A non-linear homogeneous diophantine equation of order 3 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.
 A: 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 $yz=0$. 
