, 2008

Let $a = px$, $b = py$, $c = pz$ where $p$, $x$, $y$, $z$ are pairwise coprime numbers. Then the equation will be as follows: $(pxy)^2 + (pxz)^2 + (pyz)^2 = (p^2)^3 \Leftrightarrow x^2y^2 + x^2z^2 + y^2z^2 = p^4$
$$\Leftrightarrow x^2(y^2 + z^2) = (p^2 - yz)(p^2 + yz).$$
If $p^2 = y^2 + yz + z^2$, $x^2(y^2 + z^2) = (y^2 + 2yz + z^2)(y^2 + z^2) \Rightarrow x = y + z$. Under such conditions the solution will be the following set of three $(p(y+z), py, pz)$.
We just have to show that there are infinitely many sets of three natural numbers $(p, y, z)$ for which $p^2 = y^2 + yz + z^2$ is true. Let's denote $u = \frac{y}{p}$, $v = \frac{z}{p}$, i.e. we have to show that equation $u^2 + uv + v^2 = 1$ has infinitely many solutions in rational coordinates. One point is $(1, 0)$. Let's choose rational number $k$, then besides $(1, 0)$ line $v = k(u-1)$ intersects curve (ellipse) $u^2 + uv + v^2 = 1$ at one more point. According to the Vieta theorem this point is rational.