24th Korean Mathematical Olympiad Final Round

We will prove this statement by contradiction. Assume that there are positive integers $x,y,z$ satisfying the equation $$x^2{y}^4-x^4{y}^2+4x^2{y}^2{z}^2+x^2{z}^4-y^2{z}^4=0.(1)$$ It is easy to check that $x\ne y$. If there are solutions of the equation (1), we have a solution of the equation (1) such that $\gcd (x,y)=(x,y)=1$. Now let $x,y,z$ be a positive integer solution of the equation (1) such that $(x,y)=1$. By factoring the equation (1), we get $$\begin{array}{rl}0& =x^2{y}^4-x^4{y}^2+4x^2{y}^2{z}^2+x^2{z}^4-y^2{z}^4\\ & =x^2(y^4+2y^2{z}^2+z^4)-y^2(x^4-2x^2{z}^2+z^4)\\ & =x^2(y^2+z^2{)}^2-y^2(x^2-z^2{)}^2\\ & =(x(y^2+z^2)+y(x^2-z^2))(x(y^2+z^2)-y(x^2-z^2))\\ & =((x-y)z^2+xy(x+y))((x+y)z^2+xy(y-x)).\end{array}$$ Then $$(y-x)z^2=xy(x+y)\text{or}(x+y)z^2=xy(x-y).$$ By multiplying both sides of the first equation by $y-x$ we have $$(y-x{)}^2{z}^2=xy(y^2-x^2).(2)$$ By multiplying both sides of the second equation by $x+y$ we have $$(x+y{)}^2{z}^2=xy(x^2-y^2)(3)$$ Since we can solve the equation (3) by the same way in which we solve the equation (2), we are going to solve the equation (2). Because $(x,y)=1$, $(x,y^2-x^2)=(y,y^2-x^2)=1$. Now the left hand side of the equation (2) is a perfect square of an integer. We know that $x,y$, and $y^2-x^2$ are perfect squares. So there are positive integers $a,b,c$ such that $x=a^2,y=b^2,y^2-x^2=c^2$. Therefore we have the equation $$b^4-a^4=c^2.$$ But it is well-known that there are no positive integers $a,b,c$ satisfying the above equation. It can be proved by infinite descent. Therefore, we prove the statement by contradiction. $\square$