Ukrajina 2008

Let $a=px$, $b=py$, $c=pz$ where $p$, $x$, $y$, $z$ are pairwise coprime numbers. Then the equation will be as follows: $(pxy{)}^2+(pxz{)}^2+(pyz{)}^2=(p^2{)}^3\iff x^2{y}^2+x^2{z}^2+y^2{z}^2=p^4$ $$\iff x^2(y^2+z^2)=(p^2-yz)(p^2+yz).$$ 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.