfrac{a(b-c)}{bc-1}\right)\cdot \left(\sum_{cyc}\dfrac{bc-1}{a(b-c)}\right)$$ is the square of an integer.

Proposed by Titu Andreescu.

Solution:
For simplicity, we put $x=ab, y=bc, z=ca$, so that $x+y+z=3$. We obtain
$$P=\left(\dfrac{x-z}{y-1}+\dfrac{z-y}{x-1}+\dfrac{y-x}{z-1}\right)\left(\dfrac{y-1}{x-z}+\dfrac{x-1}{z-y}+\dfrac{z-1}{y-x}\right).$$
We observe that $$\dfrac{x-1}{z-y}+\dfrac{z-1}{y-x}=\dfrac{(x-z)(-x-z+y+1)}{(z-y)(y-x)}=\dfrac{2(x-z)(y-1)}{(z-y)(y-x)}.$$
Likewise,
$$\dfrac{y-1}{x-z}+\dfrac{z-1}{y-x}=\dfrac{2(z-y)(x-1)}{(x-z)(y-x)}, \qquad \dfrac{y-1}{x-z}+\dfrac{x-1}{z-y}=\dfrac{2(y-x)(z-1)}{(x-z)(z-y)}.$$
Hence,
$$\begin{array}{lll}
P & = & 3+2\dfrac{(x-z)^2}{(z-y)(y-x)}+2\dfrac{(z-y)^2}{(x-z)(y-x)}+2\dfrac{(y-x)^2}{(z-y)(x-z)} \\ & = & 3+2\left(\dfrac{y-x}{z-y}+\dfrac{z-y}{y-x}+2+\dfrac{y-x}{x-z}+\dfrac{x-z}{y-x}+2+\dfrac{x-z}{z-y}+\dfrac{z-y}{x-z}+2\right) \\ & = & 3+2(6-3) = 9. \end{array}$$