I'm a math teacher and one of my student have come to me with several questions. One of them is the following;
Prove that there is no positive integral solution of the equation $$x^2y^4+4x^2y^2z^2+x^2z^4=x^4y^2+y^2z^4.$$
I have tried several hours but failed. Help me if you have any opinion.
Note. It can be factored so that written equivalently $$f(x,y,z)f(x,-y,z)=0$$ where $f(x,y,z)=xy(x+y)+z^2(x-y)$. So, it is equivalent to prove that $f(x,y,z)=0$ implies that $xyz=0$ over integers.
Let $f(x,y,z) = xy(x+y) +z^2(x-y)$. As you already have in the post, we need to prove that $f(x,y,z)=0$ implies $xyz=0$ over integers.
First, rewrite $f(x,y,z) = y x^2 + (y^2+z^2) x - z^2 y$ and consider $f(x,y,z)=0$ as a quadratic equation in $x$. Then the discriminant of the quadratic equation is $$ D=(y^2 + z^2)^2 + 4y^2z^2.$$ For the equation to have integer solution $x$, we must have that $D$ is a perfect square.
We prove that $D$ cannot be a perfect square if $yz\neq 0$.
This is a Diophantine equation $$x^4 + 6x^2y^2 + y^4 = z^2.$$ Then we need to prove that $xy=0$.
This equation is described in Mordell's book 'Diophantine Equations', page 17. The idea is first proving that $$ x^4 - y^4 = z^2, \ \ (x,y)=1$$ gives $xyz=0$. This was proven there by using Pythagorian triple and infinite descent.
As a corollary, we have $$ x^4 + y^4 = 2z^2, \ \ (x,y)=1$$ has only integer solutions $x^2=y^2=1$.
To prove this, note that $x, y$ are both odd and $$ z^4 - x^4y^4 = \left(\frac{x^4-y^4}{2}\right)^2.$$ Then substituting $x+y$ for $x$, and $x-y$ for $y$, we have $$ (x+y)^4 + (x-y)^4 = 2( x^4 + 6x^2y^2 + y^4) = 2z^2.$$ Thus, we see that the Diophantine equation $$ x^4 + 6x^2y^2 + y^4 = z^2$$ gives $(x+y)^2=(x-y)^2$. Therefore, $xy=0$.
Going back to the original problem, we now have that $D$ is not a perfect square if $yz\neq 0$.
Hence, any integer solution to $f(x,y,z)=0$ must satisfy $yz=0$.